lang
stringclasses 1
value | prompt
stringlengths 1.38k
11.3k
| eval_prompt
stringlengths 37
8.09k
| ground_truth
stringlengths 1
328
| unit_tests
stringclasses 145
values | task_id
stringlengths 23
25
| split
stringclasses 2
values |
|---|---|---|---|---|---|---|
python
|
Complete the code in python to solve this programming problem:
Description: The Narrator has an integer array $$$a$$$ of length $$$n$$$, but he will only tell you the size $$$n$$$ and $$$q$$$ statements, each of them being three integers $$$i, j, x$$$, which means that $$$a_i \mid a_j = x$$$, where $$$|$$$ denotes the bitwise OR operation.Find the lexicographically smallest array $$$a$$$ that satisfies all the statements.An array $$$a$$$ is lexicographically smaller than an array $$$b$$$ of the same length if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the array $$$a$$$ has a smaller element than the corresponding element in $$$b$$$.
Input Specification: In the first line you are given with two integers $$$n$$$ and $$$q$$$ ($$$1 \le n \le 10^5$$$, $$$0 \le q \le 2 \cdot 10^5$$$). In the next $$$q$$$ lines you are given with three integers $$$i$$$, $$$j$$$, and $$$x$$$ ($$$1 \le i, j \le n$$$, $$$0 \le x < 2^{30}$$$) — the statements. It is guaranteed that all $$$q$$$ statements hold for at least one array.
Output Specification: On a single line print $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$0 \le a_i < 2^{30}$$$) — array $$$a$$$.
Notes: NoteIn the first sample, these are all the arrays satisfying the statements: $$$[0, 3, 2, 2]$$$, $$$[2, 1, 0, 0]$$$, $$$[2, 1, 0, 2]$$$, $$$[2, 1, 2, 0]$$$, $$$[2, 1, 2, 2]$$$, $$$[2, 3, 0, 0]$$$, $$$[2, 3, 0, 2]$$$, $$$[2, 3, 2, 0]$$$, $$$[2, 3, 2, 2]$$$.
Code:
n,q = map(int,input().split())
graph = [set() for _ in range(n)]
start = [0xffffffff]*n
for _ in range(q):
i,j,x = map(int,input().split())
i -= 1; j -= 1
graph[i].add(j)
graph[j].add(i)
start[i] &= x
start[j] &= x
for i in range(n):
if i in graph[i]:
# TODO: Your code here
val = start[i]
for j in graph[i]:
val &= start[j]
start[i] ^= val
print(*start)
|
n,q = map(int,input().split())
graph = [set() for _ in range(n)]
start = [0xffffffff]*n
for _ in range(q):
i,j,x = map(int,input().split())
i -= 1; j -= 1
graph[i].add(j)
graph[j].add(i)
start[i] &= x
start[j] &= x
for i in range(n):
if i in graph[i]:
{{completion}}
val = start[i]
for j in graph[i]:
val &= start[j]
start[i] ^= val
print(*start)
|
continue
|
[{"input": "4 3\n1 2 3\n1 3 2\n4 1 2", "output": ["0 3 2 2"]}, {"input": "1 0", "output": ["0"]}, {"input": "2 1\n1 1 1073741823", "output": ["1073741823 0"]}]
|
block_completion_000016
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: The Narrator has an integer array $$$a$$$ of length $$$n$$$, but he will only tell you the size $$$n$$$ and $$$q$$$ statements, each of them being three integers $$$i, j, x$$$, which means that $$$a_i \mid a_j = x$$$, where $$$|$$$ denotes the bitwise OR operation.Find the lexicographically smallest array $$$a$$$ that satisfies all the statements.An array $$$a$$$ is lexicographically smaller than an array $$$b$$$ of the same length if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the array $$$a$$$ has a smaller element than the corresponding element in $$$b$$$.
Input Specification: In the first line you are given with two integers $$$n$$$ and $$$q$$$ ($$$1 \le n \le 10^5$$$, $$$0 \le q \le 2 \cdot 10^5$$$). In the next $$$q$$$ lines you are given with three integers $$$i$$$, $$$j$$$, and $$$x$$$ ($$$1 \le i, j \le n$$$, $$$0 \le x < 2^{30}$$$) — the statements. It is guaranteed that all $$$q$$$ statements hold for at least one array.
Output Specification: On a single line print $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$0 \le a_i < 2^{30}$$$) — array $$$a$$$.
Notes: NoteIn the first sample, these are all the arrays satisfying the statements: $$$[0, 3, 2, 2]$$$, $$$[2, 1, 0, 0]$$$, $$$[2, 1, 0, 2]$$$, $$$[2, 1, 2, 0]$$$, $$$[2, 1, 2, 2]$$$, $$$[2, 3, 0, 0]$$$, $$$[2, 3, 0, 2]$$$, $$$[2, 3, 2, 0]$$$, $$$[2, 3, 2, 2]$$$.
Code:
n,q = map(int,input().split())
graph = [set() for _ in range(n)]
start = [0xffffffff]*n
for _ in range(q):
i,j,x = map(int,input().split())
i -= 1; j -= 1
graph[i].add(j)
graph[j].add(i)
start[i] &= x
start[j] &= x
for i in range(n):
if i in graph[i]:
continue
val = start[i]
for j in graph[i]:
# TODO: Your code here
start[i] ^= val
print(*start)
|
n,q = map(int,input().split())
graph = [set() for _ in range(n)]
start = [0xffffffff]*n
for _ in range(q):
i,j,x = map(int,input().split())
i -= 1; j -= 1
graph[i].add(j)
graph[j].add(i)
start[i] &= x
start[j] &= x
for i in range(n):
if i in graph[i]:
continue
val = start[i]
for j in graph[i]:
{{completion}}
start[i] ^= val
print(*start)
|
val &= start[j]
|
[{"input": "4 3\n1 2 3\n1 3 2\n4 1 2", "output": ["0 3 2 2"]}, {"input": "1 0", "output": ["0"]}, {"input": "2 1\n1 1 1073741823", "output": ["1073741823 0"]}]
|
block_completion_000017
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: The Narrator has an integer array $$$a$$$ of length $$$n$$$, but he will only tell you the size $$$n$$$ and $$$q$$$ statements, each of them being three integers $$$i, j, x$$$, which means that $$$a_i \mid a_j = x$$$, where $$$|$$$ denotes the bitwise OR operation.Find the lexicographically smallest array $$$a$$$ that satisfies all the statements.An array $$$a$$$ is lexicographically smaller than an array $$$b$$$ of the same length if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the array $$$a$$$ has a smaller element than the corresponding element in $$$b$$$.
Input Specification: In the first line you are given with two integers $$$n$$$ and $$$q$$$ ($$$1 \le n \le 10^5$$$, $$$0 \le q \le 2 \cdot 10^5$$$). In the next $$$q$$$ lines you are given with three integers $$$i$$$, $$$j$$$, and $$$x$$$ ($$$1 \le i, j \le n$$$, $$$0 \le x < 2^{30}$$$) — the statements. It is guaranteed that all $$$q$$$ statements hold for at least one array.
Output Specification: On a single line print $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$0 \le a_i < 2^{30}$$$) — array $$$a$$$.
Notes: NoteIn the first sample, these are all the arrays satisfying the statements: $$$[0, 3, 2, 2]$$$, $$$[2, 1, 0, 0]$$$, $$$[2, 1, 0, 2]$$$, $$$[2, 1, 2, 0]$$$, $$$[2, 1, 2, 2]$$$, $$$[2, 3, 0, 0]$$$, $$$[2, 3, 0, 2]$$$, $$$[2, 3, 2, 0]$$$, $$$[2, 3, 2, 2]$$$.
Code:
import sys
n, Q = list(map(int, sys.stdin.readline().strip().split()))
m = [0] * n
M = [2 ** 30 - 1] * n
L = [[] for i in range (0, n)]
for q in range (0, Q):
i, j, x = list(map(int, sys.stdin.readline().strip().split()))
i -= 1
j -= 1
M[i] &= x
M[j] &= x
L[i].append((j, x))
L[j].append((i, x))
for i in range (0, n):
for (j, x) in L[i]:
if j != i:
m[i] |= x ^ M[j]
else:
# TODO: Your code here
M[i] = m[i]
print(*m)
|
import sys
n, Q = list(map(int, sys.stdin.readline().strip().split()))
m = [0] * n
M = [2 ** 30 - 1] * n
L = [[] for i in range (0, n)]
for q in range (0, Q):
i, j, x = list(map(int, sys.stdin.readline().strip().split()))
i -= 1
j -= 1
M[i] &= x
M[j] &= x
L[i].append((j, x))
L[j].append((i, x))
for i in range (0, n):
for (j, x) in L[i]:
if j != i:
m[i] |= x ^ M[j]
else:
{{completion}}
M[i] = m[i]
print(*m)
|
m[i] = x
|
[{"input": "4 3\n1 2 3\n1 3 2\n4 1 2", "output": ["0 3 2 2"]}, {"input": "1 0", "output": ["0"]}, {"input": "2 1\n1 1 1073741823", "output": ["1073741823 0"]}]
|
block_completion_000018
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: The Narrator has an integer array $$$a$$$ of length $$$n$$$, but he will only tell you the size $$$n$$$ and $$$q$$$ statements, each of them being three integers $$$i, j, x$$$, which means that $$$a_i \mid a_j = x$$$, where $$$|$$$ denotes the bitwise OR operation.Find the lexicographically smallest array $$$a$$$ that satisfies all the statements.An array $$$a$$$ is lexicographically smaller than an array $$$b$$$ of the same length if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the array $$$a$$$ has a smaller element than the corresponding element in $$$b$$$.
Input Specification: In the first line you are given with two integers $$$n$$$ and $$$q$$$ ($$$1 \le n \le 10^5$$$, $$$0 \le q \le 2 \cdot 10^5$$$). In the next $$$q$$$ lines you are given with three integers $$$i$$$, $$$j$$$, and $$$x$$$ ($$$1 \le i, j \le n$$$, $$$0 \le x < 2^{30}$$$) — the statements. It is guaranteed that all $$$q$$$ statements hold for at least one array.
Output Specification: On a single line print $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$0 \le a_i < 2^{30}$$$) — array $$$a$$$.
Notes: NoteIn the first sample, these are all the arrays satisfying the statements: $$$[0, 3, 2, 2]$$$, $$$[2, 1, 0, 0]$$$, $$$[2, 1, 0, 2]$$$, $$$[2, 1, 2, 0]$$$, $$$[2, 1, 2, 2]$$$, $$$[2, 3, 0, 0]$$$, $$$[2, 3, 0, 2]$$$, $$$[2, 3, 2, 0]$$$, $$$[2, 3, 2, 2]$$$.
Code:
from sys import stdin, stdout
input, print = stdin.buffer.readline, stdout.write
n, T = [int(x) for x in input().split()]
ans = [(1<<31)-1] * n
from collections import defaultdict
R = defaultdict(list)
for _ in range(T):
a,b, x = [int(_a) for _a in input().split()]
a -= 1
b -= 1
a,b = min(a,b), max(a,b)
ans[a] &= x
ans[b] &= x
R[a].append(b)
R[b].append(a)
for i in range(len(ans)):
for b in range(30,-1,-1):
mask = 1 << b
if mask & ans[i] == 0:
continue
can_remove = True
for j in R[i]:
if i == j or mask & ans[j] == 0:
# TODO: Your code here
if can_remove:
ans[i] ^= mask
print(" ".join(str(x)for x in ans)+"\n")
|
from sys import stdin, stdout
input, print = stdin.buffer.readline, stdout.write
n, T = [int(x) for x in input().split()]
ans = [(1<<31)-1] * n
from collections import defaultdict
R = defaultdict(list)
for _ in range(T):
a,b, x = [int(_a) for _a in input().split()]
a -= 1
b -= 1
a,b = min(a,b), max(a,b)
ans[a] &= x
ans[b] &= x
R[a].append(b)
R[b].append(a)
for i in range(len(ans)):
for b in range(30,-1,-1):
mask = 1 << b
if mask & ans[i] == 0:
continue
can_remove = True
for j in R[i]:
if i == j or mask & ans[j] == 0:
{{completion}}
if can_remove:
ans[i] ^= mask
print(" ".join(str(x)for x in ans)+"\n")
|
can_remove = False
break
|
[{"input": "4 3\n1 2 3\n1 3 2\n4 1 2", "output": ["0 3 2 2"]}, {"input": "1 0", "output": ["0"]}, {"input": "2 1\n1 1 1073741823", "output": ["1073741823 0"]}]
|
block_completion_000019
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: The Narrator has an integer array $$$a$$$ of length $$$n$$$, but he will only tell you the size $$$n$$$ and $$$q$$$ statements, each of them being three integers $$$i, j, x$$$, which means that $$$a_i \mid a_j = x$$$, where $$$|$$$ denotes the bitwise OR operation.Find the lexicographically smallest array $$$a$$$ that satisfies all the statements.An array $$$a$$$ is lexicographically smaller than an array $$$b$$$ of the same length if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the array $$$a$$$ has a smaller element than the corresponding element in $$$b$$$.
Input Specification: In the first line you are given with two integers $$$n$$$ and $$$q$$$ ($$$1 \le n \le 10^5$$$, $$$0 \le q \le 2 \cdot 10^5$$$). In the next $$$q$$$ lines you are given with three integers $$$i$$$, $$$j$$$, and $$$x$$$ ($$$1 \le i, j \le n$$$, $$$0 \le x < 2^{30}$$$) — the statements. It is guaranteed that all $$$q$$$ statements hold for at least one array.
Output Specification: On a single line print $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$0 \le a_i < 2^{30}$$$) — array $$$a$$$.
Notes: NoteIn the first sample, these are all the arrays satisfying the statements: $$$[0, 3, 2, 2]$$$, $$$[2, 1, 0, 0]$$$, $$$[2, 1, 0, 2]$$$, $$$[2, 1, 2, 0]$$$, $$$[2, 1, 2, 2]$$$, $$$[2, 3, 0, 0]$$$, $$$[2, 3, 0, 2]$$$, $$$[2, 3, 2, 0]$$$, $$$[2, 3, 2, 2]$$$.
Code:
n,q = map(int, input().split())
adj = [list() for i in range(n+1)]
val = [-1]*(n+1)
for _ in range(q):
i,j,x=map(int, input().split())
val[i] &= x
val[j] &= x
adj[i].append(j)
adj[j].append(i)
# print(*val[1:], sep=" ")
# print(*adj, sep="\n")
for a in range(1, n+1):
if val[a] == -1:
val[a] = 0
continue
t = val[a]
for b in adj[a]:
if b == a:
# TODO: Your code here
t &= val[b]
val[a] ^= t
# print(*val[1:], sep=" ")
print(*val[1:], sep=" ")
|
n,q = map(int, input().split())
adj = [list() for i in range(n+1)]
val = [-1]*(n+1)
for _ in range(q):
i,j,x=map(int, input().split())
val[i] &= x
val[j] &= x
adj[i].append(j)
adj[j].append(i)
# print(*val[1:], sep=" ")
# print(*adj, sep="\n")
for a in range(1, n+1):
if val[a] == -1:
val[a] = 0
continue
t = val[a]
for b in adj[a]:
if b == a:
{{completion}}
t &= val[b]
val[a] ^= t
# print(*val[1:], sep=" ")
print(*val[1:], sep=" ")
|
t = 0
break
|
[{"input": "4 3\n1 2 3\n1 3 2\n4 1 2", "output": ["0 3 2 2"]}, {"input": "1 0", "output": ["0"]}, {"input": "2 1\n1 1 1073741823", "output": ["1073741823 0"]}]
|
block_completion_000020
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: The Narrator has an integer array $$$a$$$ of length $$$n$$$, but he will only tell you the size $$$n$$$ and $$$q$$$ statements, each of them being three integers $$$i, j, x$$$, which means that $$$a_i \mid a_j = x$$$, where $$$|$$$ denotes the bitwise OR operation.Find the lexicographically smallest array $$$a$$$ that satisfies all the statements.An array $$$a$$$ is lexicographically smaller than an array $$$b$$$ of the same length if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the array $$$a$$$ has a smaller element than the corresponding element in $$$b$$$.
Input Specification: In the first line you are given with two integers $$$n$$$ and $$$q$$$ ($$$1 \le n \le 10^5$$$, $$$0 \le q \le 2 \cdot 10^5$$$). In the next $$$q$$$ lines you are given with three integers $$$i$$$, $$$j$$$, and $$$x$$$ ($$$1 \le i, j \le n$$$, $$$0 \le x < 2^{30}$$$) — the statements. It is guaranteed that all $$$q$$$ statements hold for at least one array.
Output Specification: On a single line print $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$0 \le a_i < 2^{30}$$$) — array $$$a$$$.
Notes: NoteIn the first sample, these are all the arrays satisfying the statements: $$$[0, 3, 2, 2]$$$, $$$[2, 1, 0, 0]$$$, $$$[2, 1, 0, 2]$$$, $$$[2, 1, 2, 0]$$$, $$$[2, 1, 2, 2]$$$, $$$[2, 3, 0, 0]$$$, $$$[2, 3, 0, 2]$$$, $$$[2, 3, 2, 0]$$$, $$$[2, 3, 2, 2]$$$.
Code:
n, q = map(int, input().strip().split())
qs = [[] for _ in range(n)]
refers_self = [False for _ in range(n)]
for _ in range(q):
i, j, x = map(int, input().strip().split())
if i==j:
refers_self[i-1] = True
qs[i-1].append((j-1, x))
qs[j-1].append((i-1, x))
a = []
for i in range(n):
if qs[i]:
ans = (2<<32) - 1
for j, x in qs[i]:
# TODO: Your code here
a.append(ans)
else:
a.append(0)
for i in range(n):
if refers_self[i]:
continue
ans = (2<<32) - 1
for j, x in qs[i]:
ans = ans & a[j]
a[i] = a[i] - (a[i]&ans)
print(*a)
|
n, q = map(int, input().strip().split())
qs = [[] for _ in range(n)]
refers_self = [False for _ in range(n)]
for _ in range(q):
i, j, x = map(int, input().strip().split())
if i==j:
refers_self[i-1] = True
qs[i-1].append((j-1, x))
qs[j-1].append((i-1, x))
a = []
for i in range(n):
if qs[i]:
ans = (2<<32) - 1
for j, x in qs[i]:
{{completion}}
a.append(ans)
else:
a.append(0)
for i in range(n):
if refers_self[i]:
continue
ans = (2<<32) - 1
for j, x in qs[i]:
ans = ans & a[j]
a[i] = a[i] - (a[i]&ans)
print(*a)
|
ans = ans & x
|
[{"input": "4 3\n1 2 3\n1 3 2\n4 1 2", "output": ["0 3 2 2"]}, {"input": "1 0", "output": ["0"]}, {"input": "2 1\n1 1 1073741823", "output": ["1073741823 0"]}]
|
block_completion_000021
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: Stanley has decided to buy a new desktop PC made by the company "Monoblock", and to solve captcha on their website, he needs to solve the following task.The awesomeness of an array is the minimum number of blocks of consecutive identical numbers in which the array could be split. For example, the awesomeness of an array $$$[1, 1, 1]$$$ is $$$1$$$; $$$[5, 7]$$$ is $$$2$$$, as it could be split into blocks $$$[5]$$$ and $$$[7]$$$; $$$[1, 7, 7, 7, 7, 7, 7, 7, 9, 9, 9, 9, 9, 9, 9, 9, 9]$$$ is 3, as it could be split into blocks $$$[1]$$$, $$$[7, 7, 7, 7, 7, 7, 7]$$$, and $$$[9, 9, 9, 9, 9, 9, 9, 9, 9]$$$. You are given an array $$$a$$$ of length $$$n$$$. There are $$$m$$$ queries of two integers $$$i$$$, $$$x$$$. A query $$$i$$$, $$$x$$$ means that from now on the $$$i$$$-th element of the array $$$a$$$ is equal to $$$x$$$.After each query print the sum of awesomeness values among all subsegments of array $$$a$$$. In other words, after each query you need to calculate $$$$$$\sum\limits_{l = 1}^n \sum\limits_{r = l}^n g(l, r),$$$$$$ where $$$g(l, r)$$$ is the awesomeness of the array $$$b = [a_l, a_{l + 1}, \ldots, a_r]$$$.
Input Specification: In the first line you are given with two integers $$$n$$$ and $$$m$$$ ($$$1 \leq n, m \leq 10^5$$$). The second line contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — the array $$$a$$$. In the next $$$m$$$ lines you are given the descriptions of queries. Each line contains two integers $$$i$$$ and $$$x$$$ ($$$1 \leq i \leq n$$$, $$$1 \leq x \leq 10^9$$$).
Output Specification: Print the answer to each query on a new line.
Notes: NoteAfter the first query $$$a$$$ is equal to $$$[1, 2, 2, 4, 5]$$$, and the answer is $$$29$$$ because we can split each of the subsegments the following way: $$$[1; 1]$$$: $$$[1]$$$, 1 block; $$$[1; 2]$$$: $$$[1] + [2]$$$, 2 blocks; $$$[1; 3]$$$: $$$[1] + [2, 2]$$$, 2 blocks; $$$[1; 4]$$$: $$$[1] + [2, 2] + [4]$$$, 3 blocks; $$$[1; 5]$$$: $$$[1] + [2, 2] + [4] + [5]$$$, 4 blocks; $$$[2; 2]$$$: $$$[2]$$$, 1 block; $$$[2; 3]$$$: $$$[2, 2]$$$, 1 block; $$$[2; 4]$$$: $$$[2, 2] + [4]$$$, 2 blocks; $$$[2; 5]$$$: $$$[2, 2] + [4] + [5]$$$, 3 blocks; $$$[3; 3]$$$: $$$[2]$$$, 1 block; $$$[3; 4]$$$: $$$[2] + [4]$$$, 2 blocks; $$$[3; 5]$$$: $$$[2] + [4] + [5]$$$, 3 blocks; $$$[4; 4]$$$: $$$[4]$$$, 1 block; $$$[4; 5]$$$: $$$[4] + [5]$$$, 2 blocks; $$$[5; 5]$$$: $$$[5]$$$, 1 block; which is $$$1 + 2 + 2 + 3 + 4 + 1 + 1 + 2 + 3 + 1 + 2 + 3 + 1 + 2 + 1 = 29$$$ in total.
Code:
from sys import stdin
input = stdin.readline
inp = lambda : list(map(int,input().split()))
def update(i , t):
global ans
if(i + 1 < n and a[i] == a[i + 1]):
ans += t * (i + 1)
else:
ans += t * (n - i) * (i + 1)
return ans
def answer():
global ans
ans = 0
for i in range(n):
update(i , 1)
for q in range(m):
i , x = inp()
i -= 1
if(i >= 0):# TODO: Your code here
update(i , -1)
a[i] = x
if(i >= 0):update(i - 1 , 1)
update(i , 1)
print(ans)
for T in range(1):
n , m = inp()
a = inp()
answer()
|
from sys import stdin
input = stdin.readline
inp = lambda : list(map(int,input().split()))
def update(i , t):
global ans
if(i + 1 < n and a[i] == a[i + 1]):
ans += t * (i + 1)
else:
ans += t * (n - i) * (i + 1)
return ans
def answer():
global ans
ans = 0
for i in range(n):
update(i , 1)
for q in range(m):
i , x = inp()
i -= 1
if(i >= 0):{{completion}}
update(i , -1)
a[i] = x
if(i >= 0):update(i - 1 , 1)
update(i , 1)
print(ans)
for T in range(1):
n , m = inp()
a = inp()
answer()
|
update(i - 1 , -1)
|
[{"input": "5 5\n1 2 3 4 5\n3 2\n4 2\n3 1\n2 1\n2 2", "output": ["29\n23\n35\n25\n35"]}]
|
block_completion_000074
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: Stanley has decided to buy a new desktop PC made by the company "Monoblock", and to solve captcha on their website, he needs to solve the following task.The awesomeness of an array is the minimum number of blocks of consecutive identical numbers in which the array could be split. For example, the awesomeness of an array $$$[1, 1, 1]$$$ is $$$1$$$; $$$[5, 7]$$$ is $$$2$$$, as it could be split into blocks $$$[5]$$$ and $$$[7]$$$; $$$[1, 7, 7, 7, 7, 7, 7, 7, 9, 9, 9, 9, 9, 9, 9, 9, 9]$$$ is 3, as it could be split into blocks $$$[1]$$$, $$$[7, 7, 7, 7, 7, 7, 7]$$$, and $$$[9, 9, 9, 9, 9, 9, 9, 9, 9]$$$. You are given an array $$$a$$$ of length $$$n$$$. There are $$$m$$$ queries of two integers $$$i$$$, $$$x$$$. A query $$$i$$$, $$$x$$$ means that from now on the $$$i$$$-th element of the array $$$a$$$ is equal to $$$x$$$.After each query print the sum of awesomeness values among all subsegments of array $$$a$$$. In other words, after each query you need to calculate $$$$$$\sum\limits_{l = 1}^n \sum\limits_{r = l}^n g(l, r),$$$$$$ where $$$g(l, r)$$$ is the awesomeness of the array $$$b = [a_l, a_{l + 1}, \ldots, a_r]$$$.
Input Specification: In the first line you are given with two integers $$$n$$$ and $$$m$$$ ($$$1 \leq n, m \leq 10^5$$$). The second line contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — the array $$$a$$$. In the next $$$m$$$ lines you are given the descriptions of queries. Each line contains two integers $$$i$$$ and $$$x$$$ ($$$1 \leq i \leq n$$$, $$$1 \leq x \leq 10^9$$$).
Output Specification: Print the answer to each query on a new line.
Notes: NoteAfter the first query $$$a$$$ is equal to $$$[1, 2, 2, 4, 5]$$$, and the answer is $$$29$$$ because we can split each of the subsegments the following way: $$$[1; 1]$$$: $$$[1]$$$, 1 block; $$$[1; 2]$$$: $$$[1] + [2]$$$, 2 blocks; $$$[1; 3]$$$: $$$[1] + [2, 2]$$$, 2 blocks; $$$[1; 4]$$$: $$$[1] + [2, 2] + [4]$$$, 3 blocks; $$$[1; 5]$$$: $$$[1] + [2, 2] + [4] + [5]$$$, 4 blocks; $$$[2; 2]$$$: $$$[2]$$$, 1 block; $$$[2; 3]$$$: $$$[2, 2]$$$, 1 block; $$$[2; 4]$$$: $$$[2, 2] + [4]$$$, 2 blocks; $$$[2; 5]$$$: $$$[2, 2] + [4] + [5]$$$, 3 blocks; $$$[3; 3]$$$: $$$[2]$$$, 1 block; $$$[3; 4]$$$: $$$[2] + [4]$$$, 2 blocks; $$$[3; 5]$$$: $$$[2] + [4] + [5]$$$, 3 blocks; $$$[4; 4]$$$: $$$[4]$$$, 1 block; $$$[4; 5]$$$: $$$[4] + [5]$$$, 2 blocks; $$$[5; 5]$$$: $$$[5]$$$, 1 block; which is $$$1 + 2 + 2 + 3 + 4 + 1 + 1 + 2 + 3 + 1 + 2 + 3 + 1 + 2 + 1 = 29$$$ in total.
Code:
from sys import stdin
input = stdin.readline
inp = lambda : list(map(int,input().split()))
def update(i , t):
global ans
if(i + 1 < n and a[i] == a[i + 1]):
ans += t * (i + 1)
else:
ans += t * (n - i) * (i + 1)
return ans
def answer():
global ans
ans = 0
for i in range(n):
update(i , 1)
for q in range(m):
i , x = inp()
i -= 1
if(i >= 0):update(i - 1 , -1)
update(i , -1)
a[i] = x
if(i >= 0):# TODO: Your code here
update(i , 1)
print(ans)
for T in range(1):
n , m = inp()
a = inp()
answer()
|
from sys import stdin
input = stdin.readline
inp = lambda : list(map(int,input().split()))
def update(i , t):
global ans
if(i + 1 < n and a[i] == a[i + 1]):
ans += t * (i + 1)
else:
ans += t * (n - i) * (i + 1)
return ans
def answer():
global ans
ans = 0
for i in range(n):
update(i , 1)
for q in range(m):
i , x = inp()
i -= 1
if(i >= 0):update(i - 1 , -1)
update(i , -1)
a[i] = x
if(i >= 0):{{completion}}
update(i , 1)
print(ans)
for T in range(1):
n , m = inp()
a = inp()
answer()
|
update(i - 1 , 1)
|
[{"input": "5 5\n1 2 3 4 5\n3 2\n4 2\n3 1\n2 1\n2 2", "output": ["29\n23\n35\n25\n35"]}]
|
block_completion_000075
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: Stanley has decided to buy a new desktop PC made by the company "Monoblock", and to solve captcha on their website, he needs to solve the following task.The awesomeness of an array is the minimum number of blocks of consecutive identical numbers in which the array could be split. For example, the awesomeness of an array $$$[1, 1, 1]$$$ is $$$1$$$; $$$[5, 7]$$$ is $$$2$$$, as it could be split into blocks $$$[5]$$$ and $$$[7]$$$; $$$[1, 7, 7, 7, 7, 7, 7, 7, 9, 9, 9, 9, 9, 9, 9, 9, 9]$$$ is 3, as it could be split into blocks $$$[1]$$$, $$$[7, 7, 7, 7, 7, 7, 7]$$$, and $$$[9, 9, 9, 9, 9, 9, 9, 9, 9]$$$. You are given an array $$$a$$$ of length $$$n$$$. There are $$$m$$$ queries of two integers $$$i$$$, $$$x$$$. A query $$$i$$$, $$$x$$$ means that from now on the $$$i$$$-th element of the array $$$a$$$ is equal to $$$x$$$.After each query print the sum of awesomeness values among all subsegments of array $$$a$$$. In other words, after each query you need to calculate $$$$$$\sum\limits_{l = 1}^n \sum\limits_{r = l}^n g(l, r),$$$$$$ where $$$g(l, r)$$$ is the awesomeness of the array $$$b = [a_l, a_{l + 1}, \ldots, a_r]$$$.
Input Specification: In the first line you are given with two integers $$$n$$$ and $$$m$$$ ($$$1 \leq n, m \leq 10^5$$$). The second line contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — the array $$$a$$$. In the next $$$m$$$ lines you are given the descriptions of queries. Each line contains two integers $$$i$$$ and $$$x$$$ ($$$1 \leq i \leq n$$$, $$$1 \leq x \leq 10^9$$$).
Output Specification: Print the answer to each query on a new line.
Notes: NoteAfter the first query $$$a$$$ is equal to $$$[1, 2, 2, 4, 5]$$$, and the answer is $$$29$$$ because we can split each of the subsegments the following way: $$$[1; 1]$$$: $$$[1]$$$, 1 block; $$$[1; 2]$$$: $$$[1] + [2]$$$, 2 blocks; $$$[1; 3]$$$: $$$[1] + [2, 2]$$$, 2 blocks; $$$[1; 4]$$$: $$$[1] + [2, 2] + [4]$$$, 3 blocks; $$$[1; 5]$$$: $$$[1] + [2, 2] + [4] + [5]$$$, 4 blocks; $$$[2; 2]$$$: $$$[2]$$$, 1 block; $$$[2; 3]$$$: $$$[2, 2]$$$, 1 block; $$$[2; 4]$$$: $$$[2, 2] + [4]$$$, 2 blocks; $$$[2; 5]$$$: $$$[2, 2] + [4] + [5]$$$, 3 blocks; $$$[3; 3]$$$: $$$[2]$$$, 1 block; $$$[3; 4]$$$: $$$[2] + [4]$$$, 2 blocks; $$$[3; 5]$$$: $$$[2] + [4] + [5]$$$, 3 blocks; $$$[4; 4]$$$: $$$[4]$$$, 1 block; $$$[4; 5]$$$: $$$[4] + [5]$$$, 2 blocks; $$$[5; 5]$$$: $$$[5]$$$, 1 block; which is $$$1 + 2 + 2 + 3 + 4 + 1 + 1 + 2 + 3 + 1 + 2 + 3 + 1 + 2 + 1 = 29$$$ in total.
Code:
import sys
input = sys.stdin.readline
n, m = map(int, input().split())
a = list(map(int, input().split()))
a.insert(0, 0)
a.append(0)
ans = 0
for i in range(1, n + 1):
# TODO: Your code here
while(m):
i, x = map(int, input().split())
ans -= (a[i] != a[i - 1]) * (n - i + 1) * (i - 1)
ans -= (a[i] != a[i + 1]) * (n - (i + 1) + 1) * i
a[i] = x
ans += (a[i] != a[i - 1]) * (n - i + 1) * (i - 1)
ans += (a[i] != a[i + 1]) * (n - (i + 1) + 1) * i
print(ans + n * (n + 1) // 2)
m -= 1
|
import sys
input = sys.stdin.readline
n, m = map(int, input().split())
a = list(map(int, input().split()))
a.insert(0, 0)
a.append(0)
ans = 0
for i in range(1, n + 1):
{{completion}}
while(m):
i, x = map(int, input().split())
ans -= (a[i] != a[i - 1]) * (n - i + 1) * (i - 1)
ans -= (a[i] != a[i + 1]) * (n - (i + 1) + 1) * i
a[i] = x
ans += (a[i] != a[i - 1]) * (n - i + 1) * (i - 1)
ans += (a[i] != a[i + 1]) * (n - (i + 1) + 1) * i
print(ans + n * (n + 1) // 2)
m -= 1
|
ans += (a[i] != a[i + 1]) * (n - (i + 1) + 1) * i
|
[{"input": "5 5\n1 2 3 4 5\n3 2\n4 2\n3 1\n2 1\n2 2", "output": ["29\n23\n35\n25\n35"]}]
|
block_completion_000076
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: Stanley has decided to buy a new desktop PC made by the company "Monoblock", and to solve captcha on their website, he needs to solve the following task.The awesomeness of an array is the minimum number of blocks of consecutive identical numbers in which the array could be split. For example, the awesomeness of an array $$$[1, 1, 1]$$$ is $$$1$$$; $$$[5, 7]$$$ is $$$2$$$, as it could be split into blocks $$$[5]$$$ and $$$[7]$$$; $$$[1, 7, 7, 7, 7, 7, 7, 7, 9, 9, 9, 9, 9, 9, 9, 9, 9]$$$ is 3, as it could be split into blocks $$$[1]$$$, $$$[7, 7, 7, 7, 7, 7, 7]$$$, and $$$[9, 9, 9, 9, 9, 9, 9, 9, 9]$$$. You are given an array $$$a$$$ of length $$$n$$$. There are $$$m$$$ queries of two integers $$$i$$$, $$$x$$$. A query $$$i$$$, $$$x$$$ means that from now on the $$$i$$$-th element of the array $$$a$$$ is equal to $$$x$$$.After each query print the sum of awesomeness values among all subsegments of array $$$a$$$. In other words, after each query you need to calculate $$$$$$\sum\limits_{l = 1}^n \sum\limits_{r = l}^n g(l, r),$$$$$$ where $$$g(l, r)$$$ is the awesomeness of the array $$$b = [a_l, a_{l + 1}, \ldots, a_r]$$$.
Input Specification: In the first line you are given with two integers $$$n$$$ and $$$m$$$ ($$$1 \leq n, m \leq 10^5$$$). The second line contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — the array $$$a$$$. In the next $$$m$$$ lines you are given the descriptions of queries. Each line contains two integers $$$i$$$ and $$$x$$$ ($$$1 \leq i \leq n$$$, $$$1 \leq x \leq 10^9$$$).
Output Specification: Print the answer to each query on a new line.
Notes: NoteAfter the first query $$$a$$$ is equal to $$$[1, 2, 2, 4, 5]$$$, and the answer is $$$29$$$ because we can split each of the subsegments the following way: $$$[1; 1]$$$: $$$[1]$$$, 1 block; $$$[1; 2]$$$: $$$[1] + [2]$$$, 2 blocks; $$$[1; 3]$$$: $$$[1] + [2, 2]$$$, 2 blocks; $$$[1; 4]$$$: $$$[1] + [2, 2] + [4]$$$, 3 blocks; $$$[1; 5]$$$: $$$[1] + [2, 2] + [4] + [5]$$$, 4 blocks; $$$[2; 2]$$$: $$$[2]$$$, 1 block; $$$[2; 3]$$$: $$$[2, 2]$$$, 1 block; $$$[2; 4]$$$: $$$[2, 2] + [4]$$$, 2 blocks; $$$[2; 5]$$$: $$$[2, 2] + [4] + [5]$$$, 3 blocks; $$$[3; 3]$$$: $$$[2]$$$, 1 block; $$$[3; 4]$$$: $$$[2] + [4]$$$, 2 blocks; $$$[3; 5]$$$: $$$[2] + [4] + [5]$$$, 3 blocks; $$$[4; 4]$$$: $$$[4]$$$, 1 block; $$$[4; 5]$$$: $$$[4] + [5]$$$, 2 blocks; $$$[5; 5]$$$: $$$[5]$$$, 1 block; which is $$$1 + 2 + 2 + 3 + 4 + 1 + 1 + 2 + 3 + 1 + 2 + 3 + 1 + 2 + 1 = 29$$$ in total.
Code:
import sys
input = sys.stdin.readline
n, m = map(int, input().split())
a = list(map(int, input().split()))
a.insert(0, 0)
a.append(0)
ans = 0
for i in range(1, n + 1):
ans += (a[i] != a[i + 1]) * (n - (i + 1) + 1) * i
while(m):
# TODO: Your code here
|
import sys
input = sys.stdin.readline
n, m = map(int, input().split())
a = list(map(int, input().split()))
a.insert(0, 0)
a.append(0)
ans = 0
for i in range(1, n + 1):
ans += (a[i] != a[i + 1]) * (n - (i + 1) + 1) * i
while(m):
{{completion}}
|
i, x = map(int, input().split())
ans -= (a[i] != a[i - 1]) * (n - i + 1) * (i - 1)
ans -= (a[i] != a[i + 1]) * (n - (i + 1) + 1) * i
a[i] = x
ans += (a[i] != a[i - 1]) * (n - i + 1) * (i - 1)
ans += (a[i] != a[i + 1]) * (n - (i + 1) + 1) * i
print(ans + n * (n + 1) // 2)
m -= 1
|
[{"input": "5 5\n1 2 3 4 5\n3 2\n4 2\n3 1\n2 1\n2 2", "output": ["29\n23\n35\n25\n35"]}]
|
block_completion_000077
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: Stanley has decided to buy a new desktop PC made by the company "Monoblock", and to solve captcha on their website, he needs to solve the following task.The awesomeness of an array is the minimum number of blocks of consecutive identical numbers in which the array could be split. For example, the awesomeness of an array $$$[1, 1, 1]$$$ is $$$1$$$; $$$[5, 7]$$$ is $$$2$$$, as it could be split into blocks $$$[5]$$$ and $$$[7]$$$; $$$[1, 7, 7, 7, 7, 7, 7, 7, 9, 9, 9, 9, 9, 9, 9, 9, 9]$$$ is 3, as it could be split into blocks $$$[1]$$$, $$$[7, 7, 7, 7, 7, 7, 7]$$$, and $$$[9, 9, 9, 9, 9, 9, 9, 9, 9]$$$. You are given an array $$$a$$$ of length $$$n$$$. There are $$$m$$$ queries of two integers $$$i$$$, $$$x$$$. A query $$$i$$$, $$$x$$$ means that from now on the $$$i$$$-th element of the array $$$a$$$ is equal to $$$x$$$.After each query print the sum of awesomeness values among all subsegments of array $$$a$$$. In other words, after each query you need to calculate $$$$$$\sum\limits_{l = 1}^n \sum\limits_{r = l}^n g(l, r),$$$$$$ where $$$g(l, r)$$$ is the awesomeness of the array $$$b = [a_l, a_{l + 1}, \ldots, a_r]$$$.
Input Specification: In the first line you are given with two integers $$$n$$$ and $$$m$$$ ($$$1 \leq n, m \leq 10^5$$$). The second line contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — the array $$$a$$$. In the next $$$m$$$ lines you are given the descriptions of queries. Each line contains two integers $$$i$$$ and $$$x$$$ ($$$1 \leq i \leq n$$$, $$$1 \leq x \leq 10^9$$$).
Output Specification: Print the answer to each query on a new line.
Notes: NoteAfter the first query $$$a$$$ is equal to $$$[1, 2, 2, 4, 5]$$$, and the answer is $$$29$$$ because we can split each of the subsegments the following way: $$$[1; 1]$$$: $$$[1]$$$, 1 block; $$$[1; 2]$$$: $$$[1] + [2]$$$, 2 blocks; $$$[1; 3]$$$: $$$[1] + [2, 2]$$$, 2 blocks; $$$[1; 4]$$$: $$$[1] + [2, 2] + [4]$$$, 3 blocks; $$$[1; 5]$$$: $$$[1] + [2, 2] + [4] + [5]$$$, 4 blocks; $$$[2; 2]$$$: $$$[2]$$$, 1 block; $$$[2; 3]$$$: $$$[2, 2]$$$, 1 block; $$$[2; 4]$$$: $$$[2, 2] + [4]$$$, 2 blocks; $$$[2; 5]$$$: $$$[2, 2] + [4] + [5]$$$, 3 blocks; $$$[3; 3]$$$: $$$[2]$$$, 1 block; $$$[3; 4]$$$: $$$[2] + [4]$$$, 2 blocks; $$$[3; 5]$$$: $$$[2] + [4] + [5]$$$, 3 blocks; $$$[4; 4]$$$: $$$[4]$$$, 1 block; $$$[4; 5]$$$: $$$[4] + [5]$$$, 2 blocks; $$$[5; 5]$$$: $$$[5]$$$, 1 block; which is $$$1 + 2 + 2 + 3 + 4 + 1 + 1 + 2 + 3 + 1 + 2 + 3 + 1 + 2 + 1 = 29$$$ in total.
Code:
import time, sys
n, m = [int(i) for i in sys.stdin.readline().split()]
a = [int(i) for i in sys.stdin.readline().split()]
t1 = time.time()
w = [(i+1)*(n-i-1) for i in range(n-1)]
c = sum([w[i] if a[i+1] != a[i] else 0 for i in range(n-1)])
for _ in range(m):
ix, x = [int(i) for i in sys.stdin.readline().split()]
ix -= 1
if ix != 0:
if a[ix] == a[ix-1] != x:
c += w[ix-1]
elif a[ix] != a[ix-1] == x:
# TODO: Your code here
if ix != n-1:
if a[ix] == a[ix+1] != x:
c += w[ix]
elif a[ix] != a[ix+1] == x:
c -= w[ix]
a[ix] = x
sys.stdout.write(str(c+(n*(n+1))//2) + '\n')
|
import time, sys
n, m = [int(i) for i in sys.stdin.readline().split()]
a = [int(i) for i in sys.stdin.readline().split()]
t1 = time.time()
w = [(i+1)*(n-i-1) for i in range(n-1)]
c = sum([w[i] if a[i+1] != a[i] else 0 for i in range(n-1)])
for _ in range(m):
ix, x = [int(i) for i in sys.stdin.readline().split()]
ix -= 1
if ix != 0:
if a[ix] == a[ix-1] != x:
c += w[ix-1]
elif a[ix] != a[ix-1] == x:
{{completion}}
if ix != n-1:
if a[ix] == a[ix+1] != x:
c += w[ix]
elif a[ix] != a[ix+1] == x:
c -= w[ix]
a[ix] = x
sys.stdout.write(str(c+(n*(n+1))//2) + '\n')
|
c -= w[ix-1]
|
[{"input": "5 5\n1 2 3 4 5\n3 2\n4 2\n3 1\n2 1\n2 2", "output": ["29\n23\n35\n25\n35"]}]
|
block_completion_000078
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: Stanley has decided to buy a new desktop PC made by the company "Monoblock", and to solve captcha on their website, he needs to solve the following task.The awesomeness of an array is the minimum number of blocks of consecutive identical numbers in which the array could be split. For example, the awesomeness of an array $$$[1, 1, 1]$$$ is $$$1$$$; $$$[5, 7]$$$ is $$$2$$$, as it could be split into blocks $$$[5]$$$ and $$$[7]$$$; $$$[1, 7, 7, 7, 7, 7, 7, 7, 9, 9, 9, 9, 9, 9, 9, 9, 9]$$$ is 3, as it could be split into blocks $$$[1]$$$, $$$[7, 7, 7, 7, 7, 7, 7]$$$, and $$$[9, 9, 9, 9, 9, 9, 9, 9, 9]$$$. You are given an array $$$a$$$ of length $$$n$$$. There are $$$m$$$ queries of two integers $$$i$$$, $$$x$$$. A query $$$i$$$, $$$x$$$ means that from now on the $$$i$$$-th element of the array $$$a$$$ is equal to $$$x$$$.After each query print the sum of awesomeness values among all subsegments of array $$$a$$$. In other words, after each query you need to calculate $$$$$$\sum\limits_{l = 1}^n \sum\limits_{r = l}^n g(l, r),$$$$$$ where $$$g(l, r)$$$ is the awesomeness of the array $$$b = [a_l, a_{l + 1}, \ldots, a_r]$$$.
Input Specification: In the first line you are given with two integers $$$n$$$ and $$$m$$$ ($$$1 \leq n, m \leq 10^5$$$). The second line contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — the array $$$a$$$. In the next $$$m$$$ lines you are given the descriptions of queries. Each line contains two integers $$$i$$$ and $$$x$$$ ($$$1 \leq i \leq n$$$, $$$1 \leq x \leq 10^9$$$).
Output Specification: Print the answer to each query on a new line.
Notes: NoteAfter the first query $$$a$$$ is equal to $$$[1, 2, 2, 4, 5]$$$, and the answer is $$$29$$$ because we can split each of the subsegments the following way: $$$[1; 1]$$$: $$$[1]$$$, 1 block; $$$[1; 2]$$$: $$$[1] + [2]$$$, 2 blocks; $$$[1; 3]$$$: $$$[1] + [2, 2]$$$, 2 blocks; $$$[1; 4]$$$: $$$[1] + [2, 2] + [4]$$$, 3 blocks; $$$[1; 5]$$$: $$$[1] + [2, 2] + [4] + [5]$$$, 4 blocks; $$$[2; 2]$$$: $$$[2]$$$, 1 block; $$$[2; 3]$$$: $$$[2, 2]$$$, 1 block; $$$[2; 4]$$$: $$$[2, 2] + [4]$$$, 2 blocks; $$$[2; 5]$$$: $$$[2, 2] + [4] + [5]$$$, 3 blocks; $$$[3; 3]$$$: $$$[2]$$$, 1 block; $$$[3; 4]$$$: $$$[2] + [4]$$$, 2 blocks; $$$[3; 5]$$$: $$$[2] + [4] + [5]$$$, 3 blocks; $$$[4; 4]$$$: $$$[4]$$$, 1 block; $$$[4; 5]$$$: $$$[4] + [5]$$$, 2 blocks; $$$[5; 5]$$$: $$$[5]$$$, 1 block; which is $$$1 + 2 + 2 + 3 + 4 + 1 + 1 + 2 + 3 + 1 + 2 + 3 + 1 + 2 + 1 = 29$$$ in total.
Code:
import time, sys
n, m = [int(i) for i in sys.stdin.readline().split()]
a = [int(i) for i in sys.stdin.readline().split()]
t1 = time.time()
w = [(i+1)*(n-i-1) for i in range(n-1)]
c = sum([w[i] if a[i+1] != a[i] else 0 for i in range(n-1)])
for _ in range(m):
ix, x = [int(i) for i in sys.stdin.readline().split()]
ix -= 1
if ix != 0:
if a[ix] == a[ix-1] != x:
c += w[ix-1]
elif a[ix] != a[ix-1] == x:
c -= w[ix-1]
if ix != n-1:
if a[ix] == a[ix+1] != x:
c += w[ix]
elif a[ix] != a[ix+1] == x:
# TODO: Your code here
a[ix] = x
sys.stdout.write(str(c+(n*(n+1))//2) + '\n')
|
import time, sys
n, m = [int(i) for i in sys.stdin.readline().split()]
a = [int(i) for i in sys.stdin.readline().split()]
t1 = time.time()
w = [(i+1)*(n-i-1) for i in range(n-1)]
c = sum([w[i] if a[i+1] != a[i] else 0 for i in range(n-1)])
for _ in range(m):
ix, x = [int(i) for i in sys.stdin.readline().split()]
ix -= 1
if ix != 0:
if a[ix] == a[ix-1] != x:
c += w[ix-1]
elif a[ix] != a[ix-1] == x:
c -= w[ix-1]
if ix != n-1:
if a[ix] == a[ix+1] != x:
c += w[ix]
elif a[ix] != a[ix+1] == x:
{{completion}}
a[ix] = x
sys.stdout.write(str(c+(n*(n+1))//2) + '\n')
|
c -= w[ix]
|
[{"input": "5 5\n1 2 3 4 5\n3 2\n4 2\n3 1\n2 1\n2 2", "output": ["29\n23\n35\n25\n35"]}]
|
block_completion_000079
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: Stanley has decided to buy a new desktop PC made by the company "Monoblock", and to solve captcha on their website, he needs to solve the following task.The awesomeness of an array is the minimum number of blocks of consecutive identical numbers in which the array could be split. For example, the awesomeness of an array $$$[1, 1, 1]$$$ is $$$1$$$; $$$[5, 7]$$$ is $$$2$$$, as it could be split into blocks $$$[5]$$$ and $$$[7]$$$; $$$[1, 7, 7, 7, 7, 7, 7, 7, 9, 9, 9, 9, 9, 9, 9, 9, 9]$$$ is 3, as it could be split into blocks $$$[1]$$$, $$$[7, 7, 7, 7, 7, 7, 7]$$$, and $$$[9, 9, 9, 9, 9, 9, 9, 9, 9]$$$. You are given an array $$$a$$$ of length $$$n$$$. There are $$$m$$$ queries of two integers $$$i$$$, $$$x$$$. A query $$$i$$$, $$$x$$$ means that from now on the $$$i$$$-th element of the array $$$a$$$ is equal to $$$x$$$.After each query print the sum of awesomeness values among all subsegments of array $$$a$$$. In other words, after each query you need to calculate $$$$$$\sum\limits_{l = 1}^n \sum\limits_{r = l}^n g(l, r),$$$$$$ where $$$g(l, r)$$$ is the awesomeness of the array $$$b = [a_l, a_{l + 1}, \ldots, a_r]$$$.
Input Specification: In the first line you are given with two integers $$$n$$$ and $$$m$$$ ($$$1 \leq n, m \leq 10^5$$$). The second line contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — the array $$$a$$$. In the next $$$m$$$ lines you are given the descriptions of queries. Each line contains two integers $$$i$$$ and $$$x$$$ ($$$1 \leq i \leq n$$$, $$$1 \leq x \leq 10^9$$$).
Output Specification: Print the answer to each query on a new line.
Notes: NoteAfter the first query $$$a$$$ is equal to $$$[1, 2, 2, 4, 5]$$$, and the answer is $$$29$$$ because we can split each of the subsegments the following way: $$$[1; 1]$$$: $$$[1]$$$, 1 block; $$$[1; 2]$$$: $$$[1] + [2]$$$, 2 blocks; $$$[1; 3]$$$: $$$[1] + [2, 2]$$$, 2 blocks; $$$[1; 4]$$$: $$$[1] + [2, 2] + [4]$$$, 3 blocks; $$$[1; 5]$$$: $$$[1] + [2, 2] + [4] + [5]$$$, 4 blocks; $$$[2; 2]$$$: $$$[2]$$$, 1 block; $$$[2; 3]$$$: $$$[2, 2]$$$, 1 block; $$$[2; 4]$$$: $$$[2, 2] + [4]$$$, 2 blocks; $$$[2; 5]$$$: $$$[2, 2] + [4] + [5]$$$, 3 blocks; $$$[3; 3]$$$: $$$[2]$$$, 1 block; $$$[3; 4]$$$: $$$[2] + [4]$$$, 2 blocks; $$$[3; 5]$$$: $$$[2] + [4] + [5]$$$, 3 blocks; $$$[4; 4]$$$: $$$[4]$$$, 1 block; $$$[4; 5]$$$: $$$[4] + [5]$$$, 2 blocks; $$$[5; 5]$$$: $$$[5]$$$, 1 block; which is $$$1 + 2 + 2 + 3 + 4 + 1 + 1 + 2 + 3 + 1 + 2 + 3 + 1 + 2 + 1 = 29$$$ in total.
Code:
import sys
import collections
inf=float('inf')
mod=10**5+7
input = lambda: sys.stdin.readline().rstrip()
inpnm = lambda: map(int,input().split())
inparr = lambda: [int(i) for i in input().split()]
inpint = lambda: int(input())
# for case in range(inpint()):
n,m=inpnm()
arr=inparr()
res=[1]
cnt=0
se=1
t=1
for i in range(1,n):
if arr[i]==arr[i-1]:
cnt+=1
res.append(res[-1]+1)
else:
# TODO: Your code here
ans=sum(res)
for q in range(m):
i,x=inpnm()
i-=1
if i!=0 and arr[i-1]!=arr[i]:
ans-=i*(n-i)
if i!=n-1 and arr[i+1]!=arr[i]:
ans-=(i+1)*(n-i-1)
arr[i]=x
if i!=0 and arr[i-1]!=x:
ans+=i*(n-i)
if i!=n-1 and arr[i+1]!=x:
ans+=(i+1)*(n-i-1)
print(ans)
|
import sys
import collections
inf=float('inf')
mod=10**5+7
input = lambda: sys.stdin.readline().rstrip()
inpnm = lambda: map(int,input().split())
inparr = lambda: [int(i) for i in input().split()]
inpint = lambda: int(input())
# for case in range(inpint()):
n,m=inpnm()
arr=inparr()
res=[1]
cnt=0
se=1
t=1
for i in range(1,n):
if arr[i]==arr[i-1]:
cnt+=1
res.append(res[-1]+1)
else:
{{completion}}
ans=sum(res)
for q in range(m):
i,x=inpnm()
i-=1
if i!=0 and arr[i-1]!=arr[i]:
ans-=i*(n-i)
if i!=n-1 and arr[i+1]!=arr[i]:
ans-=(i+1)*(n-i-1)
arr[i]=x
if i!=0 and arr[i-1]!=x:
ans+=i*(n-i)
if i!=n-1 and arr[i+1]!=x:
ans+=(i+1)*(n-i-1)
print(ans)
|
se+=1
t=res[-1]
res.append(res[-1]+se+cnt)
|
[{"input": "5 5\n1 2 3 4 5\n3 2\n4 2\n3 1\n2 1\n2 2", "output": ["29\n23\n35\n25\n35"]}]
|
block_completion_000080
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: Stanley has decided to buy a new desktop PC made by the company "Monoblock", and to solve captcha on their website, he needs to solve the following task.The awesomeness of an array is the minimum number of blocks of consecutive identical numbers in which the array could be split. For example, the awesomeness of an array $$$[1, 1, 1]$$$ is $$$1$$$; $$$[5, 7]$$$ is $$$2$$$, as it could be split into blocks $$$[5]$$$ and $$$[7]$$$; $$$[1, 7, 7, 7, 7, 7, 7, 7, 9, 9, 9, 9, 9, 9, 9, 9, 9]$$$ is 3, as it could be split into blocks $$$[1]$$$, $$$[7, 7, 7, 7, 7, 7, 7]$$$, and $$$[9, 9, 9, 9, 9, 9, 9, 9, 9]$$$. You are given an array $$$a$$$ of length $$$n$$$. There are $$$m$$$ queries of two integers $$$i$$$, $$$x$$$. A query $$$i$$$, $$$x$$$ means that from now on the $$$i$$$-th element of the array $$$a$$$ is equal to $$$x$$$.After each query print the sum of awesomeness values among all subsegments of array $$$a$$$. In other words, after each query you need to calculate $$$$$$\sum\limits_{l = 1}^n \sum\limits_{r = l}^n g(l, r),$$$$$$ where $$$g(l, r)$$$ is the awesomeness of the array $$$b = [a_l, a_{l + 1}, \ldots, a_r]$$$.
Input Specification: In the first line you are given with two integers $$$n$$$ and $$$m$$$ ($$$1 \leq n, m \leq 10^5$$$). The second line contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — the array $$$a$$$. In the next $$$m$$$ lines you are given the descriptions of queries. Each line contains two integers $$$i$$$ and $$$x$$$ ($$$1 \leq i \leq n$$$, $$$1 \leq x \leq 10^9$$$).
Output Specification: Print the answer to each query on a new line.
Notes: NoteAfter the first query $$$a$$$ is equal to $$$[1, 2, 2, 4, 5]$$$, and the answer is $$$29$$$ because we can split each of the subsegments the following way: $$$[1; 1]$$$: $$$[1]$$$, 1 block; $$$[1; 2]$$$: $$$[1] + [2]$$$, 2 blocks; $$$[1; 3]$$$: $$$[1] + [2, 2]$$$, 2 blocks; $$$[1; 4]$$$: $$$[1] + [2, 2] + [4]$$$, 3 blocks; $$$[1; 5]$$$: $$$[1] + [2, 2] + [4] + [5]$$$, 4 blocks; $$$[2; 2]$$$: $$$[2]$$$, 1 block; $$$[2; 3]$$$: $$$[2, 2]$$$, 1 block; $$$[2; 4]$$$: $$$[2, 2] + [4]$$$, 2 blocks; $$$[2; 5]$$$: $$$[2, 2] + [4] + [5]$$$, 3 blocks; $$$[3; 3]$$$: $$$[2]$$$, 1 block; $$$[3; 4]$$$: $$$[2] + [4]$$$, 2 blocks; $$$[3; 5]$$$: $$$[2] + [4] + [5]$$$, 3 blocks; $$$[4; 4]$$$: $$$[4]$$$, 1 block; $$$[4; 5]$$$: $$$[4] + [5]$$$, 2 blocks; $$$[5; 5]$$$: $$$[5]$$$, 1 block; which is $$$1 + 2 + 2 + 3 + 4 + 1 + 1 + 2 + 3 + 1 + 2 + 3 + 1 + 2 + 1 = 29$$$ in total.
Code:
def update_awesomeness(arr, i, x, curr_aws):
# TODO: Your code here
INPUT = [*open(0)]
n, m = map(int, INPUT[0].split())
arr = list(map(int, INPUT[1].split()))
tar = [0] * n
aws = (n * (n + 1)) // 2
for i, x in enumerate(arr):
aws = update_awesomeness(tar, i, x, aws)
for line in INPUT[2:]:
i, x = map(int, line.split())
aws = update_awesomeness(tar, i - 1, x, aws)
print(aws)
|
def update_awesomeness(arr, i, x, curr_aws):
{{completion}}
INPUT = [*open(0)]
n, m = map(int, INPUT[0].split())
arr = list(map(int, INPUT[1].split()))
tar = [0] * n
aws = (n * (n + 1)) // 2
for i, x in enumerate(arr):
aws = update_awesomeness(tar, i, x, aws)
for line in INPUT[2:]:
i, x = map(int, line.split())
aws = update_awesomeness(tar, i - 1, x, aws)
print(aws)
|
left_edit = (x != arr[i - 1]) - (arr[i] != arr[i - 1]) if i != 0 else 0
right_edit = (x != arr[i + 1]) - (arr[i] != arr[i + 1]) if i != n - 1 else 0
arr[i] = x
return curr_aws + left_edit * i * (n - i) + right_edit * (i + 1) * (n - i - 1)
|
[{"input": "5 5\n1 2 3 4 5\n3 2\n4 2\n3 1\n2 1\n2 2", "output": ["29\n23\n35\n25\n35"]}]
|
block_completion_000081
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: Stanley has decided to buy a new desktop PC made by the company "Monoblock", and to solve captcha on their website, he needs to solve the following task.The awesomeness of an array is the minimum number of blocks of consecutive identical numbers in which the array could be split. For example, the awesomeness of an array $$$[1, 1, 1]$$$ is $$$1$$$; $$$[5, 7]$$$ is $$$2$$$, as it could be split into blocks $$$[5]$$$ and $$$[7]$$$; $$$[1, 7, 7, 7, 7, 7, 7, 7, 9, 9, 9, 9, 9, 9, 9, 9, 9]$$$ is 3, as it could be split into blocks $$$[1]$$$, $$$[7, 7, 7, 7, 7, 7, 7]$$$, and $$$[9, 9, 9, 9, 9, 9, 9, 9, 9]$$$. You are given an array $$$a$$$ of length $$$n$$$. There are $$$m$$$ queries of two integers $$$i$$$, $$$x$$$. A query $$$i$$$, $$$x$$$ means that from now on the $$$i$$$-th element of the array $$$a$$$ is equal to $$$x$$$.After each query print the sum of awesomeness values among all subsegments of array $$$a$$$. In other words, after each query you need to calculate $$$$$$\sum\limits_{l = 1}^n \sum\limits_{r = l}^n g(l, r),$$$$$$ where $$$g(l, r)$$$ is the awesomeness of the array $$$b = [a_l, a_{l + 1}, \ldots, a_r]$$$.
Input Specification: In the first line you are given with two integers $$$n$$$ and $$$m$$$ ($$$1 \leq n, m \leq 10^5$$$). The second line contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — the array $$$a$$$. In the next $$$m$$$ lines you are given the descriptions of queries. Each line contains two integers $$$i$$$ and $$$x$$$ ($$$1 \leq i \leq n$$$, $$$1 \leq x \leq 10^9$$$).
Output Specification: Print the answer to each query on a new line.
Notes: NoteAfter the first query $$$a$$$ is equal to $$$[1, 2, 2, 4, 5]$$$, and the answer is $$$29$$$ because we can split each of the subsegments the following way: $$$[1; 1]$$$: $$$[1]$$$, 1 block; $$$[1; 2]$$$: $$$[1] + [2]$$$, 2 blocks; $$$[1; 3]$$$: $$$[1] + [2, 2]$$$, 2 blocks; $$$[1; 4]$$$: $$$[1] + [2, 2] + [4]$$$, 3 blocks; $$$[1; 5]$$$: $$$[1] + [2, 2] + [4] + [5]$$$, 4 blocks; $$$[2; 2]$$$: $$$[2]$$$, 1 block; $$$[2; 3]$$$: $$$[2, 2]$$$, 1 block; $$$[2; 4]$$$: $$$[2, 2] + [4]$$$, 2 blocks; $$$[2; 5]$$$: $$$[2, 2] + [4] + [5]$$$, 3 blocks; $$$[3; 3]$$$: $$$[2]$$$, 1 block; $$$[3; 4]$$$: $$$[2] + [4]$$$, 2 blocks; $$$[3; 5]$$$: $$$[2] + [4] + [5]$$$, 3 blocks; $$$[4; 4]$$$: $$$[4]$$$, 1 block; $$$[4; 5]$$$: $$$[4] + [5]$$$, 2 blocks; $$$[5; 5]$$$: $$$[5]$$$, 1 block; which is $$$1 + 2 + 2 + 3 + 4 + 1 + 1 + 2 + 3 + 1 + 2 + 3 + 1 + 2 + 1 = 29$$$ in total.
Code:
def update_awesomeness(arr, i, x, curr_aws):
left_edit = (x != arr[i - 1]) - (arr[i] != arr[i - 1]) if i != 0 else 0
right_edit = (x != arr[i + 1]) - (arr[i] != arr[i + 1]) if i != n - 1 else 0
arr[i] = x
return curr_aws + left_edit * i * (n - i) + right_edit * (i + 1) * (n - i - 1)
INPUT = [*open(0)]
n, m = map(int, INPUT[0].split())
arr = list(map(int, INPUT[1].split()))
tar = [0] * n
aws = (n * (n + 1)) // 2
for i, x in enumerate(arr):
# TODO: Your code here
for line in INPUT[2:]:
i, x = map(int, line.split())
aws = update_awesomeness(tar, i - 1, x, aws)
print(aws)
|
def update_awesomeness(arr, i, x, curr_aws):
left_edit = (x != arr[i - 1]) - (arr[i] != arr[i - 1]) if i != 0 else 0
right_edit = (x != arr[i + 1]) - (arr[i] != arr[i + 1]) if i != n - 1 else 0
arr[i] = x
return curr_aws + left_edit * i * (n - i) + right_edit * (i + 1) * (n - i - 1)
INPUT = [*open(0)]
n, m = map(int, INPUT[0].split())
arr = list(map(int, INPUT[1].split()))
tar = [0] * n
aws = (n * (n + 1)) // 2
for i, x in enumerate(arr):
{{completion}}
for line in INPUT[2:]:
i, x = map(int, line.split())
aws = update_awesomeness(tar, i - 1, x, aws)
print(aws)
|
aws = update_awesomeness(tar, i, x, aws)
|
[{"input": "5 5\n1 2 3 4 5\n3 2\n4 2\n3 1\n2 1\n2 2", "output": ["29\n23\n35\n25\n35"]}]
|
block_completion_000082
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: Stanley has decided to buy a new desktop PC made by the company "Monoblock", and to solve captcha on their website, he needs to solve the following task.The awesomeness of an array is the minimum number of blocks of consecutive identical numbers in which the array could be split. For example, the awesomeness of an array $$$[1, 1, 1]$$$ is $$$1$$$; $$$[5, 7]$$$ is $$$2$$$, as it could be split into blocks $$$[5]$$$ and $$$[7]$$$; $$$[1, 7, 7, 7, 7, 7, 7, 7, 9, 9, 9, 9, 9, 9, 9, 9, 9]$$$ is 3, as it could be split into blocks $$$[1]$$$, $$$[7, 7, 7, 7, 7, 7, 7]$$$, and $$$[9, 9, 9, 9, 9, 9, 9, 9, 9]$$$. You are given an array $$$a$$$ of length $$$n$$$. There are $$$m$$$ queries of two integers $$$i$$$, $$$x$$$. A query $$$i$$$, $$$x$$$ means that from now on the $$$i$$$-th element of the array $$$a$$$ is equal to $$$x$$$.After each query print the sum of awesomeness values among all subsegments of array $$$a$$$. In other words, after each query you need to calculate $$$$$$\sum\limits_{l = 1}^n \sum\limits_{r = l}^n g(l, r),$$$$$$ where $$$g(l, r)$$$ is the awesomeness of the array $$$b = [a_l, a_{l + 1}, \ldots, a_r]$$$.
Input Specification: In the first line you are given with two integers $$$n$$$ and $$$m$$$ ($$$1 \leq n, m \leq 10^5$$$). The second line contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — the array $$$a$$$. In the next $$$m$$$ lines you are given the descriptions of queries. Each line contains two integers $$$i$$$ and $$$x$$$ ($$$1 \leq i \leq n$$$, $$$1 \leq x \leq 10^9$$$).
Output Specification: Print the answer to each query on a new line.
Notes: NoteAfter the first query $$$a$$$ is equal to $$$[1, 2, 2, 4, 5]$$$, and the answer is $$$29$$$ because we can split each of the subsegments the following way: $$$[1; 1]$$$: $$$[1]$$$, 1 block; $$$[1; 2]$$$: $$$[1] + [2]$$$, 2 blocks; $$$[1; 3]$$$: $$$[1] + [2, 2]$$$, 2 blocks; $$$[1; 4]$$$: $$$[1] + [2, 2] + [4]$$$, 3 blocks; $$$[1; 5]$$$: $$$[1] + [2, 2] + [4] + [5]$$$, 4 blocks; $$$[2; 2]$$$: $$$[2]$$$, 1 block; $$$[2; 3]$$$: $$$[2, 2]$$$, 1 block; $$$[2; 4]$$$: $$$[2, 2] + [4]$$$, 2 blocks; $$$[2; 5]$$$: $$$[2, 2] + [4] + [5]$$$, 3 blocks; $$$[3; 3]$$$: $$$[2]$$$, 1 block; $$$[3; 4]$$$: $$$[2] + [4]$$$, 2 blocks; $$$[3; 5]$$$: $$$[2] + [4] + [5]$$$, 3 blocks; $$$[4; 4]$$$: $$$[4]$$$, 1 block; $$$[4; 5]$$$: $$$[4] + [5]$$$, 2 blocks; $$$[5; 5]$$$: $$$[5]$$$, 1 block; which is $$$1 + 2 + 2 + 3 + 4 + 1 + 1 + 2 + 3 + 1 + 2 + 3 + 1 + 2 + 1 = 29$$$ in total.
Code:
"""
author: Manoj
inp_start
5 5
1 2 3 4 5
3 2
4 2
3 1
2 1
2 2
inp_end
"""
n, m = list(map(int, input().split()))
li = list(map(int, input().split()))
ans = int((n*(n+1))/2)
for i in range(1, n):
if li[i]!=li[i-1]:
ans += i*(n-i)
al = []
for tc in range(m):
i, x = list(map(int, input().split()))
i -= 1
if i>0:
if li[i]!=li[i-1]:
# TODO: Your code here
if x!=li[i-1]:
ans+=i*(n-i)
if i+1<n:
if li[i]!=li[i+1]:
ans-=(i+1)*(n-i-1)
if x!=li[i+1]:
ans+=(i+1)*(n-i-1)
li[i]=x
al.append(ans)
print(*al)
|
"""
author: Manoj
inp_start
5 5
1 2 3 4 5
3 2
4 2
3 1
2 1
2 2
inp_end
"""
n, m = list(map(int, input().split()))
li = list(map(int, input().split()))
ans = int((n*(n+1))/2)
for i in range(1, n):
if li[i]!=li[i-1]:
ans += i*(n-i)
al = []
for tc in range(m):
i, x = list(map(int, input().split()))
i -= 1
if i>0:
if li[i]!=li[i-1]:
{{completion}}
if x!=li[i-1]:
ans+=i*(n-i)
if i+1<n:
if li[i]!=li[i+1]:
ans-=(i+1)*(n-i-1)
if x!=li[i+1]:
ans+=(i+1)*(n-i-1)
li[i]=x
al.append(ans)
print(*al)
|
ans-=i*(n-i)
|
[{"input": "5 5\n1 2 3 4 5\n3 2\n4 2\n3 1\n2 1\n2 2", "output": ["29\n23\n35\n25\n35"]}]
|
block_completion_000083
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: Stanley has decided to buy a new desktop PC made by the company "Monoblock", and to solve captcha on their website, he needs to solve the following task.The awesomeness of an array is the minimum number of blocks of consecutive identical numbers in which the array could be split. For example, the awesomeness of an array $$$[1, 1, 1]$$$ is $$$1$$$; $$$[5, 7]$$$ is $$$2$$$, as it could be split into blocks $$$[5]$$$ and $$$[7]$$$; $$$[1, 7, 7, 7, 7, 7, 7, 7, 9, 9, 9, 9, 9, 9, 9, 9, 9]$$$ is 3, as it could be split into blocks $$$[1]$$$, $$$[7, 7, 7, 7, 7, 7, 7]$$$, and $$$[9, 9, 9, 9, 9, 9, 9, 9, 9]$$$. You are given an array $$$a$$$ of length $$$n$$$. There are $$$m$$$ queries of two integers $$$i$$$, $$$x$$$. A query $$$i$$$, $$$x$$$ means that from now on the $$$i$$$-th element of the array $$$a$$$ is equal to $$$x$$$.After each query print the sum of awesomeness values among all subsegments of array $$$a$$$. In other words, after each query you need to calculate $$$$$$\sum\limits_{l = 1}^n \sum\limits_{r = l}^n g(l, r),$$$$$$ where $$$g(l, r)$$$ is the awesomeness of the array $$$b = [a_l, a_{l + 1}, \ldots, a_r]$$$.
Input Specification: In the first line you are given with two integers $$$n$$$ and $$$m$$$ ($$$1 \leq n, m \leq 10^5$$$). The second line contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — the array $$$a$$$. In the next $$$m$$$ lines you are given the descriptions of queries. Each line contains two integers $$$i$$$ and $$$x$$$ ($$$1 \leq i \leq n$$$, $$$1 \leq x \leq 10^9$$$).
Output Specification: Print the answer to each query on a new line.
Notes: NoteAfter the first query $$$a$$$ is equal to $$$[1, 2, 2, 4, 5]$$$, and the answer is $$$29$$$ because we can split each of the subsegments the following way: $$$[1; 1]$$$: $$$[1]$$$, 1 block; $$$[1; 2]$$$: $$$[1] + [2]$$$, 2 blocks; $$$[1; 3]$$$: $$$[1] + [2, 2]$$$, 2 blocks; $$$[1; 4]$$$: $$$[1] + [2, 2] + [4]$$$, 3 blocks; $$$[1; 5]$$$: $$$[1] + [2, 2] + [4] + [5]$$$, 4 blocks; $$$[2; 2]$$$: $$$[2]$$$, 1 block; $$$[2; 3]$$$: $$$[2, 2]$$$, 1 block; $$$[2; 4]$$$: $$$[2, 2] + [4]$$$, 2 blocks; $$$[2; 5]$$$: $$$[2, 2] + [4] + [5]$$$, 3 blocks; $$$[3; 3]$$$: $$$[2]$$$, 1 block; $$$[3; 4]$$$: $$$[2] + [4]$$$, 2 blocks; $$$[3; 5]$$$: $$$[2] + [4] + [5]$$$, 3 blocks; $$$[4; 4]$$$: $$$[4]$$$, 1 block; $$$[4; 5]$$$: $$$[4] + [5]$$$, 2 blocks; $$$[5; 5]$$$: $$$[5]$$$, 1 block; which is $$$1 + 2 + 2 + 3 + 4 + 1 + 1 + 2 + 3 + 1 + 2 + 3 + 1 + 2 + 1 = 29$$$ in total.
Code:
"""
author: Manoj
inp_start
5 5
1 2 3 4 5
3 2
4 2
3 1
2 1
2 2
inp_end
"""
n, m = list(map(int, input().split()))
li = list(map(int, input().split()))
ans = int((n*(n+1))/2)
for i in range(1, n):
if li[i]!=li[i-1]:
ans += i*(n-i)
al = []
for tc in range(m):
i, x = list(map(int, input().split()))
i -= 1
if i>0:
if li[i]!=li[i-1]:
ans-=i*(n-i)
if x!=li[i-1]:
# TODO: Your code here
if i+1<n:
if li[i]!=li[i+1]:
ans-=(i+1)*(n-i-1)
if x!=li[i+1]:
ans+=(i+1)*(n-i-1)
li[i]=x
al.append(ans)
print(*al)
|
"""
author: Manoj
inp_start
5 5
1 2 3 4 5
3 2
4 2
3 1
2 1
2 2
inp_end
"""
n, m = list(map(int, input().split()))
li = list(map(int, input().split()))
ans = int((n*(n+1))/2)
for i in range(1, n):
if li[i]!=li[i-1]:
ans += i*(n-i)
al = []
for tc in range(m):
i, x = list(map(int, input().split()))
i -= 1
if i>0:
if li[i]!=li[i-1]:
ans-=i*(n-i)
if x!=li[i-1]:
{{completion}}
if i+1<n:
if li[i]!=li[i+1]:
ans-=(i+1)*(n-i-1)
if x!=li[i+1]:
ans+=(i+1)*(n-i-1)
li[i]=x
al.append(ans)
print(*al)
|
ans+=i*(n-i)
|
[{"input": "5 5\n1 2 3 4 5\n3 2\n4 2\n3 1\n2 1\n2 2", "output": ["29\n23\n35\n25\n35"]}]
|
block_completion_000084
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: Stanley has decided to buy a new desktop PC made by the company "Monoblock", and to solve captcha on their website, he needs to solve the following task.The awesomeness of an array is the minimum number of blocks of consecutive identical numbers in which the array could be split. For example, the awesomeness of an array $$$[1, 1, 1]$$$ is $$$1$$$; $$$[5, 7]$$$ is $$$2$$$, as it could be split into blocks $$$[5]$$$ and $$$[7]$$$; $$$[1, 7, 7, 7, 7, 7, 7, 7, 9, 9, 9, 9, 9, 9, 9, 9, 9]$$$ is 3, as it could be split into blocks $$$[1]$$$, $$$[7, 7, 7, 7, 7, 7, 7]$$$, and $$$[9, 9, 9, 9, 9, 9, 9, 9, 9]$$$. You are given an array $$$a$$$ of length $$$n$$$. There are $$$m$$$ queries of two integers $$$i$$$, $$$x$$$. A query $$$i$$$, $$$x$$$ means that from now on the $$$i$$$-th element of the array $$$a$$$ is equal to $$$x$$$.After each query print the sum of awesomeness values among all subsegments of array $$$a$$$. In other words, after each query you need to calculate $$$$$$\sum\limits_{l = 1}^n \sum\limits_{r = l}^n g(l, r),$$$$$$ where $$$g(l, r)$$$ is the awesomeness of the array $$$b = [a_l, a_{l + 1}, \ldots, a_r]$$$.
Input Specification: In the first line you are given with two integers $$$n$$$ and $$$m$$$ ($$$1 \leq n, m \leq 10^5$$$). The second line contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — the array $$$a$$$. In the next $$$m$$$ lines you are given the descriptions of queries. Each line contains two integers $$$i$$$ and $$$x$$$ ($$$1 \leq i \leq n$$$, $$$1 \leq x \leq 10^9$$$).
Output Specification: Print the answer to each query on a new line.
Notes: NoteAfter the first query $$$a$$$ is equal to $$$[1, 2, 2, 4, 5]$$$, and the answer is $$$29$$$ because we can split each of the subsegments the following way: $$$[1; 1]$$$: $$$[1]$$$, 1 block; $$$[1; 2]$$$: $$$[1] + [2]$$$, 2 blocks; $$$[1; 3]$$$: $$$[1] + [2, 2]$$$, 2 blocks; $$$[1; 4]$$$: $$$[1] + [2, 2] + [4]$$$, 3 blocks; $$$[1; 5]$$$: $$$[1] + [2, 2] + [4] + [5]$$$, 4 blocks; $$$[2; 2]$$$: $$$[2]$$$, 1 block; $$$[2; 3]$$$: $$$[2, 2]$$$, 1 block; $$$[2; 4]$$$: $$$[2, 2] + [4]$$$, 2 blocks; $$$[2; 5]$$$: $$$[2, 2] + [4] + [5]$$$, 3 blocks; $$$[3; 3]$$$: $$$[2]$$$, 1 block; $$$[3; 4]$$$: $$$[2] + [4]$$$, 2 blocks; $$$[3; 5]$$$: $$$[2] + [4] + [5]$$$, 3 blocks; $$$[4; 4]$$$: $$$[4]$$$, 1 block; $$$[4; 5]$$$: $$$[4] + [5]$$$, 2 blocks; $$$[5; 5]$$$: $$$[5]$$$, 1 block; which is $$$1 + 2 + 2 + 3 + 4 + 1 + 1 + 2 + 3 + 1 + 2 + 3 + 1 + 2 + 1 = 29$$$ in total.
Code:
import sys
input=lambda:sys.stdin.readline().rstrip()
n,q=map(int,input().split())
arr=[0]+[*map(int,input().split())]+[0]
awe=0
awr=[0]*(n+1)
for i in range(1,n+1):
if arr[i]!=arr[i-1]:
awr[i]+=awr[i-1]+(i-1)
else:
# TODO: Your code here
awr[i]+=1
awe=sum(awr)
for _ in range(q):
i,x=map(int,input().split())
lprev=arr[i]!=arr[i-1]
rprev=arr[i]!=arr[i+1]
arr[i]=x
laftr=arr[i]!=arr[i-1]
raftr=arr[i]!=arr[i+1]
dl=laftr-lprev
dr=raftr-rprev
awe+=dl*(i-1)*(n-i+1)
awe+=dr*(i)*(n-i)
print(awe)
|
import sys
input=lambda:sys.stdin.readline().rstrip()
n,q=map(int,input().split())
arr=[0]+[*map(int,input().split())]+[0]
awe=0
awr=[0]*(n+1)
for i in range(1,n+1):
if arr[i]!=arr[i-1]:
awr[i]+=awr[i-1]+(i-1)
else:
{{completion}}
awr[i]+=1
awe=sum(awr)
for _ in range(q):
i,x=map(int,input().split())
lprev=arr[i]!=arr[i-1]
rprev=arr[i]!=arr[i+1]
arr[i]=x
laftr=arr[i]!=arr[i-1]
raftr=arr[i]!=arr[i+1]
dl=laftr-lprev
dr=raftr-rprev
awe+=dl*(i-1)*(n-i+1)
awe+=dr*(i)*(n-i)
print(awe)
|
awr[i]+=awr[i-1]
|
[{"input": "5 5\n1 2 3 4 5\n3 2\n4 2\n3 1\n2 1\n2 2", "output": ["29\n23\n35\n25\n35"]}]
|
block_completion_000085
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: Stanley lives in a country that consists of $$$n$$$ cities (he lives in city $$$1$$$). There are bidirectional roads between some of the cities, and you know how long it takes to ride through each of them. Additionally, there is a flight between each pair of cities, the flight between cities $$$u$$$ and $$$v$$$ takes $$$(u - v)^2$$$ time.Stanley is quite afraid of flying because of watching "Sully: Miracle on the Hudson" recently, so he can take at most $$$k$$$ flights. Stanley wants to know the minimum time of a journey to each of the $$$n$$$ cities from the city $$$1$$$.
Input Specification: In the first line of input there are three integers $$$n$$$, $$$m$$$, and $$$k$$$ ($$$2 \leq n \leq 10^{5}$$$, $$$1 \leq m \leq 10^{5}$$$, $$$1 \leq k \leq 20$$$) — the number of cities, the number of roads, and the maximal number of flights Stanley can take. The following $$$m$$$ lines describe the roads. Each contains three integers $$$u$$$, $$$v$$$, $$$w$$$ ($$$1 \leq u, v \leq n$$$, $$$u \neq v$$$, $$$1 \leq w \leq 10^{9}$$$) — the cities the road connects and the time it takes to ride through. Note that some pairs of cities may be connected by more than one road.
Output Specification: Print $$$n$$$ integers, $$$i$$$-th of which is equal to the minimum time of traveling to city $$$i$$$.
Notes: NoteIn the first sample, it takes no time to get to city 1; to get to city 2 it is possible to use a flight between 1 and 2, which will take 1 unit of time; to city 3 you can get via a road from city 1, which will take 1 unit of time. In the second sample, it also takes no time to get to city 1. To get to city 2 Stanley should use a flight between 1 and 2, which will take 1 unit of time. To get to city 3 Stanley can ride between cities 1 and 2, which will take 3 units of time, and then use a flight between 2 and 3. To get to city 4 Stanley should use a flight between 1 and 2, then take a ride from 2 to 4, which will take 5 units of time.
Code:
import sys
input=sys.stdin.readline #文字列入力はするな！！
########################################
from heapq import heappush, heappop
B=10**5+10
def dijkstra( G, dist, INF=10**11):
"""
https://tjkendev.github.io/procon-library/python/graph/dijkstra.html
O((|E|+|V|)log|V|)
V: 頂点数
G[v] = [(nod, cost)]:
頂点vから遷移可能な頂点(nod)とそのコスト(cost)
s: 始点の頂点"""
N=len(dist)
hp=[]
for i in range(N):
heappush(hp,dist[i]*B+i)
while hp:
cv=heappop(hp)
c, v = cv//B,cv%B
if dist[v] < c:
continue
for u, cost in G[v]:
if dist[v] + cost < dist[u]:
# TODO: Your code here
return dist
##################################################
#########################################
from collections import deque
class Convex_Hull_Trick():
#https://tjkendev.github.io/procon-library/python/convex_hull_trick/deque.html
#追加する傾きが単調かつqueryのxが単調
#単調性なしが良いならこちらへ(queryのxは先読み) https://judge.yosupo.jp/submission/30579
def __init__(self):
self.deq=deque()
def check(self,f1, f2, f3):
return (f2[0] - f1[0]) * (f3[1] - f2[1]) >= (f2[1] - f1[1]) * (f3[0] - f2[0])
def f(self,f1, x):
return f1[0] * x + f1[1]
# add f_i(x) = a*x + b
def add_line(self,a, b):
f1 = (a, b)
while len(self.deq) >= 2 and self.check(self.deq[-2], self.deq[-1], f1):
self.deq.pop()
self.deq.append(f1)
# min f_i(x)
def query(self,x):
while len(self.deq) >= 2 and self.f(self.deq[0], x) >= self.f(self.deq[1], x):
self.deq.popleft()
return self.f(self.deq[0], x)
##################################
n,m,k=map(int,input().split())
root=[[] for i in range(n+2)]
for i in range(m):
a,b,c=map(int,input().split())
root[a].append((b,c))
root[b].append((a,c))
dp=[10**11]*(n+1)
dp[1]=0
dp=dijkstra(root,dp)
for iii in range(k):
newdp=[10**11]*(n+1)
cht=Convex_Hull_Trick()
for i in range(1,n+1):
cht.add_line(-2*i,dp[i]+i**2)
for i in range(1,n+1):
newdp[i]=cht.query(i)+i**2
dp=newdp[:]
dp=dijkstra(root,dp)
print(*dp[1:])
|
import sys
input=sys.stdin.readline #文字列入力はするな！！
########################################
from heapq import heappush, heappop
B=10**5+10
def dijkstra( G, dist, INF=10**11):
"""
https://tjkendev.github.io/procon-library/python/graph/dijkstra.html
O((|E|+|V|)log|V|)
V: 頂点数
G[v] = [(nod, cost)]:
頂点vから遷移可能な頂点(nod)とそのコスト(cost)
s: 始点の頂点"""
N=len(dist)
hp=[]
for i in range(N):
heappush(hp,dist[i]*B+i)
while hp:
cv=heappop(hp)
c, v = cv//B,cv%B
if dist[v] < c:
continue
for u, cost in G[v]:
if dist[v] + cost < dist[u]:
{{completion}}
return dist
##################################################
#########################################
from collections import deque
class Convex_Hull_Trick():
#https://tjkendev.github.io/procon-library/python/convex_hull_trick/deque.html
#追加する傾きが単調かつqueryのxが単調
#単調性なしが良いならこちらへ(queryのxは先読み) https://judge.yosupo.jp/submission/30579
def __init__(self):
self.deq=deque()
def check(self,f1, f2, f3):
return (f2[0] - f1[0]) * (f3[1] - f2[1]) >= (f2[1] - f1[1]) * (f3[0] - f2[0])
def f(self,f1, x):
return f1[0] * x + f1[1]
# add f_i(x) = a*x + b
def add_line(self,a, b):
f1 = (a, b)
while len(self.deq) >= 2 and self.check(self.deq[-2], self.deq[-1], f1):
self.deq.pop()
self.deq.append(f1)
# min f_i(x)
def query(self,x):
while len(self.deq) >= 2 and self.f(self.deq[0], x) >= self.f(self.deq[1], x):
self.deq.popleft()
return self.f(self.deq[0], x)
##################################
n,m,k=map(int,input().split())
root=[[] for i in range(n+2)]
for i in range(m):
a,b,c=map(int,input().split())
root[a].append((b,c))
root[b].append((a,c))
dp=[10**11]*(n+1)
dp[1]=0
dp=dijkstra(root,dp)
for iii in range(k):
newdp=[10**11]*(n+1)
cht=Convex_Hull_Trick()
for i in range(1,n+1):
cht.add_line(-2*i,dp[i]+i**2)
for i in range(1,n+1):
newdp[i]=cht.query(i)+i**2
dp=newdp[:]
dp=dijkstra(root,dp)
print(*dp[1:])
|
dist[u] = dist[v] + cost
heappush(hp, dist[u]*B+u)
|
[{"input": "3 1 2\n1 3 1", "output": ["0 1 1"]}, {"input": "4 3 1\n1 2 3\n2 4 5\n3 4 7", "output": ["0 1 4 6"]}, {"input": "2 1 1\n2 1 893746473", "output": ["0 1"]}, {"input": "5 5 2\n2 1 33\n1 5 93\n5 3 48\n2 3 21\n4 2 1", "output": ["0 1 2 2 3"]}]
|
block_completion_000108
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: Stanley lives in a country that consists of $$$n$$$ cities (he lives in city $$$1$$$). There are bidirectional roads between some of the cities, and you know how long it takes to ride through each of them. Additionally, there is a flight between each pair of cities, the flight between cities $$$u$$$ and $$$v$$$ takes $$$(u - v)^2$$$ time.Stanley is quite afraid of flying because of watching "Sully: Miracle on the Hudson" recently, so he can take at most $$$k$$$ flights. Stanley wants to know the minimum time of a journey to each of the $$$n$$$ cities from the city $$$1$$$.
Input Specification: In the first line of input there are three integers $$$n$$$, $$$m$$$, and $$$k$$$ ($$$2 \leq n \leq 10^{5}$$$, $$$1 \leq m \leq 10^{5}$$$, $$$1 \leq k \leq 20$$$) — the number of cities, the number of roads, and the maximal number of flights Stanley can take. The following $$$m$$$ lines describe the roads. Each contains three integers $$$u$$$, $$$v$$$, $$$w$$$ ($$$1 \leq u, v \leq n$$$, $$$u \neq v$$$, $$$1 \leq w \leq 10^{9}$$$) — the cities the road connects and the time it takes to ride through. Note that some pairs of cities may be connected by more than one road.
Output Specification: Print $$$n$$$ integers, $$$i$$$-th of which is equal to the minimum time of traveling to city $$$i$$$.
Notes: NoteIn the first sample, it takes no time to get to city 1; to get to city 2 it is possible to use a flight between 1 and 2, which will take 1 unit of time; to city 3 you can get via a road from city 1, which will take 1 unit of time. In the second sample, it also takes no time to get to city 1. To get to city 2 Stanley should use a flight between 1 and 2, which will take 1 unit of time. To get to city 3 Stanley can ride between cities 1 and 2, which will take 3 units of time, and then use a flight between 2 and 3. To get to city 4 Stanley should use a flight between 1 and 2, then take a ride from 2 to 4, which will take 5 units of time.
Code:
import sys
input=sys.stdin.readline #文字列入力はするな！！
########################################
from heapq import heappush, heappop
B=10**5+10
def dijkstra( G, dist, INF=10**11):
"""
https://tjkendev.github.io/procon-library/python/graph/dijkstra.html
O((|E|+|V|)log|V|)
V: 頂点数
G[v] = [(nod, cost)]:
頂点vから遷移可能な頂点(nod)とそのコスト(cost)
s: 始点の頂点"""
N=len(dist)
hp=[]
for i in range(N):
heappush(hp,dist[i]*B+i)
while hp:
cv=heappop(hp)
c, v = cv//B,cv%B
if dist[v] < c:
# TODO: Your code here
for u, cost in G[v]:
if dist[v] + cost < dist[u]:
dist[u] = dist[v] + cost
heappush(hp, dist[u]*B+u)
return dist
##################################################
#########################################
from collections import deque
class Convex_Hull_Trick():
#https://tjkendev.github.io/procon-library/python/convex_hull_trick/deque.html
#追加する傾きが単調かつqueryのxが単調
#単調性なしが良いならこちらへ(queryのxは先読み) https://judge.yosupo.jp/submission/30579
def __init__(self):
self.deq=deque()
def check(self,f1, f2, f3):
return (f2[0] - f1[0]) * (f3[1] - f2[1]) >= (f2[1] - f1[1]) * (f3[0] - f2[0])
def f(self,f1, x):
return f1[0] * x + f1[1]
# add f_i(x) = a*x + b
def add_line(self,a, b):
f1 = (a, b)
while len(self.deq) >= 2 and self.check(self.deq[-2], self.deq[-1], f1):
self.deq.pop()
self.deq.append(f1)
# min f_i(x)
def query(self,x):
while len(self.deq) >= 2 and self.f(self.deq[0], x) >= self.f(self.deq[1], x):
self.deq.popleft()
return self.f(self.deq[0], x)
##################################
n,m,k=map(int,input().split())
root=[[] for i in range(n+2)]
for i in range(m):
a,b,c=map(int,input().split())
root[a].append((b,c))
root[b].append((a,c))
dp=[10**11]*(n+1)
dp[1]=0
dp=dijkstra(root,dp)
for iii in range(k):
newdp=[10**11]*(n+1)
cht=Convex_Hull_Trick()
for i in range(1,n+1):
cht.add_line(-2*i,dp[i]+i**2)
for i in range(1,n+1):
newdp[i]=cht.query(i)+i**2
dp=newdp[:]
dp=dijkstra(root,dp)
print(*dp[1:])
|
import sys
input=sys.stdin.readline #文字列入力はするな！！
########################################
from heapq import heappush, heappop
B=10**5+10
def dijkstra( G, dist, INF=10**11):
"""
https://tjkendev.github.io/procon-library/python/graph/dijkstra.html
O((|E|+|V|)log|V|)
V: 頂点数
G[v] = [(nod, cost)]:
頂点vから遷移可能な頂点(nod)とそのコスト(cost)
s: 始点の頂点"""
N=len(dist)
hp=[]
for i in range(N):
heappush(hp,dist[i]*B+i)
while hp:
cv=heappop(hp)
c, v = cv//B,cv%B
if dist[v] < c:
{{completion}}
for u, cost in G[v]:
if dist[v] + cost < dist[u]:
dist[u] = dist[v] + cost
heappush(hp, dist[u]*B+u)
return dist
##################################################
#########################################
from collections import deque
class Convex_Hull_Trick():
#https://tjkendev.github.io/procon-library/python/convex_hull_trick/deque.html
#追加する傾きが単調かつqueryのxが単調
#単調性なしが良いならこちらへ(queryのxは先読み) https://judge.yosupo.jp/submission/30579
def __init__(self):
self.deq=deque()
def check(self,f1, f2, f3):
return (f2[0] - f1[0]) * (f3[1] - f2[1]) >= (f2[1] - f1[1]) * (f3[0] - f2[0])
def f(self,f1, x):
return f1[0] * x + f1[1]
# add f_i(x) = a*x + b
def add_line(self,a, b):
f1 = (a, b)
while len(self.deq) >= 2 and self.check(self.deq[-2], self.deq[-1], f1):
self.deq.pop()
self.deq.append(f1)
# min f_i(x)
def query(self,x):
while len(self.deq) >= 2 and self.f(self.deq[0], x) >= self.f(self.deq[1], x):
self.deq.popleft()
return self.f(self.deq[0], x)
##################################
n,m,k=map(int,input().split())
root=[[] for i in range(n+2)]
for i in range(m):
a,b,c=map(int,input().split())
root[a].append((b,c))
root[b].append((a,c))
dp=[10**11]*(n+1)
dp[1]=0
dp=dijkstra(root,dp)
for iii in range(k):
newdp=[10**11]*(n+1)
cht=Convex_Hull_Trick()
for i in range(1,n+1):
cht.add_line(-2*i,dp[i]+i**2)
for i in range(1,n+1):
newdp[i]=cht.query(i)+i**2
dp=newdp[:]
dp=dijkstra(root,dp)
print(*dp[1:])
|
continue
|
[{"input": "3 1 2\n1 3 1", "output": ["0 1 1"]}, {"input": "4 3 1\n1 2 3\n2 4 5\n3 4 7", "output": ["0 1 4 6"]}, {"input": "2 1 1\n2 1 893746473", "output": ["0 1"]}, {"input": "5 5 2\n2 1 33\n1 5 93\n5 3 48\n2 3 21\n4 2 1", "output": ["0 1 2 2 3"]}]
|
block_completion_000109
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: You are walking with your dog, and now you are at the promenade. The promenade can be represented as an infinite line. Initially, you are in the point $$$0$$$ with your dog. You decided to give some freedom to your dog, so you untied her and let her run for a while. Also, you watched what your dog is doing, so you have some writings about how she ran. During the $$$i$$$-th minute, the dog position changed from her previous position by the value $$$a_i$$$ (it means, that the dog ran for $$$a_i$$$ meters during the $$$i$$$-th minute). If $$$a_i$$$ is positive, the dog ran $$$a_i$$$ meters to the right, otherwise (if $$$a_i$$$ is negative) she ran $$$a_i$$$ meters to the left.During some minutes, you were chatting with your friend, so you don't have writings about your dog movement during these minutes. These values $$$a_i$$$ equal zero.You want your dog to return to you after the end of the walk, so the destination point of the dog after $$$n$$$ minutes should be $$$0$$$.Now you are wondering: what is the maximum possible number of different integer points of the line your dog could visit on her way, if you replace every $$$0$$$ with some integer from $$$-k$$$ to $$$k$$$ (and your dog should return to $$$0$$$ after the walk)? The dog visits an integer point if she runs through that point or reaches in it at the end of any minute. Point $$$0$$$ is always visited by the dog, since she is initially there.If the dog cannot return to the point $$$0$$$ after $$$n$$$ minutes regardless of the integers you place, print -1.
Input Specification: The first line of the input contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 3000; 1 \le k \le 10^9$$$) — the number of minutes and the maximum possible speed of your dog during the minutes without records. The second line of the input contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$-10^9 \le a_i \le 10^9$$$), where $$$a_i$$$ is the number of meters your dog ran during the $$$i$$$-th minutes (to the left if $$$a_i$$$ is negative, to the right otherwise). If $$$a_i = 0$$$ then this value is unknown and can be replaced with any integer from the range $$$[-k; k]$$$.
Output Specification: If the dog cannot return to the point $$$0$$$ after $$$n$$$ minutes regardless of the set of integers you place, print -1. Otherwise, print one integer — the maximum number of different integer points your dog could visit if you fill all the unknown values optimally and the dog will return to the point $$$0$$$ at the end of the walk.
Code:
n,k=map(int,input().split())
l=list(map(int,input().split()))
ans=-2
b=l.count(0)
for y in range(n):
a=l[y:]+l[:y]
ind=[]
s=0
for i in range(n):
if a[i]==0:
# TODO: Your code here
s+=a[i]
while s>0 and len(ind)>0:
a[ind[-1]]=max(k-s,-k)
s+=(-k+a[ind[-1]])
ind=ind[:-1]
s=0
f=0
for i in range(n):
s+=a[i]
f=max(f,s)
if sum(a)==0:
ans=max(ans,f)
print(ans+1)
|
n,k=map(int,input().split())
l=list(map(int,input().split()))
ans=-2
b=l.count(0)
for y in range(n):
a=l[y:]+l[:y]
ind=[]
s=0
for i in range(n):
if a[i]==0:
{{completion}}
s+=a[i]
while s>0 and len(ind)>0:
a[ind[-1]]=max(k-s,-k)
s+=(-k+a[ind[-1]])
ind=ind[:-1]
s=0
f=0
for i in range(n):
s+=a[i]
f=max(f,s)
if sum(a)==0:
ans=max(ans,f)
print(ans+1)
|
ind+=[i]
a[i]=k
|
[{"input": "3 2\n5 0 -4", "output": ["6"]}, {"input": "6 4\n1 -2 0 3 -4 5", "output": ["7"]}, {"input": "3 1000000000\n0 0 0", "output": ["1000000001"]}, {"input": "5 9\n-7 -3 8 12 0", "output": ["-1"]}, {"input": "5 3\n-1 0 3 3 0", "output": ["7"]}, {"input": "5 4\n0 2 0 3 0", "output": ["9"]}]
|
block_completion_000198
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: You are walking with your dog, and now you are at the promenade. The promenade can be represented as an infinite line. Initially, you are in the point $$$0$$$ with your dog. You decided to give some freedom to your dog, so you untied her and let her run for a while. Also, you watched what your dog is doing, so you have some writings about how she ran. During the $$$i$$$-th minute, the dog position changed from her previous position by the value $$$a_i$$$ (it means, that the dog ran for $$$a_i$$$ meters during the $$$i$$$-th minute). If $$$a_i$$$ is positive, the dog ran $$$a_i$$$ meters to the right, otherwise (if $$$a_i$$$ is negative) she ran $$$a_i$$$ meters to the left.During some minutes, you were chatting with your friend, so you don't have writings about your dog movement during these minutes. These values $$$a_i$$$ equal zero.You want your dog to return to you after the end of the walk, so the destination point of the dog after $$$n$$$ minutes should be $$$0$$$.Now you are wondering: what is the maximum possible number of different integer points of the line your dog could visit on her way, if you replace every $$$0$$$ with some integer from $$$-k$$$ to $$$k$$$ (and your dog should return to $$$0$$$ after the walk)? The dog visits an integer point if she runs through that point or reaches in it at the end of any minute. Point $$$0$$$ is always visited by the dog, since she is initially there.If the dog cannot return to the point $$$0$$$ after $$$n$$$ minutes regardless of the integers you place, print -1.
Input Specification: The first line of the input contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 3000; 1 \le k \le 10^9$$$) — the number of minutes and the maximum possible speed of your dog during the minutes without records. The second line of the input contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$-10^9 \le a_i \le 10^9$$$), where $$$a_i$$$ is the number of meters your dog ran during the $$$i$$$-th minutes (to the left if $$$a_i$$$ is negative, to the right otherwise). If $$$a_i = 0$$$ then this value is unknown and can be replaced with any integer from the range $$$[-k; k]$$$.
Output Specification: If the dog cannot return to the point $$$0$$$ after $$$n$$$ minutes regardless of the set of integers you place, print -1. Otherwise, print one integer — the maximum number of different integer points your dog could visit if you fill all the unknown values optimally and the dog will return to the point $$$0$$$ at the end of the walk.
Code:
R=lambda:map(int,input().split())
n,k=R();n+=1
a=[0]+[*R()]
p0,p=[0]*n,[0]*n
for i in range(1,n):
p0[i]=p0[i-1]+int(a[i]==0)
p[i]=p[i-1]+a[i]
s=p[-1]
if p0[-1]*k<abs(s):
res=-1
else:
res=0
for i in range(n):
for j in range(i+1,n):
# TODO: Your code here
print(res)
|
R=lambda:map(int,input().split())
n,k=R();n+=1
a=[0]+[*R()]
p0,p=[0]*n,[0]*n
for i in range(1,n):
p0[i]=p0[i-1]+int(a[i]==0)
p[i]=p[i-1]+a[i]
s=p[-1]
if p0[-1]*k<abs(s):
res=-1
else:
res=0
for i in range(n):
for j in range(i+1,n):
{{completion}}
print(res)
|
l0=p0[j]-p0[i];r0=p0[-1]-l0
l,r=max(-l0*k, -s-r0*k),min(l0*k, -s+r0*k)
v=p[j]-p[i]
res=max(res, 1+abs(v+l), 1+abs(v+r))
|
[{"input": "3 2\n5 0 -4", "output": ["6"]}, {"input": "6 4\n1 -2 0 3 -4 5", "output": ["7"]}, {"input": "3 1000000000\n0 0 0", "output": ["1000000001"]}, {"input": "5 9\n-7 -3 8 12 0", "output": ["-1"]}, {"input": "5 3\n-1 0 3 3 0", "output": ["7"]}, {"input": "5 4\n0 2 0 3 0", "output": ["9"]}]
|
block_completion_000199
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: You are walking with your dog, and now you are at the promenade. The promenade can be represented as an infinite line. Initially, you are in the point $$$0$$$ with your dog. You decided to give some freedom to your dog, so you untied her and let her run for a while. Also, you watched what your dog is doing, so you have some writings about how she ran. During the $$$i$$$-th minute, the dog position changed from her previous position by the value $$$a_i$$$ (it means, that the dog ran for $$$a_i$$$ meters during the $$$i$$$-th minute). If $$$a_i$$$ is positive, the dog ran $$$a_i$$$ meters to the right, otherwise (if $$$a_i$$$ is negative) she ran $$$a_i$$$ meters to the left.During some minutes, you were chatting with your friend, so you don't have writings about your dog movement during these minutes. These values $$$a_i$$$ equal zero.You want your dog to return to you after the end of the walk, so the destination point of the dog after $$$n$$$ minutes should be $$$0$$$.Now you are wondering: what is the maximum possible number of different integer points of the line your dog could visit on her way, if you replace every $$$0$$$ with some integer from $$$-k$$$ to $$$k$$$ (and your dog should return to $$$0$$$ after the walk)? The dog visits an integer point if she runs through that point or reaches in it at the end of any minute. Point $$$0$$$ is always visited by the dog, since she is initially there.If the dog cannot return to the point $$$0$$$ after $$$n$$$ minutes regardless of the integers you place, print -1.
Input Specification: The first line of the input contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 3000; 1 \le k \le 10^9$$$) — the number of minutes and the maximum possible speed of your dog during the minutes without records. The second line of the input contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$-10^9 \le a_i \le 10^9$$$), where $$$a_i$$$ is the number of meters your dog ran during the $$$i$$$-th minutes (to the left if $$$a_i$$$ is negative, to the right otherwise). If $$$a_i = 0$$$ then this value is unknown and can be replaced with any integer from the range $$$[-k; k]$$$.
Output Specification: If the dog cannot return to the point $$$0$$$ after $$$n$$$ minutes regardless of the set of integers you place, print -1. Otherwise, print one integer — the maximum number of different integer points your dog could visit if you fill all the unknown values optimally and the dog will return to the point $$$0$$$ at the end of the walk.
Code:
import sys
input = sys.stdin.readline
def ProGamerMove():
n, k = map(int, input().split())
a = list(map(int, input().split()))
zeros = a.count(0)
sm = sum(a)
s1, s2 = 0, 0
c1, c2 = 0, 0
res = -2
def intersect(m1, b1, m2, b2):
l1, r1 = m1 - b1 * k, m1 + b1 * k
l2, r2 = m2 - b2 * k, m2 + b2 * k
return not (r1 < l2 or r2 < l1)
for l in range(0, n + 1):
s2, c2 = 0, 0
for r in range(0, n + 1):
if l <= r:
b1, b2, b3 = c1, c2 - c1, zeros - c2
m1, m2, m3 = s1, s2 - s1, sm - s2
# b1 + b3 == -b2
# min b1, b3
if not intersect(m1 + m3, b1 + b3, -m2, b2): # TODO: Your code here
r1 = max(m1 + m3 - (b1 + b3) * k, -m2 - b2 * k)
r2 = min(m1 + m3 + (b1 + b3) * k, -m2 + b2 * k)
res = max(res, abs(r1), abs(r2))
if r < n:
s2 += a[r]
c2 += a[r] == 0
if l < n:
s1 += a[l]
c1 += a[l] == 0
print(res + 1)
n = 1
#n = int(input())
for _ in range(0, n): ProGamerMove()
|
import sys
input = sys.stdin.readline
def ProGamerMove():
n, k = map(int, input().split())
a = list(map(int, input().split()))
zeros = a.count(0)
sm = sum(a)
s1, s2 = 0, 0
c1, c2 = 0, 0
res = -2
def intersect(m1, b1, m2, b2):
l1, r1 = m1 - b1 * k, m1 + b1 * k
l2, r2 = m2 - b2 * k, m2 + b2 * k
return not (r1 < l2 or r2 < l1)
for l in range(0, n + 1):
s2, c2 = 0, 0
for r in range(0, n + 1):
if l <= r:
b1, b2, b3 = c1, c2 - c1, zeros - c2
m1, m2, m3 = s1, s2 - s1, sm - s2
# b1 + b3 == -b2
# min b1, b3
if not intersect(m1 + m3, b1 + b3, -m2, b2): {{completion}}
r1 = max(m1 + m3 - (b1 + b3) * k, -m2 - b2 * k)
r2 = min(m1 + m3 + (b1 + b3) * k, -m2 + b2 * k)
res = max(res, abs(r1), abs(r2))
if r < n:
s2 += a[r]
c2 += a[r] == 0
if l < n:
s1 += a[l]
c1 += a[l] == 0
print(res + 1)
n = 1
#n = int(input())
for _ in range(0, n): ProGamerMove()
|
continue
|
[{"input": "3 2\n5 0 -4", "output": ["6"]}, {"input": "6 4\n1 -2 0 3 -4 5", "output": ["7"]}, {"input": "3 1000000000\n0 0 0", "output": ["1000000001"]}, {"input": "5 9\n-7 -3 8 12 0", "output": ["-1"]}, {"input": "5 3\n-1 0 3 3 0", "output": ["7"]}, {"input": "5 4\n0 2 0 3 0", "output": ["9"]}]
|
block_completion_000200
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: You are walking with your dog, and now you are at the promenade. The promenade can be represented as an infinite line. Initially, you are in the point $$$0$$$ with your dog. You decided to give some freedom to your dog, so you untied her and let her run for a while. Also, you watched what your dog is doing, so you have some writings about how she ran. During the $$$i$$$-th minute, the dog position changed from her previous position by the value $$$a_i$$$ (it means, that the dog ran for $$$a_i$$$ meters during the $$$i$$$-th minute). If $$$a_i$$$ is positive, the dog ran $$$a_i$$$ meters to the right, otherwise (if $$$a_i$$$ is negative) she ran $$$a_i$$$ meters to the left.During some minutes, you were chatting with your friend, so you don't have writings about your dog movement during these minutes. These values $$$a_i$$$ equal zero.You want your dog to return to you after the end of the walk, so the destination point of the dog after $$$n$$$ minutes should be $$$0$$$.Now you are wondering: what is the maximum possible number of different integer points of the line your dog could visit on her way, if you replace every $$$0$$$ with some integer from $$$-k$$$ to $$$k$$$ (and your dog should return to $$$0$$$ after the walk)? The dog visits an integer point if she runs through that point or reaches in it at the end of any minute. Point $$$0$$$ is always visited by the dog, since she is initially there.If the dog cannot return to the point $$$0$$$ after $$$n$$$ minutes regardless of the integers you place, print -1.
Input Specification: The first line of the input contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 3000; 1 \le k \le 10^9$$$) — the number of minutes and the maximum possible speed of your dog during the minutes without records. The second line of the input contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$-10^9 \le a_i \le 10^9$$$), where $$$a_i$$$ is the number of meters your dog ran during the $$$i$$$-th minutes (to the left if $$$a_i$$$ is negative, to the right otherwise). If $$$a_i = 0$$$ then this value is unknown and can be replaced with any integer from the range $$$[-k; k]$$$.
Output Specification: If the dog cannot return to the point $$$0$$$ after $$$n$$$ minutes regardless of the set of integers you place, print -1. Otherwise, print one integer — the maximum number of different integer points your dog could visit if you fill all the unknown values optimally and the dog will return to the point $$$0$$$ at the end of the walk.
Code:
n, k = map(int, input().split())
A = list(map(int, input().split()))
ans = 0
for i in range(n):
C = [0]*n
for j in range(n-1, -1, -1):
if A[j] == 0:
C[j] = 1
if j+1 < n:
C[j] += C[j+1]
B = A.copy()
s = sum(B)
flag = True
for j in range(n):
if B[j] == 0:
if j+1 < n:
x = C[j+1]
else:
# TODO: Your code here
B[j] = min(k, x*k-s)
if B[j] < -k:
flag = False
s += B[j]
if flag:
pos = 0
mn = 0
mx = 0
for j in range(n):
pos += B[j]
mn = min(mn, pos)
mx = max(mx, pos)
if pos == 0:
ans = max(ans, mx-mn+1)
A = A[1:]+A[0:1]
if ans != 0:
print(ans)
else:
print(-1)
|
n, k = map(int, input().split())
A = list(map(int, input().split()))
ans = 0
for i in range(n):
C = [0]*n
for j in range(n-1, -1, -1):
if A[j] == 0:
C[j] = 1
if j+1 < n:
C[j] += C[j+1]
B = A.copy()
s = sum(B)
flag = True
for j in range(n):
if B[j] == 0:
if j+1 < n:
x = C[j+1]
else:
{{completion}}
B[j] = min(k, x*k-s)
if B[j] < -k:
flag = False
s += B[j]
if flag:
pos = 0
mn = 0
mx = 0
for j in range(n):
pos += B[j]
mn = min(mn, pos)
mx = max(mx, pos)
if pos == 0:
ans = max(ans, mx-mn+1)
A = A[1:]+A[0:1]
if ans != 0:
print(ans)
else:
print(-1)
|
x = 0
|
[{"input": "3 2\n5 0 -4", "output": ["6"]}, {"input": "6 4\n1 -2 0 3 -4 5", "output": ["7"]}, {"input": "3 1000000000\n0 0 0", "output": ["1000000001"]}, {"input": "5 9\n-7 -3 8 12 0", "output": ["-1"]}, {"input": "5 3\n-1 0 3 3 0", "output": ["7"]}, {"input": "5 4\n0 2 0 3 0", "output": ["9"]}]
|
block_completion_000201
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: You are given a positive integer $$$n$$$. Since $$$n$$$ may be very large, you are given its binary representation.You should compute the number of triples $$$(a,b,c)$$$ with $$$0 \leq a,b,c \leq n$$$ such that $$$a \oplus b$$$, $$$b \oplus c$$$, and $$$a \oplus c$$$ are the sides of a non-degenerate triangle. Here, $$$\oplus$$$ denotes the bitwise XOR operation.You should output the answer modulo $$$998\,244\,353$$$.Three positive values $$$x$$$, $$$y$$$, and $$$z$$$ are the sides of a non-degenerate triangle if and only if $$$x+y>z$$$, $$$x+z>y$$$, and $$$y+z>x$$$.
Input Specification: The first and only line contains the binary representation of an integer $$$n$$$ ($$$0 < n < 2^{200\,000}$$$) without leading zeros. For example, the string 10 is the binary representation of the number $$$2$$$, while the string 1010 represents the number $$$10$$$.
Output Specification: Print one integer — the number of triples $$$(a,b,c)$$$ satisfying the conditions described in the statement modulo $$$998\,244\,353$$$.
Notes: NoteIn the first test case, $$$101_2=5$$$. The triple $$$(a, b, c) = (0, 3, 5)$$$ is valid because $$$(a\oplus b, b\oplus c, c\oplus a) = (3, 6, 5)$$$ are the sides of a non-degenerate triangle. The triple $$$(a, b, c) = (1, 2, 4)$$$ is valid because $$$(a\oplus b, b\oplus c, c\oplus a) = (3, 6, 5)$$$ are the sides of a non-degenerate triangle. The $$$6$$$ permutations of each of these two triples are all the valid triples, thus the answer is $$$12$$$.In the third test case, $$$11\,011\,111\,101\,010\,010_2=114\,514$$$. The full answer (before taking the modulo) is $$$1\,466\,408\,118\,808\,164$$$.
Code:
MOD = 998244353
TRANS = [6, 3, 7, 4, 1, 0]
s = input().strip()
dp = [0] * 7 + [1]
for c in map(int, s):
dp1 = [0] * 8
for i in range(8):
for k in TRANS:
if c:
dp1[k & i] += dp[i]
elif (k & i) == 0:
# TODO: Your code here
dp = [x % MOD for x in dp1]
n = int(s, base=2) + 1
print((n**3 + 3 * n**2 - n - 3 * sum(dp)) % MOD)
|
MOD = 998244353
TRANS = [6, 3, 7, 4, 1, 0]
s = input().strip()
dp = [0] * 7 + [1]
for c in map(int, s):
dp1 = [0] * 8
for i in range(8):
for k in TRANS:
if c:
dp1[k & i] += dp[i]
elif (k & i) == 0:
{{completion}}
dp = [x % MOD for x in dp1]
n = int(s, base=2) + 1
print((n**3 + 3 * n**2 - n - 3 * sum(dp)) % MOD)
|
dp1[i] += dp[i]
|
[{"input": "101", "output": ["12"]}, {"input": "1110", "output": ["780"]}, {"input": "11011111101010010", "output": ["141427753"]}]
|
block_completion_000281
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: You are given a positive integer $$$n$$$. Since $$$n$$$ may be very large, you are given its binary representation.You should compute the number of triples $$$(a,b,c)$$$ with $$$0 \leq a,b,c \leq n$$$ such that $$$a \oplus b$$$, $$$b \oplus c$$$, and $$$a \oplus c$$$ are the sides of a non-degenerate triangle. Here, $$$\oplus$$$ denotes the bitwise XOR operation.You should output the answer modulo $$$998\,244\,353$$$.Three positive values $$$x$$$, $$$y$$$, and $$$z$$$ are the sides of a non-degenerate triangle if and only if $$$x+y>z$$$, $$$x+z>y$$$, and $$$y+z>x$$$.
Input Specification: The first and only line contains the binary representation of an integer $$$n$$$ ($$$0 < n < 2^{200\,000}$$$) without leading zeros. For example, the string 10 is the binary representation of the number $$$2$$$, while the string 1010 represents the number $$$10$$$.
Output Specification: Print one integer — the number of triples $$$(a,b,c)$$$ satisfying the conditions described in the statement modulo $$$998\,244\,353$$$.
Notes: NoteIn the first test case, $$$101_2=5$$$. The triple $$$(a, b, c) = (0, 3, 5)$$$ is valid because $$$(a\oplus b, b\oplus c, c\oplus a) = (3, 6, 5)$$$ are the sides of a non-degenerate triangle. The triple $$$(a, b, c) = (1, 2, 4)$$$ is valid because $$$(a\oplus b, b\oplus c, c\oplus a) = (3, 6, 5)$$$ are the sides of a non-degenerate triangle. The $$$6$$$ permutations of each of these two triples are all the valid triples, thus the answer is $$$12$$$.In the third test case, $$$11\,011\,111\,101\,010\,010_2=114\,514$$$. The full answer (before taking the modulo) is $$$1\,466\,408\,118\,808\,164$$$.
Code:
MOD=998244353
TRANS=[6,3,7,4,1,0]
s=input().strip()
dp=[0]*7+[1]
for c in map(int,s):
dp1=[0]*8
for i in range(8):
for k in TRANS:
if c:
dp1[k&i]+=dp[i]
elif(k&i)==0:
# TODO: Your code here
dp=[x%MOD for x in dp1]
n=int(s,base=2)+1
print((n**3+3*n**2-n-3*sum(dp))%MOD)
|
MOD=998244353
TRANS=[6,3,7,4,1,0]
s=input().strip()
dp=[0]*7+[1]
for c in map(int,s):
dp1=[0]*8
for i in range(8):
for k in TRANS:
if c:
dp1[k&i]+=dp[i]
elif(k&i)==0:
{{completion}}
dp=[x%MOD for x in dp1]
n=int(s,base=2)+1
print((n**3+3*n**2-n-3*sum(dp))%MOD)
|
dp1[i]+=dp[i]
|
[{"input": "101", "output": ["12"]}, {"input": "1110", "output": ["780"]}, {"input": "11011111101010010", "output": ["141427753"]}]
|
block_completion_000282
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: We have an array of length $$$n$$$. Initially, each element is equal to $$$0$$$ and there is a pointer located on the first element.We can do the following two kinds of operations any number of times (possibly zero) in any order: If the pointer is not on the last element, increase the element the pointer is currently on by $$$1$$$. Then move it to the next element. If the pointer is not on the first element, decrease the element the pointer is currently on by $$$1$$$. Then move it to the previous element.But there is one additional rule. After we are done, the pointer has to be on the first element.You are given an array $$$a$$$. Determine whether it's possible to obtain $$$a$$$ after some operations or not.
Input Specification: The first line contains a single integer $$$t$$$ $$$(1\le t\le 1000)$$$ — the number of test cases. The description of the test cases follows. The first line of each test case contains a single integer $$$n$$$ $$$(1\le n\le 2 \cdot 10^5)$$$ — the size of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$-10^9 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases doesn't exceed $$$2 \cdot 10^5$$$.
Output Specification: For each test case, print "Yes" (without quotes) if it's possible to obtain $$$a$$$ after some operations, and "No" (without quotes) otherwise. You can output "Yes" and "No" in any case (for example, strings "yEs", "yes" and "Yes" will be recognized as a positive response).
Notes: NoteIn the first test case we can obtain the array after some operations, but the pointer won't be on the first element.One way of obtaining the array in the second test case is shown below.$$$\langle \underline{0}, 0, 0, 0\rangle \to \langle 1, \underline{0}, 0, 0 \rangle \to \langle \underline{1}, -1, 0, 0\rangle \to \langle 2, \underline{-1}, 0, 0\rangle \to \langle 2, 0, \underline{0}, 0\rangle \to \langle 2, \underline{0}, -1, 0\rangle \to \langle \underline{2}, -1, -1, 0\rangle$$$
Code:
I=input
for _ in [0]*int(I()):
I();p,z,zero=0,1,0
for v in I().split():
p+=int(v)
if zero and p>0:# TODO: Your code here
if p==0:zero=True
if p<0:z=0;break
print(['NO','YES'][zero and z])
|
I=input
for _ in [0]*int(I()):
I();p,z,zero=0,1,0
for v in I().split():
p+=int(v)
if zero and p>0:{{completion}}
if p==0:zero=True
if p<0:z=0;break
print(['NO','YES'][zero and z])
|
z=0;break
|
[{"input": "7\n2\n1 0\n4\n2 -1 -1 0\n4\n1 -4 3 0\n4\n1 -1 1 -1\n5\n1 2 3 4 -10\n7\n2 -1 1 -2 0 0 0\n1\n0", "output": ["No\nYes\nNo\nNo\nYes\nYes\nYes"]}]
|
block_completion_000421
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: We have an array of length $$$n$$$. Initially, each element is equal to $$$0$$$ and there is a pointer located on the first element.We can do the following two kinds of operations any number of times (possibly zero) in any order: If the pointer is not on the last element, increase the element the pointer is currently on by $$$1$$$. Then move it to the next element. If the pointer is not on the first element, decrease the element the pointer is currently on by $$$1$$$. Then move it to the previous element.But there is one additional rule. After we are done, the pointer has to be on the first element.You are given an array $$$a$$$. Determine whether it's possible to obtain $$$a$$$ after some operations or not.
Input Specification: The first line contains a single integer $$$t$$$ $$$(1\le t\le 1000)$$$ — the number of test cases. The description of the test cases follows. The first line of each test case contains a single integer $$$n$$$ $$$(1\le n\le 2 \cdot 10^5)$$$ — the size of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$-10^9 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases doesn't exceed $$$2 \cdot 10^5$$$.
Output Specification: For each test case, print "Yes" (without quotes) if it's possible to obtain $$$a$$$ after some operations, and "No" (without quotes) otherwise. You can output "Yes" and "No" in any case (for example, strings "yEs", "yes" and "Yes" will be recognized as a positive response).
Notes: NoteIn the first test case we can obtain the array after some operations, but the pointer won't be on the first element.One way of obtaining the array in the second test case is shown below.$$$\langle \underline{0}, 0, 0, 0\rangle \to \langle 1, \underline{0}, 0, 0 \rangle \to \langle \underline{1}, -1, 0, 0\rangle \to \langle 2, \underline{-1}, 0, 0\rangle \to \langle 2, 0, \underline{0}, 0\rangle \to \langle 2, \underline{0}, -1, 0\rangle \to \langle \underline{2}, -1, -1, 0\rangle$$$
Code:
I=input
for _ in [0]*int(I()):
I();p,z,zero=0,1,0
for v in I().split():
p+=int(v)
if zero and p>0:z=0;break
if p==0:# TODO: Your code here
if p<0:z=0;break
print(['NO','YES'][zero and z])
|
I=input
for _ in [0]*int(I()):
I();p,z,zero=0,1,0
for v in I().split():
p+=int(v)
if zero and p>0:z=0;break
if p==0:{{completion}}
if p<0:z=0;break
print(['NO','YES'][zero and z])
|
zero=True
|
[{"input": "7\n2\n1 0\n4\n2 -1 -1 0\n4\n1 -4 3 0\n4\n1 -1 1 -1\n5\n1 2 3 4 -10\n7\n2 -1 1 -2 0 0 0\n1\n0", "output": ["No\nYes\nNo\nNo\nYes\nYes\nYes"]}]
|
block_completion_000422
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: We have an array of length $$$n$$$. Initially, each element is equal to $$$0$$$ and there is a pointer located on the first element.We can do the following two kinds of operations any number of times (possibly zero) in any order: If the pointer is not on the last element, increase the element the pointer is currently on by $$$1$$$. Then move it to the next element. If the pointer is not on the first element, decrease the element the pointer is currently on by $$$1$$$. Then move it to the previous element.But there is one additional rule. After we are done, the pointer has to be on the first element.You are given an array $$$a$$$. Determine whether it's possible to obtain $$$a$$$ after some operations or not.
Input Specification: The first line contains a single integer $$$t$$$ $$$(1\le t\le 1000)$$$ — the number of test cases. The description of the test cases follows. The first line of each test case contains a single integer $$$n$$$ $$$(1\le n\le 2 \cdot 10^5)$$$ — the size of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$-10^9 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases doesn't exceed $$$2 \cdot 10^5$$$.
Output Specification: For each test case, print "Yes" (without quotes) if it's possible to obtain $$$a$$$ after some operations, and "No" (without quotes) otherwise. You can output "Yes" and "No" in any case (for example, strings "yEs", "yes" and "Yes" will be recognized as a positive response).
Notes: NoteIn the first test case we can obtain the array after some operations, but the pointer won't be on the first element.One way of obtaining the array in the second test case is shown below.$$$\langle \underline{0}, 0, 0, 0\rangle \to \langle 1, \underline{0}, 0, 0 \rangle \to \langle \underline{1}, -1, 0, 0\rangle \to \langle 2, \underline{-1}, 0, 0\rangle \to \langle 2, 0, \underline{0}, 0\rangle \to \langle 2, \underline{0}, -1, 0\rangle \to \langle \underline{2}, -1, -1, 0\rangle$$$
Code:
import sys
input = lambda : sys.stdin.readline().rstrip()
dx = [-1, 0, 1, 0]
dy = [0, -1, 0, 1]
def solve():
n = int(input())
arr = list(map(int, input().split()))
if sum(arr)!=0:
return 0
psum = 0
f = 0
for i in range(len(arr)):
psum += arr[i]
if psum < 0:
return 0
if psum==0:
f = 1
elif f:
# TODO: Your code here
return 1
for __ in range(int(input())):
print('Yes' if solve() else 'No')
|
import sys
input = lambda : sys.stdin.readline().rstrip()
dx = [-1, 0, 1, 0]
dy = [0, -1, 0, 1]
def solve():
n = int(input())
arr = list(map(int, input().split()))
if sum(arr)!=0:
return 0
psum = 0
f = 0
for i in range(len(arr)):
psum += arr[i]
if psum < 0:
return 0
if psum==0:
f = 1
elif f:
{{completion}}
return 1
for __ in range(int(input())):
print('Yes' if solve() else 'No')
|
return 0
|
[{"input": "7\n2\n1 0\n4\n2 -1 -1 0\n4\n1 -4 3 0\n4\n1 -1 1 -1\n5\n1 2 3 4 -10\n7\n2 -1 1 -2 0 0 0\n1\n0", "output": ["No\nYes\nNo\nNo\nYes\nYes\nYes"]}]
|
block_completion_000423
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: We have an array of length $$$n$$$. Initially, each element is equal to $$$0$$$ and there is a pointer located on the first element.We can do the following two kinds of operations any number of times (possibly zero) in any order: If the pointer is not on the last element, increase the element the pointer is currently on by $$$1$$$. Then move it to the next element. If the pointer is not on the first element, decrease the element the pointer is currently on by $$$1$$$. Then move it to the previous element.But there is one additional rule. After we are done, the pointer has to be on the first element.You are given an array $$$a$$$. Determine whether it's possible to obtain $$$a$$$ after some operations or not.
Input Specification: The first line contains a single integer $$$t$$$ $$$(1\le t\le 1000)$$$ — the number of test cases. The description of the test cases follows. The first line of each test case contains a single integer $$$n$$$ $$$(1\le n\le 2 \cdot 10^5)$$$ — the size of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$-10^9 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases doesn't exceed $$$2 \cdot 10^5$$$.
Output Specification: For each test case, print "Yes" (without quotes) if it's possible to obtain $$$a$$$ after some operations, and "No" (without quotes) otherwise. You can output "Yes" and "No" in any case (for example, strings "yEs", "yes" and "Yes" will be recognized as a positive response).
Notes: NoteIn the first test case we can obtain the array after some operations, but the pointer won't be on the first element.One way of obtaining the array in the second test case is shown below.$$$\langle \underline{0}, 0, 0, 0\rangle \to \langle 1, \underline{0}, 0, 0 \rangle \to \langle \underline{1}, -1, 0, 0\rangle \to \langle 2, \underline{-1}, 0, 0\rangle \to \langle 2, 0, \underline{0}, 0\rangle \to \langle 2, \underline{0}, -1, 0\rangle \to \langle \underline{2}, -1, -1, 0\rangle$$$
Code:
for _ in range(int(input())):
n = int(input())
a = list(map(int,input().split()))
tot = a[0]
for i in range(1, n):
if tot < 0:
break
elif tot == 0:
if a[i] != 0:
# TODO: Your code here
else:
tot += a[i]
else:
if tot == 0:
print("Yes")
continue
print("No")
|
for _ in range(int(input())):
n = int(input())
a = list(map(int,input().split()))
tot = a[0]
for i in range(1, n):
if tot < 0:
break
elif tot == 0:
if a[i] != 0:
{{completion}}
else:
tot += a[i]
else:
if tot == 0:
print("Yes")
continue
print("No")
|
break
|
[{"input": "7\n2\n1 0\n4\n2 -1 -1 0\n4\n1 -4 3 0\n4\n1 -1 1 -1\n5\n1 2 3 4 -10\n7\n2 -1 1 -2 0 0 0\n1\n0", "output": ["No\nYes\nNo\nNo\nYes\nYes\nYes"]}]
|
block_completion_000424
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: We have an array of length $$$n$$$. Initially, each element is equal to $$$0$$$ and there is a pointer located on the first element.We can do the following two kinds of operations any number of times (possibly zero) in any order: If the pointer is not on the last element, increase the element the pointer is currently on by $$$1$$$. Then move it to the next element. If the pointer is not on the first element, decrease the element the pointer is currently on by $$$1$$$. Then move it to the previous element.But there is one additional rule. After we are done, the pointer has to be on the first element.You are given an array $$$a$$$. Determine whether it's possible to obtain $$$a$$$ after some operations or not.
Input Specification: The first line contains a single integer $$$t$$$ $$$(1\le t\le 1000)$$$ — the number of test cases. The description of the test cases follows. The first line of each test case contains a single integer $$$n$$$ $$$(1\le n\le 2 \cdot 10^5)$$$ — the size of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$-10^9 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases doesn't exceed $$$2 \cdot 10^5$$$.
Output Specification: For each test case, print "Yes" (without quotes) if it's possible to obtain $$$a$$$ after some operations, and "No" (without quotes) otherwise. You can output "Yes" and "No" in any case (for example, strings "yEs", "yes" and "Yes" will be recognized as a positive response).
Notes: NoteIn the first test case we can obtain the array after some operations, but the pointer won't be on the first element.One way of obtaining the array in the second test case is shown below.$$$\langle \underline{0}, 0, 0, 0\rangle \to \langle 1, \underline{0}, 0, 0 \rangle \to \langle \underline{1}, -1, 0, 0\rangle \to \langle 2, \underline{-1}, 0, 0\rangle \to \langle 2, 0, \underline{0}, 0\rangle \to \langle 2, \underline{0}, -1, 0\rangle \to \langle \underline{2}, -1, -1, 0\rangle$$$
Code:
from sys import stdin
t = int(stdin.readline())
for h in range(t):
n = int(stdin.readline())
a = list(map(int,stdin.readline().split(' ')))
b = 0
v = True
for i in range(n):
b += a[i]
if b<0:
v = False
break
elif b==0:
for j in range(i+1,n):
if a[j] != 0:
# TODO: Your code here
break
if v and sum(a) == 0:
print('YES')
else:
print('NO')
|
from sys import stdin
t = int(stdin.readline())
for h in range(t):
n = int(stdin.readline())
a = list(map(int,stdin.readline().split(' ')))
b = 0
v = True
for i in range(n):
b += a[i]
if b<0:
v = False
break
elif b==0:
for j in range(i+1,n):
if a[j] != 0:
{{completion}}
break
if v and sum(a) == 0:
print('YES')
else:
print('NO')
|
v = False
break
|
[{"input": "7\n2\n1 0\n4\n2 -1 -1 0\n4\n1 -4 3 0\n4\n1 -1 1 -1\n5\n1 2 3 4 -10\n7\n2 -1 1 -2 0 0 0\n1\n0", "output": ["No\nYes\nNo\nNo\nYes\nYes\nYes"]}]
|
block_completion_000425
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: We have an array of length $$$n$$$. Initially, each element is equal to $$$0$$$ and there is a pointer located on the first element.We can do the following two kinds of operations any number of times (possibly zero) in any order: If the pointer is not on the last element, increase the element the pointer is currently on by $$$1$$$. Then move it to the next element. If the pointer is not on the first element, decrease the element the pointer is currently on by $$$1$$$. Then move it to the previous element.But there is one additional rule. After we are done, the pointer has to be on the first element.You are given an array $$$a$$$. Determine whether it's possible to obtain $$$a$$$ after some operations or not.
Input Specification: The first line contains a single integer $$$t$$$ $$$(1\le t\le 1000)$$$ — the number of test cases. The description of the test cases follows. The first line of each test case contains a single integer $$$n$$$ $$$(1\le n\le 2 \cdot 10^5)$$$ — the size of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$-10^9 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases doesn't exceed $$$2 \cdot 10^5$$$.
Output Specification: For each test case, print "Yes" (without quotes) if it's possible to obtain $$$a$$$ after some operations, and "No" (without quotes) otherwise. You can output "Yes" and "No" in any case (for example, strings "yEs", "yes" and "Yes" will be recognized as a positive response).
Notes: NoteIn the first test case we can obtain the array after some operations, but the pointer won't be on the first element.One way of obtaining the array in the second test case is shown below.$$$\langle \underline{0}, 0, 0, 0\rangle \to \langle 1, \underline{0}, 0, 0 \rangle \to \langle \underline{1}, -1, 0, 0\rangle \to \langle 2, \underline{-1}, 0, 0\rangle \to \langle 2, 0, \underline{0}, 0\rangle \to \langle 2, \underline{0}, -1, 0\rangle \to \langle \underline{2}, -1, -1, 0\rangle$$$
Code:
for _ in [0]*int(input()):
input()
n = list(map(int,input().split()))
s,f,m = 0,0,0
for i in n:
s+=i
if s<0:# TODO: Your code here
if s==0:f=1
if f and s>0:m=1;break
print("YNEOS"[(m or not f)::2])
|
for _ in [0]*int(input()):
input()
n = list(map(int,input().split()))
s,f,m = 0,0,0
for i in n:
s+=i
if s<0:{{completion}}
if s==0:f=1
if f and s>0:m=1;break
print("YNEOS"[(m or not f)::2])
|
m = 1;break
|
[{"input": "7\n2\n1 0\n4\n2 -1 -1 0\n4\n1 -4 3 0\n4\n1 -1 1 -1\n5\n1 2 3 4 -10\n7\n2 -1 1 -2 0 0 0\n1\n0", "output": ["No\nYes\nNo\nNo\nYes\nYes\nYes"]}]
|
block_completion_000426
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: We have an array of length $$$n$$$. Initially, each element is equal to $$$0$$$ and there is a pointer located on the first element.We can do the following two kinds of operations any number of times (possibly zero) in any order: If the pointer is not on the last element, increase the element the pointer is currently on by $$$1$$$. Then move it to the next element. If the pointer is not on the first element, decrease the element the pointer is currently on by $$$1$$$. Then move it to the previous element.But there is one additional rule. After we are done, the pointer has to be on the first element.You are given an array $$$a$$$. Determine whether it's possible to obtain $$$a$$$ after some operations or not.
Input Specification: The first line contains a single integer $$$t$$$ $$$(1\le t\le 1000)$$$ — the number of test cases. The description of the test cases follows. The first line of each test case contains a single integer $$$n$$$ $$$(1\le n\le 2 \cdot 10^5)$$$ — the size of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$-10^9 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases doesn't exceed $$$2 \cdot 10^5$$$.
Output Specification: For each test case, print "Yes" (without quotes) if it's possible to obtain $$$a$$$ after some operations, and "No" (without quotes) otherwise. You can output "Yes" and "No" in any case (for example, strings "yEs", "yes" and "Yes" will be recognized as a positive response).
Notes: NoteIn the first test case we can obtain the array after some operations, but the pointer won't be on the first element.One way of obtaining the array in the second test case is shown below.$$$\langle \underline{0}, 0, 0, 0\rangle \to \langle 1, \underline{0}, 0, 0 \rangle \to \langle \underline{1}, -1, 0, 0\rangle \to \langle 2, \underline{-1}, 0, 0\rangle \to \langle 2, 0, \underline{0}, 0\rangle \to \langle 2, \underline{0}, -1, 0\rangle \to \langle \underline{2}, -1, -1, 0\rangle$$$
Code:
for _ in [0]*int(input()):
input()
n = list(map(int,input().split()))
s,f,m = 0,0,0
for i in n:
s+=i
if s<0:m = 1;break
if s==0:# TODO: Your code here
if f and s>0:m=1;break
print("YNEOS"[(m or not f)::2])
|
for _ in [0]*int(input()):
input()
n = list(map(int,input().split()))
s,f,m = 0,0,0
for i in n:
s+=i
if s<0:m = 1;break
if s==0:{{completion}}
if f and s>0:m=1;break
print("YNEOS"[(m or not f)::2])
|
f=1
|
[{"input": "7\n2\n1 0\n4\n2 -1 -1 0\n4\n1 -4 3 0\n4\n1 -1 1 -1\n5\n1 2 3 4 -10\n7\n2 -1 1 -2 0 0 0\n1\n0", "output": ["No\nYes\nNo\nNo\nYes\nYes\nYes"]}]
|
block_completion_000427
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: We have an array of length $$$n$$$. Initially, each element is equal to $$$0$$$ and there is a pointer located on the first element.We can do the following two kinds of operations any number of times (possibly zero) in any order: If the pointer is not on the last element, increase the element the pointer is currently on by $$$1$$$. Then move it to the next element. If the pointer is not on the first element, decrease the element the pointer is currently on by $$$1$$$. Then move it to the previous element.But there is one additional rule. After we are done, the pointer has to be on the first element.You are given an array $$$a$$$. Determine whether it's possible to obtain $$$a$$$ after some operations or not.
Input Specification: The first line contains a single integer $$$t$$$ $$$(1\le t\le 1000)$$$ — the number of test cases. The description of the test cases follows. The first line of each test case contains a single integer $$$n$$$ $$$(1\le n\le 2 \cdot 10^5)$$$ — the size of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$-10^9 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases doesn't exceed $$$2 \cdot 10^5$$$.
Output Specification: For each test case, print "Yes" (without quotes) if it's possible to obtain $$$a$$$ after some operations, and "No" (without quotes) otherwise. You can output "Yes" and "No" in any case (for example, strings "yEs", "yes" and "Yes" will be recognized as a positive response).
Notes: NoteIn the first test case we can obtain the array after some operations, but the pointer won't be on the first element.One way of obtaining the array in the second test case is shown below.$$$\langle \underline{0}, 0, 0, 0\rangle \to \langle 1, \underline{0}, 0, 0 \rangle \to \langle \underline{1}, -1, 0, 0\rangle \to \langle 2, \underline{-1}, 0, 0\rangle \to \langle 2, 0, \underline{0}, 0\rangle \to \langle 2, \underline{0}, -1, 0\rangle \to \langle \underline{2}, -1, -1, 0\rangle$$$
Code:
if __name__ == '__main__':
t = int(input())
for _ in range(t):
n = int(input())
a = [int(i) for i in input().split()]
x = a[0]
ok = True
for v in a[1:]:
if x < 0:
# TODO: Your code here
if x == 0 and v != 0:
ok = False
break
x += v
print("yes" if ok and x == 0 else "no")
|
if __name__ == '__main__':
t = int(input())
for _ in range(t):
n = int(input())
a = [int(i) for i in input().split()]
x = a[0]
ok = True
for v in a[1:]:
if x < 0:
{{completion}}
if x == 0 and v != 0:
ok = False
break
x += v
print("yes" if ok and x == 0 else "no")
|
ok = False
break
|
[{"input": "7\n2\n1 0\n4\n2 -1 -1 0\n4\n1 -4 3 0\n4\n1 -1 1 -1\n5\n1 2 3 4 -10\n7\n2 -1 1 -2 0 0 0\n1\n0", "output": ["No\nYes\nNo\nNo\nYes\nYes\nYes"]}]
|
block_completion_000428
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: We have an array of length $$$n$$$. Initially, each element is equal to $$$0$$$ and there is a pointer located on the first element.We can do the following two kinds of operations any number of times (possibly zero) in any order: If the pointer is not on the last element, increase the element the pointer is currently on by $$$1$$$. Then move it to the next element. If the pointer is not on the first element, decrease the element the pointer is currently on by $$$1$$$. Then move it to the previous element.But there is one additional rule. After we are done, the pointer has to be on the first element.You are given an array $$$a$$$. Determine whether it's possible to obtain $$$a$$$ after some operations or not.
Input Specification: The first line contains a single integer $$$t$$$ $$$(1\le t\le 1000)$$$ — the number of test cases. The description of the test cases follows. The first line of each test case contains a single integer $$$n$$$ $$$(1\le n\le 2 \cdot 10^5)$$$ — the size of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$-10^9 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases doesn't exceed $$$2 \cdot 10^5$$$.
Output Specification: For each test case, print "Yes" (without quotes) if it's possible to obtain $$$a$$$ after some operations, and "No" (without quotes) otherwise. You can output "Yes" and "No" in any case (for example, strings "yEs", "yes" and "Yes" will be recognized as a positive response).
Notes: NoteIn the first test case we can obtain the array after some operations, but the pointer won't be on the first element.One way of obtaining the array in the second test case is shown below.$$$\langle \underline{0}, 0, 0, 0\rangle \to \langle 1, \underline{0}, 0, 0 \rangle \to \langle \underline{1}, -1, 0, 0\rangle \to \langle 2, \underline{-1}, 0, 0\rangle \to \langle 2, 0, \underline{0}, 0\rangle \to \langle 2, \underline{0}, -1, 0\rangle \to \langle \underline{2}, -1, -1, 0\rangle$$$
Code:
if __name__ == '__main__':
t = int(input())
for _ in range(t):
n = int(input())
a = [int(i) for i in input().split()]
x = a[0]
ok = True
for v in a[1:]:
if x < 0:
ok = False
break
if x == 0 and v != 0:
# TODO: Your code here
x += v
print("yes" if ok and x == 0 else "no")
|
if __name__ == '__main__':
t = int(input())
for _ in range(t):
n = int(input())
a = [int(i) for i in input().split()]
x = a[0]
ok = True
for v in a[1:]:
if x < 0:
ok = False
break
if x == 0 and v != 0:
{{completion}}
x += v
print("yes" if ok and x == 0 else "no")
|
ok = False
break
|
[{"input": "7\n2\n1 0\n4\n2 -1 -1 0\n4\n1 -4 3 0\n4\n1 -1 1 -1\n5\n1 2 3 4 -10\n7\n2 -1 1 -2 0 0 0\n1\n0", "output": ["No\nYes\nNo\nNo\nYes\nYes\nYes"]}]
|
block_completion_000429
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: We have an array of length $$$n$$$. Initially, each element is equal to $$$0$$$ and there is a pointer located on the first element.We can do the following two kinds of operations any number of times (possibly zero) in any order: If the pointer is not on the last element, increase the element the pointer is currently on by $$$1$$$. Then move it to the next element. If the pointer is not on the first element, decrease the element the pointer is currently on by $$$1$$$. Then move it to the previous element.But there is one additional rule. After we are done, the pointer has to be on the first element.You are given an array $$$a$$$. Determine whether it's possible to obtain $$$a$$$ after some operations or not.
Input Specification: The first line contains a single integer $$$t$$$ $$$(1\le t\le 1000)$$$ — the number of test cases. The description of the test cases follows. The first line of each test case contains a single integer $$$n$$$ $$$(1\le n\le 2 \cdot 10^5)$$$ — the size of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$-10^9 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases doesn't exceed $$$2 \cdot 10^5$$$.
Output Specification: For each test case, print "Yes" (without quotes) if it's possible to obtain $$$a$$$ after some operations, and "No" (without quotes) otherwise. You can output "Yes" and "No" in any case (for example, strings "yEs", "yes" and "Yes" will be recognized as a positive response).
Notes: NoteIn the first test case we can obtain the array after some operations, but the pointer won't be on the first element.One way of obtaining the array in the second test case is shown below.$$$\langle \underline{0}, 0, 0, 0\rangle \to \langle 1, \underline{0}, 0, 0 \rangle \to \langle \underline{1}, -1, 0, 0\rangle \to \langle 2, \underline{-1}, 0, 0\rangle \to \langle 2, 0, \underline{0}, 0\rangle \to \langle 2, \underline{0}, -1, 0\rangle \to \langle \underline{2}, -1, -1, 0\rangle$$$
Code:
input = __import__('sys').stdin.readline
def solve():
n = int(input())
allzeros = False
total = 0
for x in map(int, input().split()):
total += x
if total < 0 or total != 0 and allzeros:
# TODO: Your code here
allzeros = allzeros or total == 0
print('YES' if total == 0 else 'NO')
for _ in range(int(input())):
solve()
|
input = __import__('sys').stdin.readline
def solve():
n = int(input())
allzeros = False
total = 0
for x in map(int, input().split()):
total += x
if total < 0 or total != 0 and allzeros:
{{completion}}
allzeros = allzeros or total == 0
print('YES' if total == 0 else 'NO')
for _ in range(int(input())):
solve()
|
print('No')
return
|
[{"input": "7\n2\n1 0\n4\n2 -1 -1 0\n4\n1 -4 3 0\n4\n1 -1 1 -1\n5\n1 2 3 4 -10\n7\n2 -1 1 -2 0 0 0\n1\n0", "output": ["No\nYes\nNo\nNo\nYes\nYes\nYes"]}]
|
block_completion_000430
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: We have an array of length $$$n$$$. Initially, each element is equal to $$$0$$$ and there is a pointer located on the first element.We can do the following two kinds of operations any number of times (possibly zero) in any order: If the pointer is not on the last element, increase the element the pointer is currently on by $$$1$$$. Then move it to the next element. If the pointer is not on the first element, decrease the element the pointer is currently on by $$$1$$$. Then move it to the previous element.But there is one additional rule. After we are done, the pointer has to be on the first element.You are given an array $$$a$$$. Determine whether it's possible to obtain $$$a$$$ after some operations or not.
Input Specification: The first line contains a single integer $$$t$$$ $$$(1\le t\le 1000)$$$ — the number of test cases. The description of the test cases follows. The first line of each test case contains a single integer $$$n$$$ $$$(1\le n\le 2 \cdot 10^5)$$$ — the size of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$-10^9 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases doesn't exceed $$$2 \cdot 10^5$$$.
Output Specification: For each test case, print "Yes" (without quotes) if it's possible to obtain $$$a$$$ after some operations, and "No" (without quotes) otherwise. You can output "Yes" and "No" in any case (for example, strings "yEs", "yes" and "Yes" will be recognized as a positive response).
Notes: NoteIn the first test case we can obtain the array after some operations, but the pointer won't be on the first element.One way of obtaining the array in the second test case is shown below.$$$\langle \underline{0}, 0, 0, 0\rangle \to \langle 1, \underline{0}, 0, 0 \rangle \to \langle \underline{1}, -1, 0, 0\rangle \to \langle 2, \underline{-1}, 0, 0\rangle \to \langle 2, 0, \underline{0}, 0\rangle \to \langle 2, \underline{0}, -1, 0\rangle \to \langle \underline{2}, -1, -1, 0\rangle$$$
Code:
import sys
input=sys.stdin.readline
I = lambda : list(map(int,input().split()))
t,=I()
for _ in range(t):
n, = I()
l = I()
pos = 0
if sum(l)!=0 or l[-1]>0:
pos=1
else:
pref = l[0]
seen = 0
if pref<0:
pos=1
if pref==0:
seen = 1
for i in range(1,n):
pref+=l[i]
if pref<0:
pos=1
break
elif pref==0:
seen = 1
else:
if seen:
# TODO: Your code here
print("YNeos"[pos::2])
|
import sys
input=sys.stdin.readline
I = lambda : list(map(int,input().split()))
t,=I()
for _ in range(t):
n, = I()
l = I()
pos = 0
if sum(l)!=0 or l[-1]>0:
pos=1
else:
pref = l[0]
seen = 0
if pref<0:
pos=1
if pref==0:
seen = 1
for i in range(1,n):
pref+=l[i]
if pref<0:
pos=1
break
elif pref==0:
seen = 1
else:
if seen:
{{completion}}
print("YNeos"[pos::2])
|
pos=1
break
|
[{"input": "7\n2\n1 0\n4\n2 -1 -1 0\n4\n1 -4 3 0\n4\n1 -1 1 -1\n5\n1 2 3 4 -10\n7\n2 -1 1 -2 0 0 0\n1\n0", "output": ["No\nYes\nNo\nNo\nYes\nYes\nYes"]}]
|
block_completion_000431
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: We have an array of length $$$n$$$. Initially, each element is equal to $$$0$$$ and there is a pointer located on the first element.We can do the following two kinds of operations any number of times (possibly zero) in any order: If the pointer is not on the last element, increase the element the pointer is currently on by $$$1$$$. Then move it to the next element. If the pointer is not on the first element, decrease the element the pointer is currently on by $$$1$$$. Then move it to the previous element.But there is one additional rule. After we are done, the pointer has to be on the first element.You are given an array $$$a$$$. Determine whether it's possible to obtain $$$a$$$ after some operations or not.
Input Specification: The first line contains a single integer $$$t$$$ $$$(1\le t\le 1000)$$$ — the number of test cases. The description of the test cases follows. The first line of each test case contains a single integer $$$n$$$ $$$(1\le n\le 2 \cdot 10^5)$$$ — the size of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$-10^9 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases doesn't exceed $$$2 \cdot 10^5$$$.
Output Specification: For each test case, print "Yes" (without quotes) if it's possible to obtain $$$a$$$ after some operations, and "No" (without quotes) otherwise. You can output "Yes" and "No" in any case (for example, strings "yEs", "yes" and "Yes" will be recognized as a positive response).
Notes: NoteIn the first test case we can obtain the array after some operations, but the pointer won't be on the first element.One way of obtaining the array in the second test case is shown below.$$$\langle \underline{0}, 0, 0, 0\rangle \to \langle 1, \underline{0}, 0, 0 \rangle \to \langle \underline{1}, -1, 0, 0\rangle \to \langle 2, \underline{-1}, 0, 0\rangle \to \langle 2, 0, \underline{0}, 0\rangle \to \langle 2, \underline{0}, -1, 0\rangle \to \langle \underline{2}, -1, -1, 0\rangle$$$
Code:
for t in range(int(input())):
n=int(input())
a=list(map(int,input().split()))
i=n-1
while(a[i]==0 and i!=0):
i-=1
while(i>0):
if a[i]>=0:
print("NO")
break
a[i-1]+=a[i]
i-=1
else:
if a[i]==0:
print("YES")
else:
# TODO: Your code here
|
for t in range(int(input())):
n=int(input())
a=list(map(int,input().split()))
i=n-1
while(a[i]==0 and i!=0):
i-=1
while(i>0):
if a[i]>=0:
print("NO")
break
a[i-1]+=a[i]
i-=1
else:
if a[i]==0:
print("YES")
else:
{{completion}}
|
print("NO")
|
[{"input": "7\n2\n1 0\n4\n2 -1 -1 0\n4\n1 -4 3 0\n4\n1 -1 1 -1\n5\n1 2 3 4 -10\n7\n2 -1 1 -2 0 0 0\n1\n0", "output": ["No\nYes\nNo\nNo\nYes\nYes\nYes"]}]
|
block_completion_000432
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: We have an array of length $$$n$$$. Initially, each element is equal to $$$0$$$ and there is a pointer located on the first element.We can do the following two kinds of operations any number of times (possibly zero) in any order: If the pointer is not on the last element, increase the element the pointer is currently on by $$$1$$$. Then move it to the next element. If the pointer is not on the first element, decrease the element the pointer is currently on by $$$1$$$. Then move it to the previous element.But there is one additional rule. After we are done, the pointer has to be on the first element.You are given an array $$$a$$$. Determine whether it's possible to obtain $$$a$$$ after some operations or not.
Input Specification: The first line contains a single integer $$$t$$$ $$$(1\le t\le 1000)$$$ — the number of test cases. The description of the test cases follows. The first line of each test case contains a single integer $$$n$$$ $$$(1\le n\le 2 \cdot 10^5)$$$ — the size of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$-10^9 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases doesn't exceed $$$2 \cdot 10^5$$$.
Output Specification: For each test case, print "Yes" (without quotes) if it's possible to obtain $$$a$$$ after some operations, and "No" (without quotes) otherwise. You can output "Yes" and "No" in any case (for example, strings "yEs", "yes" and "Yes" will be recognized as a positive response).
Notes: NoteIn the first test case we can obtain the array after some operations, but the pointer won't be on the first element.One way of obtaining the array in the second test case is shown below.$$$\langle \underline{0}, 0, 0, 0\rangle \to \langle 1, \underline{0}, 0, 0 \rangle \to \langle \underline{1}, -1, 0, 0\rangle \to \langle 2, \underline{-1}, 0, 0\rangle \to \langle 2, 0, \underline{0}, 0\rangle \to \langle 2, \underline{0}, -1, 0\rangle \to \langle \underline{2}, -1, -1, 0\rangle$$$
Code:
for i in range(int(input())):
n=int(input())
c=[int(j) for j in input().split()]
k=1
if [0]*n==c:
print('Yes')
else:
g=0
while c[-1]==0:
c.pop()
while len(c)-1:
if g<=c[-1]:
# TODO: Your code here
g=g-c.pop()
print(['No','Yes'][g==c[0] and k])
|
for i in range(int(input())):
n=int(input())
c=[int(j) for j in input().split()]
k=1
if [0]*n==c:
print('Yes')
else:
g=0
while c[-1]==0:
c.pop()
while len(c)-1:
if g<=c[-1]:
{{completion}}
g=g-c.pop()
print(['No','Yes'][g==c[0] and k])
|
k=0
break
|
[{"input": "7\n2\n1 0\n4\n2 -1 -1 0\n4\n1 -4 3 0\n4\n1 -1 1 -1\n5\n1 2 3 4 -10\n7\n2 -1 1 -2 0 0 0\n1\n0", "output": ["No\nYes\nNo\nNo\nYes\nYes\nYes"]}]
|
block_completion_000433
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: AmShZ has traveled to Italy from Iran for the Thom Yorke concert. There are $$$n$$$ cities in Italy indexed from $$$1$$$ to $$$n$$$ and $$$m$$$ directed roads indexed from $$$1$$$ to $$$m$$$. Initially, Keshi is located in the city $$$1$$$ and wants to go to AmShZ's house in the city $$$n$$$. Since Keshi doesn't know the map of Italy, AmShZ helps him to see each other as soon as possible.In the beginning of each day, AmShZ can send one of the following two messages to Keshi: AmShZ sends the index of one road to Keshi as a blocked road. Then Keshi will understand that he should never use that road and he will remain in his current city for the day. AmShZ tells Keshi to move. Then, Keshi will randomly choose one of the cities reachable from his current city and move there. (city $$$B$$$ is reachable from city $$$A$$$ if there's an out-going road from city $$$A$$$ to city $$$B$$$ which hasn't become blocked yet). If there are no such cities, Keshi will remain in his current city.Note that AmShZ always knows Keshi's current location. AmShZ and Keshi want to find the smallest possible integer $$$d$$$ for which they can make sure that they will see each other after at most $$$d$$$ days. Help them find $$$d$$$.
Input Specification: The first line of the input contains two integers $$$n$$$ and $$$m$$$ $$$(2 \le n \le 2 \cdot 10^5, 1 \le m \le 2 \cdot 10^5)$$$ — the number of cities and roads correspondingly. The $$$i$$$-th line of the following $$$m$$$ lines contains two integers $$$v_i$$$ and $$$u_i$$$ $$$(1 \le v_i , u_i \le n,v_i \neq u_i)$$$, denoting a directed road going from city $$$v_i$$$ to city $$$u_i$$$. It is guaranteed that there is at least one route from city $$$1$$$ to city $$$n$$$. Note that there may be more than one road between a pair of cities in each direction.
Output Specification: Output the smallest possible integer $$$d$$$ to make sure that AmShZ and Keshi will see each other after at most $$$d$$$ days.
Notes: NoteIn the first sample, it's enough for AmShZ to send the second type of message.In the second sample, on the first day, AmShZ blocks the first road. So the only reachable city from city $$$1$$$ will be city $$$4$$$. Hence on the second day, AmShZ can tell Keshi to move and Keshi will arrive at AmShZ's house.It's also possible for AmShZ to tell Keshi to move for two days.
Code:
import heapq as hq
INF = 1001001001
N, M = map(int, input().split())
G = [[] for _ in range(N)]
d = [0] * N
for _ in range(M):
U, V = map(int, input().split())
G[V - 1].append(U - 1)
d[U - 1] += 1
dists = [INF] * N
dists[N - 1] = 0
queue = [(0, N - 1)]
while queue:
dist, V = hq.heappop(queue)
if dists[V] < dist: continue
for v in G[V]:
if dist + d[v] < dists[v]:
# TODO: Your code here
d[v] -= 1
print(dists[0])
|
import heapq as hq
INF = 1001001001
N, M = map(int, input().split())
G = [[] for _ in range(N)]
d = [0] * N
for _ in range(M):
U, V = map(int, input().split())
G[V - 1].append(U - 1)
d[U - 1] += 1
dists = [INF] * N
dists[N - 1] = 0
queue = [(0, N - 1)]
while queue:
dist, V = hq.heappop(queue)
if dists[V] < dist: continue
for v in G[V]:
if dist + d[v] < dists[v]:
{{completion}}
d[v] -= 1
print(dists[0])
|
dists[v] = dist + d[v]
hq.heappush(queue, (dist + d[v], v))
|
[{"input": "2 1\n1 2", "output": ["1"]}, {"input": "4 4\n1 2\n1 4\n2 4\n1 4", "output": ["2"]}, {"input": "5 7\n1 2\n2 3\n3 5\n1 4\n4 3\n4 5\n3 1", "output": ["4"]}]
|
block_completion_000469
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: AmShZ has traveled to Italy from Iran for the Thom Yorke concert. There are $$$n$$$ cities in Italy indexed from $$$1$$$ to $$$n$$$ and $$$m$$$ directed roads indexed from $$$1$$$ to $$$m$$$. Initially, Keshi is located in the city $$$1$$$ and wants to go to AmShZ's house in the city $$$n$$$. Since Keshi doesn't know the map of Italy, AmShZ helps him to see each other as soon as possible.In the beginning of each day, AmShZ can send one of the following two messages to Keshi: AmShZ sends the index of one road to Keshi as a blocked road. Then Keshi will understand that he should never use that road and he will remain in his current city for the day. AmShZ tells Keshi to move. Then, Keshi will randomly choose one of the cities reachable from his current city and move there. (city $$$B$$$ is reachable from city $$$A$$$ if there's an out-going road from city $$$A$$$ to city $$$B$$$ which hasn't become blocked yet). If there are no such cities, Keshi will remain in his current city.Note that AmShZ always knows Keshi's current location. AmShZ and Keshi want to find the smallest possible integer $$$d$$$ for which they can make sure that they will see each other after at most $$$d$$$ days. Help them find $$$d$$$.
Input Specification: The first line of the input contains two integers $$$n$$$ and $$$m$$$ $$$(2 \le n \le 2 \cdot 10^5, 1 \le m \le 2 \cdot 10^5)$$$ — the number of cities and roads correspondingly. The $$$i$$$-th line of the following $$$m$$$ lines contains two integers $$$v_i$$$ and $$$u_i$$$ $$$(1 \le v_i , u_i \le n,v_i \neq u_i)$$$, denoting a directed road going from city $$$v_i$$$ to city $$$u_i$$$. It is guaranteed that there is at least one route from city $$$1$$$ to city $$$n$$$. Note that there may be more than one road between a pair of cities in each direction.
Output Specification: Output the smallest possible integer $$$d$$$ to make sure that AmShZ and Keshi will see each other after at most $$$d$$$ days.
Notes: NoteIn the first sample, it's enough for AmShZ to send the second type of message.In the second sample, on the first day, AmShZ blocks the first road. So the only reachable city from city $$$1$$$ will be city $$$4$$$. Hence on the second day, AmShZ can tell Keshi to move and Keshi will arrive at AmShZ's house.It's also possible for AmShZ to tell Keshi to move for two days.
Code:
from heapq import*;I=input;R=lambda:map(int,I().split())
n,m=R();g,q,vis=[[] for _ in range(n)],[(0,n-1)],[0]*n
d,out=[m+1]*n,[0]*n;d[-1]=0
for _ in range(m):u,v=R();u,v=u-1,v-1;g[v].append(u);out[u]+=1
while q:
_,u=heappop(q)
if vis[u]:continue
vis[u]=1
for v in g[u]:
if d[u]+out[v]<d[v]:# TODO: Your code here
out[v]-=1
print(d[0])
|
from heapq import*;I=input;R=lambda:map(int,I().split())
n,m=R();g,q,vis=[[] for _ in range(n)],[(0,n-1)],[0]*n
d,out=[m+1]*n,[0]*n;d[-1]=0
for _ in range(m):u,v=R();u,v=u-1,v-1;g[v].append(u);out[u]+=1
while q:
_,u=heappop(q)
if vis[u]:continue
vis[u]=1
for v in g[u]:
if d[u]+out[v]<d[v]:{{completion}}
out[v]-=1
print(d[0])
|
d[v]=d[u]+out[v];heappush(q,(d[v],v))
|
[{"input": "2 1\n1 2", "output": ["1"]}, {"input": "4 4\n1 2\n1 4\n2 4\n1 4", "output": ["2"]}, {"input": "5 7\n1 2\n2 3\n3 5\n1 4\n4 3\n4 5\n3 1", "output": ["4"]}]
|
block_completion_000470
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: AmShZ has traveled to Italy from Iran for the Thom Yorke concert. There are $$$n$$$ cities in Italy indexed from $$$1$$$ to $$$n$$$ and $$$m$$$ directed roads indexed from $$$1$$$ to $$$m$$$. Initially, Keshi is located in the city $$$1$$$ and wants to go to AmShZ's house in the city $$$n$$$. Since Keshi doesn't know the map of Italy, AmShZ helps him to see each other as soon as possible.In the beginning of each day, AmShZ can send one of the following two messages to Keshi: AmShZ sends the index of one road to Keshi as a blocked road. Then Keshi will understand that he should never use that road and he will remain in his current city for the day. AmShZ tells Keshi to move. Then, Keshi will randomly choose one of the cities reachable from his current city and move there. (city $$$B$$$ is reachable from city $$$A$$$ if there's an out-going road from city $$$A$$$ to city $$$B$$$ which hasn't become blocked yet). If there are no such cities, Keshi will remain in his current city.Note that AmShZ always knows Keshi's current location. AmShZ and Keshi want to find the smallest possible integer $$$d$$$ for which they can make sure that they will see each other after at most $$$d$$$ days. Help them find $$$d$$$.
Input Specification: The first line of the input contains two integers $$$n$$$ and $$$m$$$ $$$(2 \le n \le 2 \cdot 10^5, 1 \le m \le 2 \cdot 10^5)$$$ — the number of cities and roads correspondingly. The $$$i$$$-th line of the following $$$m$$$ lines contains two integers $$$v_i$$$ and $$$u_i$$$ $$$(1 \le v_i , u_i \le n,v_i \neq u_i)$$$, denoting a directed road going from city $$$v_i$$$ to city $$$u_i$$$. It is guaranteed that there is at least one route from city $$$1$$$ to city $$$n$$$. Note that there may be more than one road between a pair of cities in each direction.
Output Specification: Output the smallest possible integer $$$d$$$ to make sure that AmShZ and Keshi will see each other after at most $$$d$$$ days.
Notes: NoteIn the first sample, it's enough for AmShZ to send the second type of message.In the second sample, on the first day, AmShZ blocks the first road. So the only reachable city from city $$$1$$$ will be city $$$4$$$. Hence on the second day, AmShZ can tell Keshi to move and Keshi will arrive at AmShZ's house.It's also possible for AmShZ to tell Keshi to move for two days.
Code:
import sys
input=sys.stdin.readline #文字列入力はするな！！
from heapq import *
n,m=map(int,input().split())
root=[[] for i in range(n+2)]
rootinv=[[] for i in range(n+2)]
no=[0]*(n+2)
for i in range(m):
u,v=map(int,input().split())
root[u].append(v)
rootinv[v].append(u)
no[u]+=1
dp=[10**18]*(n+3)
dp[n]=0
hp=[(0,n)]
while hp:
c,x=heappop(hp)
if dp[x]<c:continue
for y in rootinv[x]:
no[y]-=1
cost=no[y]+1
if dp[y]>dp[x]+cost:
# TODO: Your code here
print(dp[1])
|
import sys
input=sys.stdin.readline #文字列入力はするな！！
from heapq import *
n,m=map(int,input().split())
root=[[] for i in range(n+2)]
rootinv=[[] for i in range(n+2)]
no=[0]*(n+2)
for i in range(m):
u,v=map(int,input().split())
root[u].append(v)
rootinv[v].append(u)
no[u]+=1
dp=[10**18]*(n+3)
dp[n]=0
hp=[(0,n)]
while hp:
c,x=heappop(hp)
if dp[x]<c:continue
for y in rootinv[x]:
no[y]-=1
cost=no[y]+1
if dp[y]>dp[x]+cost:
{{completion}}
print(dp[1])
|
dp[y]=dp[x]+cost
heappush(hp,(dp[y],y))
|
[{"input": "2 1\n1 2", "output": ["1"]}, {"input": "4 4\n1 2\n1 4\n2 4\n1 4", "output": ["2"]}, {"input": "5 7\n1 2\n2 3\n3 5\n1 4\n4 3\n4 5\n3 1", "output": ["4"]}]
|
block_completion_000471
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: AmShZ has traveled to Italy from Iran for the Thom Yorke concert. There are $$$n$$$ cities in Italy indexed from $$$1$$$ to $$$n$$$ and $$$m$$$ directed roads indexed from $$$1$$$ to $$$m$$$. Initially, Keshi is located in the city $$$1$$$ and wants to go to AmShZ's house in the city $$$n$$$. Since Keshi doesn't know the map of Italy, AmShZ helps him to see each other as soon as possible.In the beginning of each day, AmShZ can send one of the following two messages to Keshi: AmShZ sends the index of one road to Keshi as a blocked road. Then Keshi will understand that he should never use that road and he will remain in his current city for the day. AmShZ tells Keshi to move. Then, Keshi will randomly choose one of the cities reachable from his current city and move there. (city $$$B$$$ is reachable from city $$$A$$$ if there's an out-going road from city $$$A$$$ to city $$$B$$$ which hasn't become blocked yet). If there are no such cities, Keshi will remain in his current city.Note that AmShZ always knows Keshi's current location. AmShZ and Keshi want to find the smallest possible integer $$$d$$$ for which they can make sure that they will see each other after at most $$$d$$$ days. Help them find $$$d$$$.
Input Specification: The first line of the input contains two integers $$$n$$$ and $$$m$$$ $$$(2 \le n \le 2 \cdot 10^5, 1 \le m \le 2 \cdot 10^5)$$$ — the number of cities and roads correspondingly. The $$$i$$$-th line of the following $$$m$$$ lines contains two integers $$$v_i$$$ and $$$u_i$$$ $$$(1 \le v_i , u_i \le n,v_i \neq u_i)$$$, denoting a directed road going from city $$$v_i$$$ to city $$$u_i$$$. It is guaranteed that there is at least one route from city $$$1$$$ to city $$$n$$$. Note that there may be more than one road between a pair of cities in each direction.
Output Specification: Output the smallest possible integer $$$d$$$ to make sure that AmShZ and Keshi will see each other after at most $$$d$$$ days.
Notes: NoteIn the first sample, it's enough for AmShZ to send the second type of message.In the second sample, on the first day, AmShZ blocks the first road. So the only reachable city from city $$$1$$$ will be city $$$4$$$. Hence on the second day, AmShZ can tell Keshi to move and Keshi will arrive at AmShZ's house.It's also possible for AmShZ to tell Keshi to move for two days.
Code:
import sys, heapq
input=sys.stdin.readline
n,m=map(int,input().split())
iadj=[{} for _ in range(n)] # inverted road
nadj=[0]*n
dist=[n+1]*n
cost=[float("inf")]*n
visit=[0]*n
for _ in range(m):
v,u=map(int,input().split())
v-=1
u-=1
pi=iadj[u].setdefault(v,0)
iadj[u][v]=1+pi # road from v->u
nadj[v]+=1
q = [(0,n-1)]
cost[n-1]=0
while q:
d,v=heapq.heappop(q)
if visit[v]:
continue
visit[v]=1
for u in iadj[v]:
nadj[u]-=iadj[v][u]
if cost[v] + nadj[u] + 1 < cost[u]:
# TODO: Your code here
print(cost[0])
|
import sys, heapq
input=sys.stdin.readline
n,m=map(int,input().split())
iadj=[{} for _ in range(n)] # inverted road
nadj=[0]*n
dist=[n+1]*n
cost=[float("inf")]*n
visit=[0]*n
for _ in range(m):
v,u=map(int,input().split())
v-=1
u-=1
pi=iadj[u].setdefault(v,0)
iadj[u][v]=1+pi # road from v->u
nadj[v]+=1
q = [(0,n-1)]
cost[n-1]=0
while q:
d,v=heapq.heappop(q)
if visit[v]:
continue
visit[v]=1
for u in iadj[v]:
nadj[u]-=iadj[v][u]
if cost[v] + nadj[u] + 1 < cost[u]:
{{completion}}
print(cost[0])
|
cost[u]=cost[v]+nadj[u] + 1
heapq.heappush(q, (cost[u], u))
|
[{"input": "2 1\n1 2", "output": ["1"]}, {"input": "4 4\n1 2\n1 4\n2 4\n1 4", "output": ["2"]}, {"input": "5 7\n1 2\n2 3\n3 5\n1 4\n4 3\n4 5\n3 1", "output": ["4"]}]
|
block_completion_000472
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: Let's call an array $$$a$$$ of $$$m$$$ integers $$$a_1, a_2, \ldots, a_m$$$ Decinc if $$$a$$$ can be made increasing by removing a decreasing subsequence (possibly empty) from it. For example, if $$$a = [3, 2, 4, 1, 5]$$$, we can remove the decreasing subsequence $$$[a_1, a_4]$$$ from $$$a$$$ and obtain $$$a = [2, 4, 5]$$$, which is increasing.You are given a permutation $$$p$$$ of numbers from $$$1$$$ to $$$n$$$. Find the number of pairs of integers $$$(l, r)$$$ with $$$1 \le l \le r \le n$$$ such that $$$p[l \ldots r]$$$ (the subarray of $$$p$$$ from $$$l$$$ to $$$r$$$) is a Decinc array.
Input Specification: The first line contains a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$) — the size of $$$p$$$. The second line contains $$$n$$$ integers $$$p_1, p_2, \ldots, p_n$$$ ($$$1 \le p_i \le n$$$, all $$$p_i$$$ are distinct) — elements of the permutation.
Output Specification: Output the number of pairs of integers $$$(l, r)$$$ such that $$$p[l \ldots r]$$$ (the subarray of $$$p$$$ from $$$l$$$ to $$$r$$$) is a Decinc array. $$$(1 \le l \le r \le n)$$$
Notes: NoteIn the first sample, all subarrays are Decinc.In the second sample, all subarrays except $$$p[1 \ldots 6]$$$ and $$$p[2 \ldots 6]$$$ are Decinc.
Code:
input = __import__('sys').stdin.readline
n = int(input())
a = list(map(int, input().split())) + [n+1]
ans = 0
cache = {}
for i in range(n):
u = 0
d = n+1
keys = []
j = i
while j+1 <= n:
key = (j, u, d)
v = cache.get(key, -1)
if v != -1:
j = v
break
keys.append(key)
# greedy
if u < a[j] < d: # if can insert to both
if a[j] < a[j+1]:
u = max(u, a[j])
elif a[j] > a[j+1]:
# TODO: Your code here
elif u < a[j]: # if only can insert to increasing subsequence
u = a[j]
elif d > a[j]: # if only can insert to decreasing subsequence
d = a[j]
else:
break
j += 1
for key in keys:
cache[key] = j
ans += j - i
# print(f'at {i} max {j} ans {ans}', u, d)
# print(f'count={len(cache)}')
print(ans)
|
input = __import__('sys').stdin.readline
n = int(input())
a = list(map(int, input().split())) + [n+1]
ans = 0
cache = {}
for i in range(n):
u = 0
d = n+1
keys = []
j = i
while j+1 <= n:
key = (j, u, d)
v = cache.get(key, -1)
if v != -1:
j = v
break
keys.append(key)
# greedy
if u < a[j] < d: # if can insert to both
if a[j] < a[j+1]:
u = max(u, a[j])
elif a[j] > a[j+1]:
{{completion}}
elif u < a[j]: # if only can insert to increasing subsequence
u = a[j]
elif d > a[j]: # if only can insert to decreasing subsequence
d = a[j]
else:
break
j += 1
for key in keys:
cache[key] = j
ans += j - i
# print(f'at {i} max {j} ans {ans}', u, d)
# print(f'count={len(cache)}')
print(ans)
|
d = min(d, a[j])
|
[{"input": "3\n2 3 1", "output": ["6"]}, {"input": "6\n4 5 2 6 1 3", "output": ["19"]}, {"input": "10\n7 10 1 8 3 9 2 4 6 5", "output": ["39"]}]
|
block_completion_000487
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: Let's call an array $$$a$$$ of $$$m$$$ integers $$$a_1, a_2, \ldots, a_m$$$ Decinc if $$$a$$$ can be made increasing by removing a decreasing subsequence (possibly empty) from it. For example, if $$$a = [3, 2, 4, 1, 5]$$$, we can remove the decreasing subsequence $$$[a_1, a_4]$$$ from $$$a$$$ and obtain $$$a = [2, 4, 5]$$$, which is increasing.You are given a permutation $$$p$$$ of numbers from $$$1$$$ to $$$n$$$. Find the number of pairs of integers $$$(l, r)$$$ with $$$1 \le l \le r \le n$$$ such that $$$p[l \ldots r]$$$ (the subarray of $$$p$$$ from $$$l$$$ to $$$r$$$) is a Decinc array.
Input Specification: The first line contains a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$) — the size of $$$p$$$. The second line contains $$$n$$$ integers $$$p_1, p_2, \ldots, p_n$$$ ($$$1 \le p_i \le n$$$, all $$$p_i$$$ are distinct) — elements of the permutation.
Output Specification: Output the number of pairs of integers $$$(l, r)$$$ such that $$$p[l \ldots r]$$$ (the subarray of $$$p$$$ from $$$l$$$ to $$$r$$$) is a Decinc array. $$$(1 \le l \le r \le n)$$$
Notes: NoteIn the first sample, all subarrays are Decinc.In the second sample, all subarrays except $$$p[1 \ldots 6]$$$ and $$$p[2 \ldots 6]$$$ are Decinc.
Code:
input = __import__('sys').stdin.readline
n = int(input())
a = list(map(int, input().split())) + [n+1]
cache = {}
def check(i, u, d):
keys = []
j = i
while j+1 <= n:
key = (j, u, d)
v = cache.get(key, -1)
if v != -1:
j = v
break
keys.append(key)
if u < a[j] < d: # if can insert to both
if a[j] < a[j+1]:
u = max(u, a[j])
elif a[j] > a[j+1]:
# TODO: Your code here
elif u < a[j]: # if only can insert to increasing subsequence
u = a[j]
elif d > a[j]: # if only can insert to decreasing subsequence
d = a[j]
else:
break
j += 1
for key in keys:
cache[key] = j
return j
ans = 0
for i in range(n):
u = 0
d = n+1
j = check(i, u, d)
ans += j - i
# print(f'at {i} max {j} ans {ans}', u, d)
# print(f'count={count}')
print(ans)
|
input = __import__('sys').stdin.readline
n = int(input())
a = list(map(int, input().split())) + [n+1]
cache = {}
def check(i, u, d):
keys = []
j = i
while j+1 <= n:
key = (j, u, d)
v = cache.get(key, -1)
if v != -1:
j = v
break
keys.append(key)
if u < a[j] < d: # if can insert to both
if a[j] < a[j+1]:
u = max(u, a[j])
elif a[j] > a[j+1]:
{{completion}}
elif u < a[j]: # if only can insert to increasing subsequence
u = a[j]
elif d > a[j]: # if only can insert to decreasing subsequence
d = a[j]
else:
break
j += 1
for key in keys:
cache[key] = j
return j
ans = 0
for i in range(n):
u = 0
d = n+1
j = check(i, u, d)
ans += j - i
# print(f'at {i} max {j} ans {ans}', u, d)
# print(f'count={count}')
print(ans)
|
d = min(d, a[j])
|
[{"input": "3\n2 3 1", "output": ["6"]}, {"input": "6\n4 5 2 6 1 3", "output": ["19"]}, {"input": "10\n7 10 1 8 3 9 2 4 6 5", "output": ["39"]}]
|
block_completion_000488
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: The store sells $$$n$$$ items, the price of the $$$i$$$-th item is $$$p_i$$$. The store's management is going to hold a promotion: if a customer purchases at least $$$x$$$ items, $$$y$$$ cheapest of them are free.The management has not yet decided on the exact values of $$$x$$$ and $$$y$$$. Therefore, they ask you to process $$$q$$$ queries: for the given values of $$$x$$$ and $$$y$$$, determine the maximum total value of items received for free, if a customer makes one purchase.Note that all queries are independent; they don't affect the store's stock.
Input Specification: The first line contains two integers $$$n$$$ and $$$q$$$ ($$$1 \le n, q \le 2 \cdot 10^5$$$) — the number of items in the store and the number of queries, respectively. The second line contains $$$n$$$ integers $$$p_1, p_2, \dots, p_n$$$ ($$$1 \le p_i \le 10^6$$$), where $$$p_i$$$ — the price of the $$$i$$$-th item. The following $$$q$$$ lines contain two integers $$$x_i$$$ and $$$y_i$$$ each ($$$1 \le y_i \le x_i \le n$$$) — the values of the parameters $$$x$$$ and $$$y$$$ in the $$$i$$$-th query.
Output Specification: For each query, print a single integer — the maximum total value of items received for free for one purchase.
Notes: NoteIn the first query, a customer can buy three items worth $$$5, 3, 5$$$, the two cheapest of them are $$$3 + 5 = 8$$$.In the second query, a customer can buy two items worth $$$5$$$ and $$$5$$$, the cheapest of them is $$$5$$$.In the third query, a customer has to buy all the items to receive the three cheapest of them for free; their total price is $$$1 + 2 + 3 = 6$$$.
Code:
#from niumeng
from itertools import accumulate
I=input;R=lambda:map(int,I().split())
n,q=R();a=sorted(R())[::-1];p=[0]+list(accumulate(a))
for _ in range(q):
# TODO: Your code here
|
#from niumeng
from itertools import accumulate
I=input;R=lambda:map(int,I().split())
n,q=R();a=sorted(R())[::-1];p=[0]+list(accumulate(a))
for _ in range(q):
{{completion}}
|
x,y=R();print(p[x]-p[x-y])
|
[{"input": "5 3\n5 3 1 5 2\n3 2\n1 1\n5 3", "output": ["8\n5\n6"]}]
|
block_completion_000509
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: The store sells $$$n$$$ items, the price of the $$$i$$$-th item is $$$p_i$$$. The store's management is going to hold a promotion: if a customer purchases at least $$$x$$$ items, $$$y$$$ cheapest of them are free.The management has not yet decided on the exact values of $$$x$$$ and $$$y$$$. Therefore, they ask you to process $$$q$$$ queries: for the given values of $$$x$$$ and $$$y$$$, determine the maximum total value of items received for free, if a customer makes one purchase.Note that all queries are independent; they don't affect the store's stock.
Input Specification: The first line contains two integers $$$n$$$ and $$$q$$$ ($$$1 \le n, q \le 2 \cdot 10^5$$$) — the number of items in the store and the number of queries, respectively. The second line contains $$$n$$$ integers $$$p_1, p_2, \dots, p_n$$$ ($$$1 \le p_i \le 10^6$$$), where $$$p_i$$$ — the price of the $$$i$$$-th item. The following $$$q$$$ lines contain two integers $$$x_i$$$ and $$$y_i$$$ each ($$$1 \le y_i \le x_i \le n$$$) — the values of the parameters $$$x$$$ and $$$y$$$ in the $$$i$$$-th query.
Output Specification: For each query, print a single integer — the maximum total value of items received for free for one purchase.
Notes: NoteIn the first query, a customer can buy three items worth $$$5, 3, 5$$$, the two cheapest of them are $$$3 + 5 = 8$$$.In the second query, a customer can buy two items worth $$$5$$$ and $$$5$$$, the cheapest of them is $$$5$$$.In the third query, a customer has to buy all the items to receive the three cheapest of them for free; their total price is $$$1 + 2 + 3 = 6$$$.
Code:
n, q = [int(x) for x in input().split()]
prices = [int(price) for price in input().split(" ")]
prices.sort(reverse=True)
for i in range(1, len(prices)):
# TODO: Your code here
while q:
# 5 5 3 2 1
# 5 10 13 15 16
x, y = [int(x) for x in input().split()]
l = 0 if x == y else prices[x - y - 1]
print(prices[x-1] - l)
q -= 1
|
n, q = [int(x) for x in input().split()]
prices = [int(price) for price in input().split(" ")]
prices.sort(reverse=True)
for i in range(1, len(prices)):
{{completion}}
while q:
# 5 5 3 2 1
# 5 10 13 15 16
x, y = [int(x) for x in input().split()]
l = 0 if x == y else prices[x - y - 1]
print(prices[x-1] - l)
q -= 1
|
prices[i] += prices[i-1]
|
[{"input": "5 3\n5 3 1 5 2\n3 2\n1 1\n5 3", "output": ["8\n5\n6"]}]
|
block_completion_000510
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: The store sells $$$n$$$ items, the price of the $$$i$$$-th item is $$$p_i$$$. The store's management is going to hold a promotion: if a customer purchases at least $$$x$$$ items, $$$y$$$ cheapest of them are free.The management has not yet decided on the exact values of $$$x$$$ and $$$y$$$. Therefore, they ask you to process $$$q$$$ queries: for the given values of $$$x$$$ and $$$y$$$, determine the maximum total value of items received for free, if a customer makes one purchase.Note that all queries are independent; they don't affect the store's stock.
Input Specification: The first line contains two integers $$$n$$$ and $$$q$$$ ($$$1 \le n, q \le 2 \cdot 10^5$$$) — the number of items in the store and the number of queries, respectively. The second line contains $$$n$$$ integers $$$p_1, p_2, \dots, p_n$$$ ($$$1 \le p_i \le 10^6$$$), where $$$p_i$$$ — the price of the $$$i$$$-th item. The following $$$q$$$ lines contain two integers $$$x_i$$$ and $$$y_i$$$ each ($$$1 \le y_i \le x_i \le n$$$) — the values of the parameters $$$x$$$ and $$$y$$$ in the $$$i$$$-th query.
Output Specification: For each query, print a single integer — the maximum total value of items received for free for one purchase.
Notes: NoteIn the first query, a customer can buy three items worth $$$5, 3, 5$$$, the two cheapest of them are $$$3 + 5 = 8$$$.In the second query, a customer can buy two items worth $$$5$$$ and $$$5$$$, the cheapest of them is $$$5$$$.In the third query, a customer has to buy all the items to receive the three cheapest of them for free; their total price is $$$1 + 2 + 3 = 6$$$.
Code:
n, q = [int(x) for x in input().split()]
prices = [int(price) for price in input().split(" ")]
prices.sort(reverse=True)
for i in range(1, len(prices)):
prices[i] += prices[i-1]
while q:
# 5 5 3 2 1
# 5 10 13 15 16
# TODO: Your code here
|
n, q = [int(x) for x in input().split()]
prices = [int(price) for price in input().split(" ")]
prices.sort(reverse=True)
for i in range(1, len(prices)):
prices[i] += prices[i-1]
while q:
# 5 5 3 2 1
# 5 10 13 15 16
{{completion}}
|
x, y = [int(x) for x in input().split()]
l = 0 if x == y else prices[x - y - 1]
print(prices[x-1] - l)
q -= 1
|
[{"input": "5 3\n5 3 1 5 2\n3 2\n1 1\n5 3", "output": ["8\n5\n6"]}]
|
block_completion_000511
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: The store sells $$$n$$$ items, the price of the $$$i$$$-th item is $$$p_i$$$. The store's management is going to hold a promotion: if a customer purchases at least $$$x$$$ items, $$$y$$$ cheapest of them are free.The management has not yet decided on the exact values of $$$x$$$ and $$$y$$$. Therefore, they ask you to process $$$q$$$ queries: for the given values of $$$x$$$ and $$$y$$$, determine the maximum total value of items received for free, if a customer makes one purchase.Note that all queries are independent; they don't affect the store's stock.
Input Specification: The first line contains two integers $$$n$$$ and $$$q$$$ ($$$1 \le n, q \le 2 \cdot 10^5$$$) — the number of items in the store and the number of queries, respectively. The second line contains $$$n$$$ integers $$$p_1, p_2, \dots, p_n$$$ ($$$1 \le p_i \le 10^6$$$), where $$$p_i$$$ — the price of the $$$i$$$-th item. The following $$$q$$$ lines contain two integers $$$x_i$$$ and $$$y_i$$$ each ($$$1 \le y_i \le x_i \le n$$$) — the values of the parameters $$$x$$$ and $$$y$$$ in the $$$i$$$-th query.
Output Specification: For each query, print a single integer — the maximum total value of items received for free for one purchase.
Notes: NoteIn the first query, a customer can buy three items worth $$$5, 3, 5$$$, the two cheapest of them are $$$3 + 5 = 8$$$.In the second query, a customer can buy two items worth $$$5$$$ and $$$5$$$, the cheapest of them is $$$5$$$.In the third query, a customer has to buy all the items to receive the three cheapest of them for free; their total price is $$$1 + 2 + 3 = 6$$$.
Code:
n,q=map(int,input().split())
a=[0]
for x in sorted(map(int,input().split()))[::-1]:# TODO: Your code here
for _ in[0]*q:x,y=map(int,input().split());print(a[x]-a[x-y])
|
n,q=map(int,input().split())
a=[0]
for x in sorted(map(int,input().split()))[::-1]:{{completion}}
for _ in[0]*q:x,y=map(int,input().split());print(a[x]-a[x-y])
|
a+=a[-1]+x,
|
[{"input": "5 3\n5 3 1 5 2\n3 2\n1 1\n5 3", "output": ["8\n5\n6"]}]
|
block_completion_000512
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: The store sells $$$n$$$ items, the price of the $$$i$$$-th item is $$$p_i$$$. The store's management is going to hold a promotion: if a customer purchases at least $$$x$$$ items, $$$y$$$ cheapest of them are free.The management has not yet decided on the exact values of $$$x$$$ and $$$y$$$. Therefore, they ask you to process $$$q$$$ queries: for the given values of $$$x$$$ and $$$y$$$, determine the maximum total value of items received for free, if a customer makes one purchase.Note that all queries are independent; they don't affect the store's stock.
Input Specification: The first line contains two integers $$$n$$$ and $$$q$$$ ($$$1 \le n, q \le 2 \cdot 10^5$$$) — the number of items in the store and the number of queries, respectively. The second line contains $$$n$$$ integers $$$p_1, p_2, \dots, p_n$$$ ($$$1 \le p_i \le 10^6$$$), where $$$p_i$$$ — the price of the $$$i$$$-th item. The following $$$q$$$ lines contain two integers $$$x_i$$$ and $$$y_i$$$ each ($$$1 \le y_i \le x_i \le n$$$) — the values of the parameters $$$x$$$ and $$$y$$$ in the $$$i$$$-th query.
Output Specification: For each query, print a single integer — the maximum total value of items received for free for one purchase.
Notes: NoteIn the first query, a customer can buy three items worth $$$5, 3, 5$$$, the two cheapest of them are $$$3 + 5 = 8$$$.In the second query, a customer can buy two items worth $$$5$$$ and $$$5$$$, the cheapest of them is $$$5$$$.In the third query, a customer has to buy all the items to receive the three cheapest of them for free; their total price is $$$1 + 2 + 3 = 6$$$.
Code:
n,q=map(int,input().split())
a=[0]
for x in sorted(map(int,input().split()))[::-1]:a+=a[-1]+x,
for _ in[0]*q:# TODO: Your code here
|
n,q=map(int,input().split())
a=[0]
for x in sorted(map(int,input().split()))[::-1]:a+=a[-1]+x,
for _ in[0]*q:{{completion}}
|
x,y=map(int,input().split());print(a[x]-a[x-y])
|
[{"input": "5 3\n5 3 1 5 2\n3 2\n1 1\n5 3", "output": ["8\n5\n6"]}]
|
block_completion_000513
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: The store sells $$$n$$$ items, the price of the $$$i$$$-th item is $$$p_i$$$. The store's management is going to hold a promotion: if a customer purchases at least $$$x$$$ items, $$$y$$$ cheapest of them are free.The management has not yet decided on the exact values of $$$x$$$ and $$$y$$$. Therefore, they ask you to process $$$q$$$ queries: for the given values of $$$x$$$ and $$$y$$$, determine the maximum total value of items received for free, if a customer makes one purchase.Note that all queries are independent; they don't affect the store's stock.
Input Specification: The first line contains two integers $$$n$$$ and $$$q$$$ ($$$1 \le n, q \le 2 \cdot 10^5$$$) — the number of items in the store and the number of queries, respectively. The second line contains $$$n$$$ integers $$$p_1, p_2, \dots, p_n$$$ ($$$1 \le p_i \le 10^6$$$), where $$$p_i$$$ — the price of the $$$i$$$-th item. The following $$$q$$$ lines contain two integers $$$x_i$$$ and $$$y_i$$$ each ($$$1 \le y_i \le x_i \le n$$$) — the values of the parameters $$$x$$$ and $$$y$$$ in the $$$i$$$-th query.
Output Specification: For each query, print a single integer — the maximum total value of items received for free for one purchase.
Notes: NoteIn the first query, a customer can buy three items worth $$$5, 3, 5$$$, the two cheapest of them are $$$3 + 5 = 8$$$.In the second query, a customer can buy two items worth $$$5$$$ and $$$5$$$, the cheapest of them is $$$5$$$.In the third query, a customer has to buy all the items to receive the three cheapest of them for free; their total price is $$$1 + 2 + 3 = 6$$$.
Code:
f=open(0)
R=lambda:map(int,next(f).split())
n,q=R();p=[0]
for w in sorted(R()): # TODO: Your code here
for _ in " "*q: x, y=R();print(p[n-x+y]-p[n-x])
|
f=open(0)
R=lambda:map(int,next(f).split())
n,q=R();p=[0]
for w in sorted(R()): {{completion}}
for _ in " "*q: x, y=R();print(p[n-x+y]-p[n-x])
|
p+=p[-1]+w,
|
[{"input": "5 3\n5 3 1 5 2\n3 2\n1 1\n5 3", "output": ["8\n5\n6"]}]
|
block_completion_000514
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: The store sells $$$n$$$ items, the price of the $$$i$$$-th item is $$$p_i$$$. The store's management is going to hold a promotion: if a customer purchases at least $$$x$$$ items, $$$y$$$ cheapest of them are free.The management has not yet decided on the exact values of $$$x$$$ and $$$y$$$. Therefore, they ask you to process $$$q$$$ queries: for the given values of $$$x$$$ and $$$y$$$, determine the maximum total value of items received for free, if a customer makes one purchase.Note that all queries are independent; they don't affect the store's stock.
Input Specification: The first line contains two integers $$$n$$$ and $$$q$$$ ($$$1 \le n, q \le 2 \cdot 10^5$$$) — the number of items in the store and the number of queries, respectively. The second line contains $$$n$$$ integers $$$p_1, p_2, \dots, p_n$$$ ($$$1 \le p_i \le 10^6$$$), where $$$p_i$$$ — the price of the $$$i$$$-th item. The following $$$q$$$ lines contain two integers $$$x_i$$$ and $$$y_i$$$ each ($$$1 \le y_i \le x_i \le n$$$) — the values of the parameters $$$x$$$ and $$$y$$$ in the $$$i$$$-th query.
Output Specification: For each query, print a single integer — the maximum total value of items received for free for one purchase.
Notes: NoteIn the first query, a customer can buy three items worth $$$5, 3, 5$$$, the two cheapest of them are $$$3 + 5 = 8$$$.In the second query, a customer can buy two items worth $$$5$$$ and $$$5$$$, the cheapest of them is $$$5$$$.In the third query, a customer has to buy all the items to receive the three cheapest of them for free; their total price is $$$1 + 2 + 3 = 6$$$.
Code:
f=open(0)
R=lambda:map(int,next(f).split())
n,q=R();p=[0]
for w in sorted(R()): p+=p[-1]+w,
for _ in " "*q: # TODO: Your code here
|
f=open(0)
R=lambda:map(int,next(f).split())
n,q=R();p=[0]
for w in sorted(R()): p+=p[-1]+w,
for _ in " "*q: {{completion}}
|
x, y=R();print(p[n-x+y]-p[n-x])
|
[{"input": "5 3\n5 3 1 5 2\n3 2\n1 1\n5 3", "output": ["8\n5\n6"]}]
|
block_completion_000515
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: The store sells $$$n$$$ items, the price of the $$$i$$$-th item is $$$p_i$$$. The store's management is going to hold a promotion: if a customer purchases at least $$$x$$$ items, $$$y$$$ cheapest of them are free.The management has not yet decided on the exact values of $$$x$$$ and $$$y$$$. Therefore, they ask you to process $$$q$$$ queries: for the given values of $$$x$$$ and $$$y$$$, determine the maximum total value of items received for free, if a customer makes one purchase.Note that all queries are independent; they don't affect the store's stock.
Input Specification: The first line contains two integers $$$n$$$ and $$$q$$$ ($$$1 \le n, q \le 2 \cdot 10^5$$$) — the number of items in the store and the number of queries, respectively. The second line contains $$$n$$$ integers $$$p_1, p_2, \dots, p_n$$$ ($$$1 \le p_i \le 10^6$$$), where $$$p_i$$$ — the price of the $$$i$$$-th item. The following $$$q$$$ lines contain two integers $$$x_i$$$ and $$$y_i$$$ each ($$$1 \le y_i \le x_i \le n$$$) — the values of the parameters $$$x$$$ and $$$y$$$ in the $$$i$$$-th query.
Output Specification: For each query, print a single integer — the maximum total value of items received for free for one purchase.
Notes: NoteIn the first query, a customer can buy three items worth $$$5, 3, 5$$$, the two cheapest of them are $$$3 + 5 = 8$$$.In the second query, a customer can buy two items worth $$$5$$$ and $$$5$$$, the cheapest of them is $$$5$$$.In the third query, a customer has to buy all the items to receive the three cheapest of them for free; their total price is $$$1 + 2 + 3 = 6$$$.
Code:
from sys import stdin
# t = int(stdin.readline().rstrip())
# while t>0:
# t-=1
n,q = map(int,stdin.readline().split())
l = list(map(int,stdin.readline().split()))
l.sort()
for i in range(1,n):
l[i] += l[i-1]
# print(l)
for i in range(q):
x,y = map(int,stdin.readline().split())
actual = n-x+y-1
val = l[actual]
if n-x > 0:
# TODO: Your code here
print(val)
|
from sys import stdin
# t = int(stdin.readline().rstrip())
# while t>0:
# t-=1
n,q = map(int,stdin.readline().split())
l = list(map(int,stdin.readline().split()))
l.sort()
for i in range(1,n):
l[i] += l[i-1]
# print(l)
for i in range(q):
x,y = map(int,stdin.readline().split())
actual = n-x+y-1
val = l[actual]
if n-x > 0:
{{completion}}
print(val)
|
val -= l[n-x-1]
|
[{"input": "5 3\n5 3 1 5 2\n3 2\n1 1\n5 3", "output": ["8\n5\n6"]}]
|
block_completion_000516
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: The store sells $$$n$$$ items, the price of the $$$i$$$-th item is $$$p_i$$$. The store's management is going to hold a promotion: if a customer purchases at least $$$x$$$ items, $$$y$$$ cheapest of them are free.The management has not yet decided on the exact values of $$$x$$$ and $$$y$$$. Therefore, they ask you to process $$$q$$$ queries: for the given values of $$$x$$$ and $$$y$$$, determine the maximum total value of items received for free, if a customer makes one purchase.Note that all queries are independent; they don't affect the store's stock.
Input Specification: The first line contains two integers $$$n$$$ and $$$q$$$ ($$$1 \le n, q \le 2 \cdot 10^5$$$) — the number of items in the store and the number of queries, respectively. The second line contains $$$n$$$ integers $$$p_1, p_2, \dots, p_n$$$ ($$$1 \le p_i \le 10^6$$$), where $$$p_i$$$ — the price of the $$$i$$$-th item. The following $$$q$$$ lines contain two integers $$$x_i$$$ and $$$y_i$$$ each ($$$1 \le y_i \le x_i \le n$$$) — the values of the parameters $$$x$$$ and $$$y$$$ in the $$$i$$$-th query.
Output Specification: For each query, print a single integer — the maximum total value of items received for free for one purchase.
Notes: NoteIn the first query, a customer can buy three items worth $$$5, 3, 5$$$, the two cheapest of them are $$$3 + 5 = 8$$$.In the second query, a customer can buy two items worth $$$5$$$ and $$$5$$$, the cheapest of them is $$$5$$$.In the third query, a customer has to buy all the items to receive the three cheapest of them for free; their total price is $$$1 + 2 + 3 = 6$$$.
Code:
Y=lambda:map(int,input().split())
O=[];n,q=Y();p=sorted(Y())[::-1];s=[0]
for i in p:# TODO: Your code here
for _ in[0]*q:x,y=Y();O+=[str(s[x]-s[x-y])]
print('\n'.join(O))
|
Y=lambda:map(int,input().split())
O=[];n,q=Y();p=sorted(Y())[::-1];s=[0]
for i in p:{{completion}}
for _ in[0]*q:x,y=Y();O+=[str(s[x]-s[x-y])]
print('\n'.join(O))
|
s+=[s[-1]+i]
|
[{"input": "5 3\n5 3 1 5 2\n3 2\n1 1\n5 3", "output": ["8\n5\n6"]}]
|
block_completion_000517
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: The store sells $$$n$$$ items, the price of the $$$i$$$-th item is $$$p_i$$$. The store's management is going to hold a promotion: if a customer purchases at least $$$x$$$ items, $$$y$$$ cheapest of them are free.The management has not yet decided on the exact values of $$$x$$$ and $$$y$$$. Therefore, they ask you to process $$$q$$$ queries: for the given values of $$$x$$$ and $$$y$$$, determine the maximum total value of items received for free, if a customer makes one purchase.Note that all queries are independent; they don't affect the store's stock.
Input Specification: The first line contains two integers $$$n$$$ and $$$q$$$ ($$$1 \le n, q \le 2 \cdot 10^5$$$) — the number of items in the store and the number of queries, respectively. The second line contains $$$n$$$ integers $$$p_1, p_2, \dots, p_n$$$ ($$$1 \le p_i \le 10^6$$$), where $$$p_i$$$ — the price of the $$$i$$$-th item. The following $$$q$$$ lines contain two integers $$$x_i$$$ and $$$y_i$$$ each ($$$1 \le y_i \le x_i \le n$$$) — the values of the parameters $$$x$$$ and $$$y$$$ in the $$$i$$$-th query.
Output Specification: For each query, print a single integer — the maximum total value of items received for free for one purchase.
Notes: NoteIn the first query, a customer can buy three items worth $$$5, 3, 5$$$, the two cheapest of them are $$$3 + 5 = 8$$$.In the second query, a customer can buy two items worth $$$5$$$ and $$$5$$$, the cheapest of them is $$$5$$$.In the third query, a customer has to buy all the items to receive the three cheapest of them for free; their total price is $$$1 + 2 + 3 = 6$$$.
Code:
Y=lambda:map(int,input().split())
O=[];n,q=Y();p=sorted(Y())[::-1];s=[0]
for i in p:s+=[s[-1]+i]
for _ in[0]*q:# TODO: Your code here
print('\n'.join(O))
|
Y=lambda:map(int,input().split())
O=[];n,q=Y();p=sorted(Y())[::-1];s=[0]
for i in p:s+=[s[-1]+i]
for _ in[0]*q:{{completion}}
print('\n'.join(O))
|
x,y=Y();O+=[str(s[x]-s[x-y])]
|
[{"input": "5 3\n5 3 1 5 2\n3 2\n1 1\n5 3", "output": ["8\n5\n6"]}]
|
block_completion_000518
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: The store sells $$$n$$$ items, the price of the $$$i$$$-th item is $$$p_i$$$. The store's management is going to hold a promotion: if a customer purchases at least $$$x$$$ items, $$$y$$$ cheapest of them are free.The management has not yet decided on the exact values of $$$x$$$ and $$$y$$$. Therefore, they ask you to process $$$q$$$ queries: for the given values of $$$x$$$ and $$$y$$$, determine the maximum total value of items received for free, if a customer makes one purchase.Note that all queries are independent; they don't affect the store's stock.
Input Specification: The first line contains two integers $$$n$$$ and $$$q$$$ ($$$1 \le n, q \le 2 \cdot 10^5$$$) — the number of items in the store and the number of queries, respectively. The second line contains $$$n$$$ integers $$$p_1, p_2, \dots, p_n$$$ ($$$1 \le p_i \le 10^6$$$), where $$$p_i$$$ — the price of the $$$i$$$-th item. The following $$$q$$$ lines contain two integers $$$x_i$$$ and $$$y_i$$$ each ($$$1 \le y_i \le x_i \le n$$$) — the values of the parameters $$$x$$$ and $$$y$$$ in the $$$i$$$-th query.
Output Specification: For each query, print a single integer — the maximum total value of items received for free for one purchase.
Notes: NoteIn the first query, a customer can buy three items worth $$$5, 3, 5$$$, the two cheapest of them are $$$3 + 5 = 8$$$.In the second query, a customer can buy two items worth $$$5$$$ and $$$5$$$, the cheapest of them is $$$5$$$.In the third query, a customer has to buy all the items to receive the three cheapest of them for free; their total price is $$$1 + 2 + 3 = 6$$$.
Code:
arr=[int(i) for i in input().split()]
ans=[]
prices=[int(i) for i in input().split()]
prices.sort(reverse=True)
for i in range(1,arr[0]):
prices[i]=prices[i]+prices[i-1]
for i in range(arr[1]):
xy=[int(i) for i in input().split()]
if(xy[0]==xy[1]):
ans.append(prices[xy[0]-1])
else:
# TODO: Your code here
for ele in ans:
print(ele)
|
arr=[int(i) for i in input().split()]
ans=[]
prices=[int(i) for i in input().split()]
prices.sort(reverse=True)
for i in range(1,arr[0]):
prices[i]=prices[i]+prices[i-1]
for i in range(arr[1]):
xy=[int(i) for i in input().split()]
if(xy[0]==xy[1]):
ans.append(prices[xy[0]-1])
else:
{{completion}}
for ele in ans:
print(ele)
|
ans.append(prices[xy[0]-1]-prices[xy[0]-xy[1]-1])
|
[{"input": "5 3\n5 3 1 5 2\n3 2\n1 1\n5 3", "output": ["8\n5\n6"]}]
|
block_completion_000519
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: The store sells $$$n$$$ items, the price of the $$$i$$$-th item is $$$p_i$$$. The store's management is going to hold a promotion: if a customer purchases at least $$$x$$$ items, $$$y$$$ cheapest of them are free.The management has not yet decided on the exact values of $$$x$$$ and $$$y$$$. Therefore, they ask you to process $$$q$$$ queries: for the given values of $$$x$$$ and $$$y$$$, determine the maximum total value of items received for free, if a customer makes one purchase.Note that all queries are independent; they don't affect the store's stock.
Input Specification: The first line contains two integers $$$n$$$ and $$$q$$$ ($$$1 \le n, q \le 2 \cdot 10^5$$$) — the number of items in the store and the number of queries, respectively. The second line contains $$$n$$$ integers $$$p_1, p_2, \dots, p_n$$$ ($$$1 \le p_i \le 10^6$$$), where $$$p_i$$$ — the price of the $$$i$$$-th item. The following $$$q$$$ lines contain two integers $$$x_i$$$ and $$$y_i$$$ each ($$$1 \le y_i \le x_i \le n$$$) — the values of the parameters $$$x$$$ and $$$y$$$ in the $$$i$$$-th query.
Output Specification: For each query, print a single integer — the maximum total value of items received for free for one purchase.
Notes: NoteIn the first query, a customer can buy three items worth $$$5, 3, 5$$$, the two cheapest of them are $$$3 + 5 = 8$$$.In the second query, a customer can buy two items worth $$$5$$$ and $$$5$$$, the cheapest of them is $$$5$$$.In the third query, a customer has to buy all the items to receive the three cheapest of them for free; their total price is $$$1 + 2 + 3 = 6$$$.
Code:
import sys
n, p = map(int, sys.stdin.readline().split())
l = map(int, sys.stdin.readline().split())
l = sorted(l, reverse=True)
for i in range(n-1, 0, -1):
l[i-1] += l[i]
for _ in range(p):
xi, yi = map(int, sys.stdin.readline().split())
a = n-xi
b = a+yi
if a == 0:
print(l[-b])
else:
# TODO: Your code here
|
import sys
n, p = map(int, sys.stdin.readline().split())
l = map(int, sys.stdin.readline().split())
l = sorted(l, reverse=True)
for i in range(n-1, 0, -1):
l[i-1] += l[i]
for _ in range(p):
xi, yi = map(int, sys.stdin.readline().split())
a = n-xi
b = a+yi
if a == 0:
print(l[-b])
else:
{{completion}}
|
print(l[-b]-l[-a])
|
[{"input": "5 3\n5 3 1 5 2\n3 2\n1 1\n5 3", "output": ["8\n5\n6"]}]
|
block_completion_000520
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: The store sells $$$n$$$ items, the price of the $$$i$$$-th item is $$$p_i$$$. The store's management is going to hold a promotion: if a customer purchases at least $$$x$$$ items, $$$y$$$ cheapest of them are free.The management has not yet decided on the exact values of $$$x$$$ and $$$y$$$. Therefore, they ask you to process $$$q$$$ queries: for the given values of $$$x$$$ and $$$y$$$, determine the maximum total value of items received for free, if a customer makes one purchase.Note that all queries are independent; they don't affect the store's stock.
Input Specification: The first line contains two integers $$$n$$$ and $$$q$$$ ($$$1 \le n, q \le 2 \cdot 10^5$$$) — the number of items in the store and the number of queries, respectively. The second line contains $$$n$$$ integers $$$p_1, p_2, \dots, p_n$$$ ($$$1 \le p_i \le 10^6$$$), where $$$p_i$$$ — the price of the $$$i$$$-th item. The following $$$q$$$ lines contain two integers $$$x_i$$$ and $$$y_i$$$ each ($$$1 \le y_i \le x_i \le n$$$) — the values of the parameters $$$x$$$ and $$$y$$$ in the $$$i$$$-th query.
Output Specification: For each query, print a single integer — the maximum total value of items received for free for one purchase.
Notes: NoteIn the first query, a customer can buy three items worth $$$5, 3, 5$$$, the two cheapest of them are $$$3 + 5 = 8$$$.In the second query, a customer can buy two items worth $$$5$$$ and $$$5$$$, the cheapest of them is $$$5$$$.In the third query, a customer has to buy all the items to receive the three cheapest of them for free; their total price is $$$1 + 2 + 3 = 6$$$.
Code:
## cf does not have numpy so with lists
r=open(0)
g=lambda:map(int,next(r).split())
n,q=g()
a=[0]
for x in sorted(g())[::-1]:# TODO: Your code here
for b in[0]*q:x,y=g();print(a[x]-a[x-y])
|
## cf does not have numpy so with lists
r=open(0)
g=lambda:map(int,next(r).split())
n,q=g()
a=[0]
for x in sorted(g())[::-1]:{{completion}}
for b in[0]*q:x,y=g();print(a[x]-a[x-y])
|
a+=a[-1]+x,
|
[{"input": "5 3\n5 3 1 5 2\n3 2\n1 1\n5 3", "output": ["8\n5\n6"]}]
|
block_completion_000521
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: The store sells $$$n$$$ items, the price of the $$$i$$$-th item is $$$p_i$$$. The store's management is going to hold a promotion: if a customer purchases at least $$$x$$$ items, $$$y$$$ cheapest of them are free.The management has not yet decided on the exact values of $$$x$$$ and $$$y$$$. Therefore, they ask you to process $$$q$$$ queries: for the given values of $$$x$$$ and $$$y$$$, determine the maximum total value of items received for free, if a customer makes one purchase.Note that all queries are independent; they don't affect the store's stock.
Input Specification: The first line contains two integers $$$n$$$ and $$$q$$$ ($$$1 \le n, q \le 2 \cdot 10^5$$$) — the number of items in the store and the number of queries, respectively. The second line contains $$$n$$$ integers $$$p_1, p_2, \dots, p_n$$$ ($$$1 \le p_i \le 10^6$$$), where $$$p_i$$$ — the price of the $$$i$$$-th item. The following $$$q$$$ lines contain two integers $$$x_i$$$ and $$$y_i$$$ each ($$$1 \le y_i \le x_i \le n$$$) — the values of the parameters $$$x$$$ and $$$y$$$ in the $$$i$$$-th query.
Output Specification: For each query, print a single integer — the maximum total value of items received for free for one purchase.
Notes: NoteIn the first query, a customer can buy three items worth $$$5, 3, 5$$$, the two cheapest of them are $$$3 + 5 = 8$$$.In the second query, a customer can buy two items worth $$$5$$$ and $$$5$$$, the cheapest of them is $$$5$$$.In the third query, a customer has to buy all the items to receive the three cheapest of them for free; their total price is $$$1 + 2 + 3 = 6$$$.
Code:
## cf does not have numpy so with lists
r=open(0)
g=lambda:map(int,next(r).split())
n,q=g()
a=[0]
for x in sorted(g())[::-1]:a+=a[-1]+x,
for b in[0]*q:# TODO: Your code here
|
## cf does not have numpy so with lists
r=open(0)
g=lambda:map(int,next(r).split())
n,q=g()
a=[0]
for x in sorted(g())[::-1]:a+=a[-1]+x,
for b in[0]*q:{{completion}}
|
x,y=g();print(a[x]-a[x-y])
|
[{"input": "5 3\n5 3 1 5 2\n3 2\n1 1\n5 3", "output": ["8\n5\n6"]}]
|
block_completion_000522
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: The store sells $$$n$$$ items, the price of the $$$i$$$-th item is $$$p_i$$$. The store's management is going to hold a promotion: if a customer purchases at least $$$x$$$ items, $$$y$$$ cheapest of them are free.The management has not yet decided on the exact values of $$$x$$$ and $$$y$$$. Therefore, they ask you to process $$$q$$$ queries: for the given values of $$$x$$$ and $$$y$$$, determine the maximum total value of items received for free, if a customer makes one purchase.Note that all queries are independent; they don't affect the store's stock.
Input Specification: The first line contains two integers $$$n$$$ and $$$q$$$ ($$$1 \le n, q \le 2 \cdot 10^5$$$) — the number of items in the store and the number of queries, respectively. The second line contains $$$n$$$ integers $$$p_1, p_2, \dots, p_n$$$ ($$$1 \le p_i \le 10^6$$$), where $$$p_i$$$ — the price of the $$$i$$$-th item. The following $$$q$$$ lines contain two integers $$$x_i$$$ and $$$y_i$$$ each ($$$1 \le y_i \le x_i \le n$$$) — the values of the parameters $$$x$$$ and $$$y$$$ in the $$$i$$$-th query.
Output Specification: For each query, print a single integer — the maximum total value of items received for free for one purchase.
Notes: NoteIn the first query, a customer can buy three items worth $$$5, 3, 5$$$, the two cheapest of them are $$$3 + 5 = 8$$$.In the second query, a customer can buy two items worth $$$5$$$ and $$$5$$$, the cheapest of them is $$$5$$$.In the third query, a customer has to buy all the items to receive the three cheapest of them for free; their total price is $$$1 + 2 + 3 = 6$$$.
Code:
(n, q) = map(int, input().split())
arr = list(map(int, input().split()))
arr.sort(reverse=True)
for i in range (1,n):
arr[i] = arr[i] + arr[i-1]
for trial in range(q):
(x, y) = map(int, input().split())
if (x==y):
print (arr[x-1])
else:
# TODO: Your code here
|
(n, q) = map(int, input().split())
arr = list(map(int, input().split()))
arr.sort(reverse=True)
for i in range (1,n):
arr[i] = arr[i] + arr[i-1]
for trial in range(q):
(x, y) = map(int, input().split())
if (x==y):
print (arr[x-1])
else:
{{completion}}
|
print (arr[x-1] - arr[x-y-1])
|
[{"input": "5 3\n5 3 1 5 2\n3 2\n1 1\n5 3", "output": ["8\n5\n6"]}]
|
block_completion_000523
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: You are given $$$n$$$ points on the plane, the coordinates of the $$$i$$$-th point are $$$(x_i, y_i)$$$. No two points have the same coordinates.The distance between points $$$i$$$ and $$$j$$$ is defined as $$$d(i,j) = |x_i - x_j| + |y_i - y_j|$$$.For each point, you have to choose a color, represented by an integer from $$$1$$$ to $$$n$$$. For every ordered triple of different points $$$(a,b,c)$$$, the following constraints should be met: if $$$a$$$, $$$b$$$ and $$$c$$$ have the same color, then $$$d(a,b) = d(a,c) = d(b,c)$$$; if $$$a$$$ and $$$b$$$ have the same color, and the color of $$$c$$$ is different from the color of $$$a$$$, then $$$d(a,b) < d(a,c)$$$ and $$$d(a,b) < d(b,c)$$$. Calculate the number of different ways to choose the colors that meet these constraints.
Input Specification: The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the number of points. Then $$$n$$$ lines follow. The $$$i$$$-th of them contains two integers $$$x_i$$$ and $$$y_i$$$ ($$$0 \le x_i, y_i \le 10^8$$$). No two points have the same coordinates (i. e. if $$$i \ne j$$$, then either $$$x_i \ne x_j$$$ or $$$y_i \ne y_j$$$).
Output Specification: Print one integer — the number of ways to choose the colors for the points. Since it can be large, print it modulo $$$998244353$$$.
Notes: NoteIn the first test, the following ways to choose the colors are suitable: $$$[1, 1, 1]$$$; $$$[2, 2, 2]$$$; $$$[3, 3, 3]$$$; $$$[1, 2, 3]$$$; $$$[1, 3, 2]$$$; $$$[2, 1, 3]$$$; $$$[2, 3, 1]$$$; $$$[3, 1, 2]$$$; $$$[3, 2, 1]$$$.
Code:
from collections import deque
def solve():
n = int(input())
MOD = 998244353
arr = [list(map(int, input().split())) for i in range(n)]
dis = [[10**9]*n for i in range(n)]
for i in range(n):
for j in range(n):
if i != j:
dis[i][j] = abs(arr[i][0]-arr[j][0]) + abs(arr[i][1]-arr[j][1])
g = [[] for i in range(n)]
adj = [[0]*n for i in range(n)]
for i in range(n):
m = min(dis[i])
for j in range(n):
if dis[i][j] == m:
g[i].append(j)
adj[i][j] = 1
gr = []
for i in range(n):
cur = []
queue = deque()
queue.append(i)
v = [0] * n
v[i] = 1
while queue:
x = queue.popleft()
cur.append(x)
for y in g[x]:
if v[y] == 0:
# TODO: Your code here
ok = 1
for x in cur:
for y in cur:
if x != y and adj[x][y] == 0:
ok = 0
break
if ok:
if min(cur) == i:
gr.append(len(cur))
else:
gr.append(1)
dp = [0]*(n+1)
dp[0] = 1
for a in gr:
dp1 = [0]*(n+1)
for i in range(n):
dp1[i+1] = (dp1[i+1]+dp[i])%MOD
if a > 1 and i+a <= n:
dp1[i+a] = (dp1[i+a]+dp[i])%MOD
dp = dp1
ans = 0
k = n
for i in range(1, n+1):
ans = (ans+dp[i]*k)%MOD
k = k*(n-i)%MOD
return ans
import sys
input = lambda: sys.stdin.readline().rstrip()
print(solve())
|
from collections import deque
def solve():
n = int(input())
MOD = 998244353
arr = [list(map(int, input().split())) for i in range(n)]
dis = [[10**9]*n for i in range(n)]
for i in range(n):
for j in range(n):
if i != j:
dis[i][j] = abs(arr[i][0]-arr[j][0]) + abs(arr[i][1]-arr[j][1])
g = [[] for i in range(n)]
adj = [[0]*n for i in range(n)]
for i in range(n):
m = min(dis[i])
for j in range(n):
if dis[i][j] == m:
g[i].append(j)
adj[i][j] = 1
gr = []
for i in range(n):
cur = []
queue = deque()
queue.append(i)
v = [0] * n
v[i] = 1
while queue:
x = queue.popleft()
cur.append(x)
for y in g[x]:
if v[y] == 0:
{{completion}}
ok = 1
for x in cur:
for y in cur:
if x != y and adj[x][y] == 0:
ok = 0
break
if ok:
if min(cur) == i:
gr.append(len(cur))
else:
gr.append(1)
dp = [0]*(n+1)
dp[0] = 1
for a in gr:
dp1 = [0]*(n+1)
for i in range(n):
dp1[i+1] = (dp1[i+1]+dp[i])%MOD
if a > 1 and i+a <= n:
dp1[i+a] = (dp1[i+a]+dp[i])%MOD
dp = dp1
ans = 0
k = n
for i in range(1, n+1):
ans = (ans+dp[i]*k)%MOD
k = k*(n-i)%MOD
return ans
import sys
input = lambda: sys.stdin.readline().rstrip()
print(solve())
|
v[y] = 1
queue.append(y)
|
[{"input": "3\n1 0\n3 0\n2 1", "output": ["9"]}, {"input": "5\n1 2\n2 4\n3 4\n4 4\n1 3", "output": ["240"]}, {"input": "4\n1 0\n3 0\n2 1\n2 0", "output": ["24"]}]
|
block_completion_000541
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: You are given $$$n$$$ points on the plane, the coordinates of the $$$i$$$-th point are $$$(x_i, y_i)$$$. No two points have the same coordinates.The distance between points $$$i$$$ and $$$j$$$ is defined as $$$d(i,j) = |x_i - x_j| + |y_i - y_j|$$$.For each point, you have to choose a color, represented by an integer from $$$1$$$ to $$$n$$$. For every ordered triple of different points $$$(a,b,c)$$$, the following constraints should be met: if $$$a$$$, $$$b$$$ and $$$c$$$ have the same color, then $$$d(a,b) = d(a,c) = d(b,c)$$$; if $$$a$$$ and $$$b$$$ have the same color, and the color of $$$c$$$ is different from the color of $$$a$$$, then $$$d(a,b) < d(a,c)$$$ and $$$d(a,b) < d(b,c)$$$. Calculate the number of different ways to choose the colors that meet these constraints.
Input Specification: The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the number of points. Then $$$n$$$ lines follow. The $$$i$$$-th of them contains two integers $$$x_i$$$ and $$$y_i$$$ ($$$0 \le x_i, y_i \le 10^8$$$). No two points have the same coordinates (i. e. if $$$i \ne j$$$, then either $$$x_i \ne x_j$$$ or $$$y_i \ne y_j$$$).
Output Specification: Print one integer — the number of ways to choose the colors for the points. Since it can be large, print it modulo $$$998244353$$$.
Notes: NoteIn the first test, the following ways to choose the colors are suitable: $$$[1, 1, 1]$$$; $$$[2, 2, 2]$$$; $$$[3, 3, 3]$$$; $$$[1, 2, 3]$$$; $$$[1, 3, 2]$$$; $$$[2, 1, 3]$$$; $$$[2, 3, 1]$$$; $$$[3, 1, 2]$$$; $$$[3, 2, 1]$$$.
Code:
from collections import deque
def solve():
n = int(input())
MOD = 998244353
arr = [list(map(int, input().split())) for i in range(n)]
dis = [[10**9]*n for i in range(n)]
for i in range(n):
for j in range(n):
if i != j:
dis[i][j] = abs(arr[i][0]-arr[j][0]) + abs(arr[i][1]-arr[j][1])
g = [[] for i in range(n)]
adj = [[0]*n for i in range(n)]
for i in range(n):
m = min(dis[i])
for j in range(n):
if dis[i][j] == m:
g[i].append(j)
adj[i][j] = 1
gr = []
for i in range(n):
cur = []
queue = deque()
queue.append(i)
v = [0] * n
v[i] = 1
while queue:
x = queue.popleft()
cur.append(x)
for y in g[x]:
if v[y] == 0:
v[y] = 1
queue.append(y)
ok = 1
for x in cur:
for y in cur:
if x != y and adj[x][y] == 0:
# TODO: Your code here
if ok:
if min(cur) == i:
gr.append(len(cur))
else:
gr.append(1)
dp = [0]*(n+1)
dp[0] = 1
for a in gr:
dp1 = [0]*(n+1)
for i in range(n):
dp1[i+1] = (dp1[i+1]+dp[i])%MOD
if a > 1 and i+a <= n:
dp1[i+a] = (dp1[i+a]+dp[i])%MOD
dp = dp1
ans = 0
k = n
for i in range(1, n+1):
ans = (ans+dp[i]*k)%MOD
k = k*(n-i)%MOD
return ans
import sys
input = lambda: sys.stdin.readline().rstrip()
print(solve())
|
from collections import deque
def solve():
n = int(input())
MOD = 998244353
arr = [list(map(int, input().split())) for i in range(n)]
dis = [[10**9]*n for i in range(n)]
for i in range(n):
for j in range(n):
if i != j:
dis[i][j] = abs(arr[i][0]-arr[j][0]) + abs(arr[i][1]-arr[j][1])
g = [[] for i in range(n)]
adj = [[0]*n for i in range(n)]
for i in range(n):
m = min(dis[i])
for j in range(n):
if dis[i][j] == m:
g[i].append(j)
adj[i][j] = 1
gr = []
for i in range(n):
cur = []
queue = deque()
queue.append(i)
v = [0] * n
v[i] = 1
while queue:
x = queue.popleft()
cur.append(x)
for y in g[x]:
if v[y] == 0:
v[y] = 1
queue.append(y)
ok = 1
for x in cur:
for y in cur:
if x != y and adj[x][y] == 0:
{{completion}}
if ok:
if min(cur) == i:
gr.append(len(cur))
else:
gr.append(1)
dp = [0]*(n+1)
dp[0] = 1
for a in gr:
dp1 = [0]*(n+1)
for i in range(n):
dp1[i+1] = (dp1[i+1]+dp[i])%MOD
if a > 1 and i+a <= n:
dp1[i+a] = (dp1[i+a]+dp[i])%MOD
dp = dp1
ans = 0
k = n
for i in range(1, n+1):
ans = (ans+dp[i]*k)%MOD
k = k*(n-i)%MOD
return ans
import sys
input = lambda: sys.stdin.readline().rstrip()
print(solve())
|
ok = 0
break
|
[{"input": "3\n1 0\n3 0\n2 1", "output": ["9"]}, {"input": "5\n1 2\n2 4\n3 4\n4 4\n1 3", "output": ["240"]}, {"input": "4\n1 0\n3 0\n2 1\n2 0", "output": ["24"]}]
|
block_completion_000542
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: You are given $$$n$$$ points on the plane, the coordinates of the $$$i$$$-th point are $$$(x_i, y_i)$$$. No two points have the same coordinates.The distance between points $$$i$$$ and $$$j$$$ is defined as $$$d(i,j) = |x_i - x_j| + |y_i - y_j|$$$.For each point, you have to choose a color, represented by an integer from $$$1$$$ to $$$n$$$. For every ordered triple of different points $$$(a,b,c)$$$, the following constraints should be met: if $$$a$$$, $$$b$$$ and $$$c$$$ have the same color, then $$$d(a,b) = d(a,c) = d(b,c)$$$; if $$$a$$$ and $$$b$$$ have the same color, and the color of $$$c$$$ is different from the color of $$$a$$$, then $$$d(a,b) < d(a,c)$$$ and $$$d(a,b) < d(b,c)$$$. Calculate the number of different ways to choose the colors that meet these constraints.
Input Specification: The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the number of points. Then $$$n$$$ lines follow. The $$$i$$$-th of them contains two integers $$$x_i$$$ and $$$y_i$$$ ($$$0 \le x_i, y_i \le 10^8$$$). No two points have the same coordinates (i. e. if $$$i \ne j$$$, then either $$$x_i \ne x_j$$$ or $$$y_i \ne y_j$$$).
Output Specification: Print one integer — the number of ways to choose the colors for the points. Since it can be large, print it modulo $$$998244353$$$.
Notes: NoteIn the first test, the following ways to choose the colors are suitable: $$$[1, 1, 1]$$$; $$$[2, 2, 2]$$$; $$$[3, 3, 3]$$$; $$$[1, 2, 3]$$$; $$$[1, 3, 2]$$$; $$$[2, 1, 3]$$$; $$$[2, 3, 1]$$$; $$$[3, 1, 2]$$$; $$$[3, 2, 1]$$$.
Code:
from math import perm, comb
import sys
input = sys.stdin.readline
M = 998244353
n = int(input())
x, y = [0]*n, [0]*n
for i in range(n):
x[i], y[i] = map(int, input().split())
# print(x, y)
dist = [[] for _ in range(n)]
for i in range(n):
for j in range(n):
dist[i].append(abs(x[i] - x[j]) + abs(y[i] - y[j]))
# print(dist)
mindist, nbr = [M] * n, [[] for _ in range(n)]
for i in range(n):
for j in range(n):
if i == j:
continue
if dist[i][j] < mindist[i]:
mindist[i] = dist[i][j]
nbr[i] = [j]
elif dist[i][j] == mindist[i]:
nbr[i].append(j)
# print(mindist, nbr)
grp = [0] * n
for i in range(n):
if grp[i] > 0:
continue
if len(nbr[i]) > 3:
grp[i] = 1
continue
checknbr = [False] * n
checknbr[i] = True
for j in nbr[i]:
checknbr[j] = True
check = False
for j in nbr[i]:
if len(nbr[j]) != len(nbr[i]):
check = True
break
for c in nbr[j]:
if not checknbr[c]:
# TODO: Your code here
if check:
grp[i] = 1
else:
grp[i] = len(nbr[i]) + 1
for j in nbr[i]:
grp[j] = grp[i]
# print(grp)
cnt = [0] * 5
for i in grp:
cnt[i] += 1
cnt[2] //= 2
cnt[3] //= 3
cnt[4] //= 4
# print(cnt)
ans = 0
for i in range(cnt[2] + 1):
for j in range(cnt[3] + 1):
for k in range(cnt[4] + 1):
z = (comb(cnt[2], i) * comb(cnt[3], j) * comb(cnt[4], k)) % M
z *= perm(n, n - i - 2*j - 3*k)
z %= M
ans = (ans + z) % M
print(ans)
|
from math import perm, comb
import sys
input = sys.stdin.readline
M = 998244353
n = int(input())
x, y = [0]*n, [0]*n
for i in range(n):
x[i], y[i] = map(int, input().split())
# print(x, y)
dist = [[] for _ in range(n)]
for i in range(n):
for j in range(n):
dist[i].append(abs(x[i] - x[j]) + abs(y[i] - y[j]))
# print(dist)
mindist, nbr = [M] * n, [[] for _ in range(n)]
for i in range(n):
for j in range(n):
if i == j:
continue
if dist[i][j] < mindist[i]:
mindist[i] = dist[i][j]
nbr[i] = [j]
elif dist[i][j] == mindist[i]:
nbr[i].append(j)
# print(mindist, nbr)
grp = [0] * n
for i in range(n):
if grp[i] > 0:
continue
if len(nbr[i]) > 3:
grp[i] = 1
continue
checknbr = [False] * n
checknbr[i] = True
for j in nbr[i]:
checknbr[j] = True
check = False
for j in nbr[i]:
if len(nbr[j]) != len(nbr[i]):
check = True
break
for c in nbr[j]:
if not checknbr[c]:
{{completion}}
if check:
grp[i] = 1
else:
grp[i] = len(nbr[i]) + 1
for j in nbr[i]:
grp[j] = grp[i]
# print(grp)
cnt = [0] * 5
for i in grp:
cnt[i] += 1
cnt[2] //= 2
cnt[3] //= 3
cnt[4] //= 4
# print(cnt)
ans = 0
for i in range(cnt[2] + 1):
for j in range(cnt[3] + 1):
for k in range(cnt[4] + 1):
z = (comb(cnt[2], i) * comb(cnt[3], j) * comb(cnt[4], k)) % M
z *= perm(n, n - i - 2*j - 3*k)
z %= M
ans = (ans + z) % M
print(ans)
|
check = True
break
|
[{"input": "3\n1 0\n3 0\n2 1", "output": ["9"]}, {"input": "5\n1 2\n2 4\n3 4\n4 4\n1 3", "output": ["240"]}, {"input": "4\n1 0\n3 0\n2 1\n2 0", "output": ["24"]}]
|
block_completion_000543
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: You are given $$$n$$$ points on the plane, the coordinates of the $$$i$$$-th point are $$$(x_i, y_i)$$$. No two points have the same coordinates.The distance between points $$$i$$$ and $$$j$$$ is defined as $$$d(i,j) = |x_i - x_j| + |y_i - y_j|$$$.For each point, you have to choose a color, represented by an integer from $$$1$$$ to $$$n$$$. For every ordered triple of different points $$$(a,b,c)$$$, the following constraints should be met: if $$$a$$$, $$$b$$$ and $$$c$$$ have the same color, then $$$d(a,b) = d(a,c) = d(b,c)$$$; if $$$a$$$ and $$$b$$$ have the same color, and the color of $$$c$$$ is different from the color of $$$a$$$, then $$$d(a,b) < d(a,c)$$$ and $$$d(a,b) < d(b,c)$$$. Calculate the number of different ways to choose the colors that meet these constraints.
Input Specification: The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the number of points. Then $$$n$$$ lines follow. The $$$i$$$-th of them contains two integers $$$x_i$$$ and $$$y_i$$$ ($$$0 \le x_i, y_i \le 10^8$$$). No two points have the same coordinates (i. e. if $$$i \ne j$$$, then either $$$x_i \ne x_j$$$ or $$$y_i \ne y_j$$$).
Output Specification: Print one integer — the number of ways to choose the colors for the points. Since it can be large, print it modulo $$$998244353$$$.
Notes: NoteIn the first test, the following ways to choose the colors are suitable: $$$[1, 1, 1]$$$; $$$[2, 2, 2]$$$; $$$[3, 3, 3]$$$; $$$[1, 2, 3]$$$; $$$[1, 3, 2]$$$; $$$[2, 1, 3]$$$; $$$[2, 3, 1]$$$; $$$[3, 1, 2]$$$; $$$[3, 2, 1]$$$.
Code:
from math import perm, comb
import sys
input = sys.stdin.readline
M = 998244353
n = int(input())
x, y = [0]*n, [0]*n
for i in range(n):
x[i], y[i] = map(int, input().split())
# print(x, y)
dist = [[] for _ in range(n)]
for i in range(n):
for j in range(n):
dist[i].append(abs(x[i] - x[j]) + abs(y[i] - y[j]))
# print(dist)
mindist, nbr = [M] * n, [[] for _ in range(n)]
for i in range(n):
for j in range(n):
if i == j:
continue
if dist[i][j] < mindist[i]:
mindist[i] = dist[i][j]
nbr[i] = [j]
elif dist[i][j] == mindist[i]:
# TODO: Your code here
# print(mindist, nbr)
grp = [0] * n
for i in range(n):
if grp[i] > 0:
continue
if len(nbr[i]) > 3:
grp[i] = 1
continue
checknbr = [False] * n
checknbr[i] = True
for j in nbr[i]:
checknbr[j] = True
check = False
for j in nbr[i]:
if len(nbr[j]) != len(nbr[i]):
check = True
break
for c in nbr[j]:
if not checknbr[c]:
check = True
break
if check:
grp[i] = 1
else:
grp[i] = len(nbr[i]) + 1
for j in nbr[i]:
grp[j] = grp[i]
# print(grp)
cnt = [0] * 5
for i in grp:
cnt[i] += 1
cnt[2] //= 2
cnt[3] //= 3
cnt[4] //= 4
# print(cnt)
ans = 0
for i in range(cnt[2] + 1):
for j in range(cnt[3] + 1):
for k in range(cnt[4] + 1):
z = (comb(cnt[2], i) * comb(cnt[3], j) * comb(cnt[4], k)) % M
z *= perm(n, n - i - 2*j - 3*k)
z %= M
ans = (ans + z) % M
print(ans)
|
from math import perm, comb
import sys
input = sys.stdin.readline
M = 998244353
n = int(input())
x, y = [0]*n, [0]*n
for i in range(n):
x[i], y[i] = map(int, input().split())
# print(x, y)
dist = [[] for _ in range(n)]
for i in range(n):
for j in range(n):
dist[i].append(abs(x[i] - x[j]) + abs(y[i] - y[j]))
# print(dist)
mindist, nbr = [M] * n, [[] for _ in range(n)]
for i in range(n):
for j in range(n):
if i == j:
continue
if dist[i][j] < mindist[i]:
mindist[i] = dist[i][j]
nbr[i] = [j]
elif dist[i][j] == mindist[i]:
{{completion}}
# print(mindist, nbr)
grp = [0] * n
for i in range(n):
if grp[i] > 0:
continue
if len(nbr[i]) > 3:
grp[i] = 1
continue
checknbr = [False] * n
checknbr[i] = True
for j in nbr[i]:
checknbr[j] = True
check = False
for j in nbr[i]:
if len(nbr[j]) != len(nbr[i]):
check = True
break
for c in nbr[j]:
if not checknbr[c]:
check = True
break
if check:
grp[i] = 1
else:
grp[i] = len(nbr[i]) + 1
for j in nbr[i]:
grp[j] = grp[i]
# print(grp)
cnt = [0] * 5
for i in grp:
cnt[i] += 1
cnt[2] //= 2
cnt[3] //= 3
cnt[4] //= 4
# print(cnt)
ans = 0
for i in range(cnt[2] + 1):
for j in range(cnt[3] + 1):
for k in range(cnt[4] + 1):
z = (comb(cnt[2], i) * comb(cnt[3], j) * comb(cnt[4], k)) % M
z *= perm(n, n - i - 2*j - 3*k)
z %= M
ans = (ans + z) % M
print(ans)
|
nbr[i].append(j)
|
[{"input": "3\n1 0\n3 0\n2 1", "output": ["9"]}, {"input": "5\n1 2\n2 4\n3 4\n4 4\n1 3", "output": ["240"]}, {"input": "4\n1 0\n3 0\n2 1\n2 0", "output": ["24"]}]
|
block_completion_000544
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: You are given $$$n$$$ points on the plane, the coordinates of the $$$i$$$-th point are $$$(x_i, y_i)$$$. No two points have the same coordinates.The distance between points $$$i$$$ and $$$j$$$ is defined as $$$d(i,j) = |x_i - x_j| + |y_i - y_j|$$$.For each point, you have to choose a color, represented by an integer from $$$1$$$ to $$$n$$$. For every ordered triple of different points $$$(a,b,c)$$$, the following constraints should be met: if $$$a$$$, $$$b$$$ and $$$c$$$ have the same color, then $$$d(a,b) = d(a,c) = d(b,c)$$$; if $$$a$$$ and $$$b$$$ have the same color, and the color of $$$c$$$ is different from the color of $$$a$$$, then $$$d(a,b) < d(a,c)$$$ and $$$d(a,b) < d(b,c)$$$. Calculate the number of different ways to choose the colors that meet these constraints.
Input Specification: The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the number of points. Then $$$n$$$ lines follow. The $$$i$$$-th of them contains two integers $$$x_i$$$ and $$$y_i$$$ ($$$0 \le x_i, y_i \le 10^8$$$). No two points have the same coordinates (i. e. if $$$i \ne j$$$, then either $$$x_i \ne x_j$$$ or $$$y_i \ne y_j$$$).
Output Specification: Print one integer — the number of ways to choose the colors for the points. Since it can be large, print it modulo $$$998244353$$$.
Notes: NoteIn the first test, the following ways to choose the colors are suitable: $$$[1, 1, 1]$$$; $$$[2, 2, 2]$$$; $$$[3, 3, 3]$$$; $$$[1, 2, 3]$$$; $$$[1, 3, 2]$$$; $$$[2, 1, 3]$$$; $$$[2, 3, 1]$$$; $$$[3, 1, 2]$$$; $$$[3, 2, 1]$$$.
Code:
input = __import__('sys').stdin.readline
MOD = 998244353
fact = [1]
invfact = [1]
for i in range(1, 101):
fact.append(fact[-1] * i % MOD)
invfact.append(pow(fact[-1], MOD-2, MOD))
def C(n, k):
if k < 0 or k > n:
return 0
return fact[n] * invfact[k] % MOD * invfact[n-k] % MOD
def P(n, k):
if k < 0 or k > n:
return 0
return fact[n] * invfact[n-k] % MOD
n = int(input())
coords = []
for _ in range(n):
x, y = map(int, input().split())
coords.append((x, y))
min_dist = [10**9] * n
dist = [[-1] * n for _ in range(n)]
for u in range(n):
for v in range(n):
dist[u][v] = abs(coords[u][0] - coords[v][0]) + abs(coords[u][1] - coords[v][1])
if u != v:
min_dist[u] = min(min_dist[u], dist[u][v])
cnt = [0, 0, 0, 0, 0]
vis = [False]*n
for u in sorted(range(n), key=lambda x: min_dist[x]):
if vis[u]:
continue
vis[u] = True
seen = [False]*n
seen[u] = True
ptr = 0
found = [u]
while ptr < len(found):
v = found[ptr]
ptr += 1
for w in range(n):
if not seen[w] and dist[v][w] == min_dist[v]:
# TODO: Your code here
ok = all(dist[found[i]][found[j]] == min_dist[u] for i in range(len(found)) for j in range(i+1, len(found)))
if len(found) == 1 or not ok:
cnt[1] += 1
else:
# print('found', found, ok)
cnt[len(found)] += 1
for u in found:
vis[u] = True
# print('cnt', cnt[1:])
ans = 0
for two in range(cnt[2] + 1):
for three in range(cnt[3] + 1):
for four in range(cnt[4] + 1):
ans += P(n, n - two - 2*three - 3*four) * C(cnt[2], two) % MOD \
* C(cnt[3], three) % MOD \
* C(cnt[4], four) % MOD
if ans >= MOD:
ans -= MOD
# print(f'add P({n},{n - two - 2*three - 3*four})*C({cnt[2]},{two})*C({cnt[3]},{three})*C({cnt[4]},{four}) {ans}')
print(ans)
|
input = __import__('sys').stdin.readline
MOD = 998244353
fact = [1]
invfact = [1]
for i in range(1, 101):
fact.append(fact[-1] * i % MOD)
invfact.append(pow(fact[-1], MOD-2, MOD))
def C(n, k):
if k < 0 or k > n:
return 0
return fact[n] * invfact[k] % MOD * invfact[n-k] % MOD
def P(n, k):
if k < 0 or k > n:
return 0
return fact[n] * invfact[n-k] % MOD
n = int(input())
coords = []
for _ in range(n):
x, y = map(int, input().split())
coords.append((x, y))
min_dist = [10**9] * n
dist = [[-1] * n for _ in range(n)]
for u in range(n):
for v in range(n):
dist[u][v] = abs(coords[u][0] - coords[v][0]) + abs(coords[u][1] - coords[v][1])
if u != v:
min_dist[u] = min(min_dist[u], dist[u][v])
cnt = [0, 0, 0, 0, 0]
vis = [False]*n
for u in sorted(range(n), key=lambda x: min_dist[x]):
if vis[u]:
continue
vis[u] = True
seen = [False]*n
seen[u] = True
ptr = 0
found = [u]
while ptr < len(found):
v = found[ptr]
ptr += 1
for w in range(n):
if not seen[w] and dist[v][w] == min_dist[v]:
{{completion}}
ok = all(dist[found[i]][found[j]] == min_dist[u] for i in range(len(found)) for j in range(i+1, len(found)))
if len(found) == 1 or not ok:
cnt[1] += 1
else:
# print('found', found, ok)
cnt[len(found)] += 1
for u in found:
vis[u] = True
# print('cnt', cnt[1:])
ans = 0
for two in range(cnt[2] + 1):
for three in range(cnt[3] + 1):
for four in range(cnt[4] + 1):
ans += P(n, n - two - 2*three - 3*four) * C(cnt[2], two) % MOD \
* C(cnt[3], three) % MOD \
* C(cnt[4], four) % MOD
if ans >= MOD:
ans -= MOD
# print(f'add P({n},{n - two - 2*three - 3*four})*C({cnt[2]},{two})*C({cnt[3]},{three})*C({cnt[4]},{four}) {ans}')
print(ans)
|
seen[w] = True
found.append(w)
|
[{"input": "3\n1 0\n3 0\n2 1", "output": ["9"]}, {"input": "5\n1 2\n2 4\n3 4\n4 4\n1 3", "output": ["240"]}, {"input": "4\n1 0\n3 0\n2 1\n2 0", "output": ["24"]}]
|
block_completion_000545
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: You are given $$$n$$$ points on the plane, the coordinates of the $$$i$$$-th point are $$$(x_i, y_i)$$$. No two points have the same coordinates.The distance between points $$$i$$$ and $$$j$$$ is defined as $$$d(i,j) = |x_i - x_j| + |y_i - y_j|$$$.For each point, you have to choose a color, represented by an integer from $$$1$$$ to $$$n$$$. For every ordered triple of different points $$$(a,b,c)$$$, the following constraints should be met: if $$$a$$$, $$$b$$$ and $$$c$$$ have the same color, then $$$d(a,b) = d(a,c) = d(b,c)$$$; if $$$a$$$ and $$$b$$$ have the same color, and the color of $$$c$$$ is different from the color of $$$a$$$, then $$$d(a,b) < d(a,c)$$$ and $$$d(a,b) < d(b,c)$$$. Calculate the number of different ways to choose the colors that meet these constraints.
Input Specification: The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the number of points. Then $$$n$$$ lines follow. The $$$i$$$-th of them contains two integers $$$x_i$$$ and $$$y_i$$$ ($$$0 \le x_i, y_i \le 10^8$$$). No two points have the same coordinates (i. e. if $$$i \ne j$$$, then either $$$x_i \ne x_j$$$ or $$$y_i \ne y_j$$$).
Output Specification: Print one integer — the number of ways to choose the colors for the points. Since it can be large, print it modulo $$$998244353$$$.
Notes: NoteIn the first test, the following ways to choose the colors are suitable: $$$[1, 1, 1]$$$; $$$[2, 2, 2]$$$; $$$[3, 3, 3]$$$; $$$[1, 2, 3]$$$; $$$[1, 3, 2]$$$; $$$[2, 1, 3]$$$; $$$[2, 3, 1]$$$; $$$[3, 1, 2]$$$; $$$[3, 2, 1]$$$.
Code:
input = __import__('sys').stdin.readline
MOD = 998244353
fact = [1]
invfact = [1]
for i in range(1, 101):
fact.append(fact[-1] * i % MOD)
invfact.append(pow(fact[-1], MOD-2, MOD))
def C(n, k):
if k < 0 or k > n:
return 0
return fact[n] * invfact[k] % MOD * invfact[n-k] % MOD
def P(n, k):
if k < 0 or k > n:
return 0
return fact[n] * invfact[n-k] % MOD
n = int(input())
coords = []
for _ in range(n):
x, y = map(int, input().split())
coords.append((x, y))
min_dist = [10**9] * n
dist = [[-1] * n for _ in range(n)]
for u in range(n):
for v in range(n):
dist[u][v] = abs(coords[u][0] - coords[v][0]) + abs(coords[u][1] - coords[v][1])
if u != v:
min_dist[u] = min(min_dist[u], dist[u][v])
cnt = [0, 0, 0, 0, 0]
vis = [False]*n
for u in sorted(range(n), key=lambda x: min_dist[x]):
if vis[u]:
continue
vis[u] = True
seen = [False]*n
seen[u] = True
ptr = 0
found = [u]
while ptr < len(found):
v = found[ptr]
ptr += 1
for w in range(n):
if not seen[w] and dist[v][w] == min_dist[v]:
seen[w] = True
found.append(w)
ok = all(dist[found[i]][found[j]] == min_dist[u] for i in range(len(found)) for j in range(i+1, len(found)))
if len(found) == 1 or not ok:
cnt[1] += 1
else:
# print('found', found, ok)
cnt[len(found)] += 1
for u in found:
vis[u] = True
# print('cnt', cnt[1:])
ans = 0
for two in range(cnt[2] + 1):
for three in range(cnt[3] + 1):
for four in range(cnt[4] + 1):
ans += P(n, n - two - 2*three - 3*four) * C(cnt[2], two) % MOD \
* C(cnt[3], three) % MOD \
* C(cnt[4], four) % MOD
if ans >= MOD:
# TODO: Your code here
# print(f'add P({n},{n - two - 2*three - 3*four})*C({cnt[2]},{two})*C({cnt[3]},{three})*C({cnt[4]},{four}) {ans}')
print(ans)
|
input = __import__('sys').stdin.readline
MOD = 998244353
fact = [1]
invfact = [1]
for i in range(1, 101):
fact.append(fact[-1] * i % MOD)
invfact.append(pow(fact[-1], MOD-2, MOD))
def C(n, k):
if k < 0 or k > n:
return 0
return fact[n] * invfact[k] % MOD * invfact[n-k] % MOD
def P(n, k):
if k < 0 or k > n:
return 0
return fact[n] * invfact[n-k] % MOD
n = int(input())
coords = []
for _ in range(n):
x, y = map(int, input().split())
coords.append((x, y))
min_dist = [10**9] * n
dist = [[-1] * n for _ in range(n)]
for u in range(n):
for v in range(n):
dist[u][v] = abs(coords[u][0] - coords[v][0]) + abs(coords[u][1] - coords[v][1])
if u != v:
min_dist[u] = min(min_dist[u], dist[u][v])
cnt = [0, 0, 0, 0, 0]
vis = [False]*n
for u in sorted(range(n), key=lambda x: min_dist[x]):
if vis[u]:
continue
vis[u] = True
seen = [False]*n
seen[u] = True
ptr = 0
found = [u]
while ptr < len(found):
v = found[ptr]
ptr += 1
for w in range(n):
if not seen[w] and dist[v][w] == min_dist[v]:
seen[w] = True
found.append(w)
ok = all(dist[found[i]][found[j]] == min_dist[u] for i in range(len(found)) for j in range(i+1, len(found)))
if len(found) == 1 or not ok:
cnt[1] += 1
else:
# print('found', found, ok)
cnt[len(found)] += 1
for u in found:
vis[u] = True
# print('cnt', cnt[1:])
ans = 0
for two in range(cnt[2] + 1):
for three in range(cnt[3] + 1):
for four in range(cnt[4] + 1):
ans += P(n, n - two - 2*three - 3*four) * C(cnt[2], two) % MOD \
* C(cnt[3], three) % MOD \
* C(cnt[4], four) % MOD
if ans >= MOD:
{{completion}}
# print(f'add P({n},{n - two - 2*three - 3*four})*C({cnt[2]},{two})*C({cnt[3]},{three})*C({cnt[4]},{four}) {ans}')
print(ans)
|
ans -= MOD
|
[{"input": "3\n1 0\n3 0\n2 1", "output": ["9"]}, {"input": "5\n1 2\n2 4\n3 4\n4 4\n1 3", "output": ["240"]}, {"input": "4\n1 0\n3 0\n2 1\n2 0", "output": ["24"]}]
|
block_completion_000546
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: You are given $$$n$$$ points on the plane, the coordinates of the $$$i$$$-th point are $$$(x_i, y_i)$$$. No two points have the same coordinates.The distance between points $$$i$$$ and $$$j$$$ is defined as $$$d(i,j) = |x_i - x_j| + |y_i - y_j|$$$.For each point, you have to choose a color, represented by an integer from $$$1$$$ to $$$n$$$. For every ordered triple of different points $$$(a,b,c)$$$, the following constraints should be met: if $$$a$$$, $$$b$$$ and $$$c$$$ have the same color, then $$$d(a,b) = d(a,c) = d(b,c)$$$; if $$$a$$$ and $$$b$$$ have the same color, and the color of $$$c$$$ is different from the color of $$$a$$$, then $$$d(a,b) < d(a,c)$$$ and $$$d(a,b) < d(b,c)$$$. Calculate the number of different ways to choose the colors that meet these constraints.
Input Specification: The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the number of points. Then $$$n$$$ lines follow. The $$$i$$$-th of them contains two integers $$$x_i$$$ and $$$y_i$$$ ($$$0 \le x_i, y_i \le 10^8$$$). No two points have the same coordinates (i. e. if $$$i \ne j$$$, then either $$$x_i \ne x_j$$$ or $$$y_i \ne y_j$$$).
Output Specification: Print one integer — the number of ways to choose the colors for the points. Since it can be large, print it modulo $$$998244353$$$.
Notes: NoteIn the first test, the following ways to choose the colors are suitable: $$$[1, 1, 1]$$$; $$$[2, 2, 2]$$$; $$$[3, 3, 3]$$$; $$$[1, 2, 3]$$$; $$$[1, 3, 2]$$$; $$$[2, 1, 3]$$$; $$$[2, 3, 1]$$$; $$$[3, 1, 2]$$$; $$$[3, 2, 1]$$$.
Code:
#############################
#############
cnb_max=10**5
mod=998244353
#############
kai=[1]*(cnb_max+1)
rkai=[1]*(cnb_max+1)
for i in range(cnb_max):
kai[i+1]=kai[i]*(i+1)%mod
rkai[cnb_max]=pow(kai[cnb_max],mod-2,mod)
for i in range(cnb_max):
rkai[cnb_max-1-i]=rkai[cnb_max-i]*(cnb_max-i)%mod
def cnb(x,y):
if y>x:
return 0
if x<0:return 0
if y<0:return 0
return (kai[x]*rkai[y]%mod)*rkai[x-y]%mod
def inv(n):
return kai[n-1]*rkai[n]%mod
##################################
n=int(input())
x=[]
y=[]
for i in range(n):
a,b=map(int,input().split())
x.append(a)
y.append(b)
ok=[[0]*n for i in range(n)]
tto=[0]*n
def dist(i,j):
return abs(x[i]-x[j])+abs(y[i]-y[j])
for i in range(n):
mi=10**18
for j in range(n):
if i==j:continue
mi=min(mi,dist(i,j))
for j in range(n):
if i==j:continue
if mi==dist(i,j):
ok[i][j]=1
tto[i]+=1
s=[]
for a in range(n):
for b in range(a+1,n):
for c in range(b+1,n):
for d in range(c+1,n):
nod=[a,b,c,d]
flag=1
for i in nod:
for j in nod:
if i==j:# TODO: Your code here
flag&=ok[i][j]
if tto[i]!=3:flag=0
if flag:s.append(4)
for a in range(n):
for b in range(a+1,n):
for c in range(b+1,n):
nod=[a,b,c]
flag=1
for i in nod:
for j in nod:
if i==j:continue
flag&=ok[i][j]
if tto[i]!=2:flag=0
if flag:s.append(3)
for a in range(n):
for b in range(a+1,n):
nod=[a,b]
flag=1
for i in nod:
for j in nod:
if i==j:continue
flag&=ok[i][j]
if tto[i]!=1:flag=0
if flag:s.append(2)
dp=[0]*(n+1)
dp[n-sum(s)]=1
for cnt in s:
newdp=[0]*(n+1)
for i in range(n+1):
dp[i]%=mod
if i+cnt<=n:newdp[i+cnt]+=dp[i]
if i+1<=n:newdp[i+1]+=dp[i]
dp=newdp[:]
ans=0
for k in range(n+1):
ans+=dp[k]*cnb(n,k)*kai[k]
ans%=mod
print(ans)
|
#############################
#############
cnb_max=10**5
mod=998244353
#############
kai=[1]*(cnb_max+1)
rkai=[1]*(cnb_max+1)
for i in range(cnb_max):
kai[i+1]=kai[i]*(i+1)%mod
rkai[cnb_max]=pow(kai[cnb_max],mod-2,mod)
for i in range(cnb_max):
rkai[cnb_max-1-i]=rkai[cnb_max-i]*(cnb_max-i)%mod
def cnb(x,y):
if y>x:
return 0
if x<0:return 0
if y<0:return 0
return (kai[x]*rkai[y]%mod)*rkai[x-y]%mod
def inv(n):
return kai[n-1]*rkai[n]%mod
##################################
n=int(input())
x=[]
y=[]
for i in range(n):
a,b=map(int,input().split())
x.append(a)
y.append(b)
ok=[[0]*n for i in range(n)]
tto=[0]*n
def dist(i,j):
return abs(x[i]-x[j])+abs(y[i]-y[j])
for i in range(n):
mi=10**18
for j in range(n):
if i==j:continue
mi=min(mi,dist(i,j))
for j in range(n):
if i==j:continue
if mi==dist(i,j):
ok[i][j]=1
tto[i]+=1
s=[]
for a in range(n):
for b in range(a+1,n):
for c in range(b+1,n):
for d in range(c+1,n):
nod=[a,b,c,d]
flag=1
for i in nod:
for j in nod:
if i==j:{{completion}}
flag&=ok[i][j]
if tto[i]!=3:flag=0
if flag:s.append(4)
for a in range(n):
for b in range(a+1,n):
for c in range(b+1,n):
nod=[a,b,c]
flag=1
for i in nod:
for j in nod:
if i==j:continue
flag&=ok[i][j]
if tto[i]!=2:flag=0
if flag:s.append(3)
for a in range(n):
for b in range(a+1,n):
nod=[a,b]
flag=1
for i in nod:
for j in nod:
if i==j:continue
flag&=ok[i][j]
if tto[i]!=1:flag=0
if flag:s.append(2)
dp=[0]*(n+1)
dp[n-sum(s)]=1
for cnt in s:
newdp=[0]*(n+1)
for i in range(n+1):
dp[i]%=mod
if i+cnt<=n:newdp[i+cnt]+=dp[i]
if i+1<=n:newdp[i+1]+=dp[i]
dp=newdp[:]
ans=0
for k in range(n+1):
ans+=dp[k]*cnb(n,k)*kai[k]
ans%=mod
print(ans)
|
continue
|
[{"input": "3\n1 0\n3 0\n2 1", "output": ["9"]}, {"input": "5\n1 2\n2 4\n3 4\n4 4\n1 3", "output": ["240"]}, {"input": "4\n1 0\n3 0\n2 1\n2 0", "output": ["24"]}]
|
block_completion_000547
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: You are given $$$n$$$ points on the plane, the coordinates of the $$$i$$$-th point are $$$(x_i, y_i)$$$. No two points have the same coordinates.The distance between points $$$i$$$ and $$$j$$$ is defined as $$$d(i,j) = |x_i - x_j| + |y_i - y_j|$$$.For each point, you have to choose a color, represented by an integer from $$$1$$$ to $$$n$$$. For every ordered triple of different points $$$(a,b,c)$$$, the following constraints should be met: if $$$a$$$, $$$b$$$ and $$$c$$$ have the same color, then $$$d(a,b) = d(a,c) = d(b,c)$$$; if $$$a$$$ and $$$b$$$ have the same color, and the color of $$$c$$$ is different from the color of $$$a$$$, then $$$d(a,b) < d(a,c)$$$ and $$$d(a,b) < d(b,c)$$$. Calculate the number of different ways to choose the colors that meet these constraints.
Input Specification: The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the number of points. Then $$$n$$$ lines follow. The $$$i$$$-th of them contains two integers $$$x_i$$$ and $$$y_i$$$ ($$$0 \le x_i, y_i \le 10^8$$$). No two points have the same coordinates (i. e. if $$$i \ne j$$$, then either $$$x_i \ne x_j$$$ or $$$y_i \ne y_j$$$).
Output Specification: Print one integer — the number of ways to choose the colors for the points. Since it can be large, print it modulo $$$998244353$$$.
Notes: NoteIn the first test, the following ways to choose the colors are suitable: $$$[1, 1, 1]$$$; $$$[2, 2, 2]$$$; $$$[3, 3, 3]$$$; $$$[1, 2, 3]$$$; $$$[1, 3, 2]$$$; $$$[2, 1, 3]$$$; $$$[2, 3, 1]$$$; $$$[3, 1, 2]$$$; $$$[3, 2, 1]$$$.
Code:
#############################
#############
cnb_max=10**5
mod=998244353
#############
kai=[1]*(cnb_max+1)
rkai=[1]*(cnb_max+1)
for i in range(cnb_max):
kai[i+1]=kai[i]*(i+1)%mod
rkai[cnb_max]=pow(kai[cnb_max],mod-2,mod)
for i in range(cnb_max):
rkai[cnb_max-1-i]=rkai[cnb_max-i]*(cnb_max-i)%mod
def cnb(x,y):
if y>x:
return 0
if x<0:return 0
if y<0:return 0
return (kai[x]*rkai[y]%mod)*rkai[x-y]%mod
def inv(n):
return kai[n-1]*rkai[n]%mod
##################################
n=int(input())
x=[]
y=[]
for i in range(n):
a,b=map(int,input().split())
x.append(a)
y.append(b)
ok=[[0]*n for i in range(n)]
tto=[0]*n
def dist(i,j):
return abs(x[i]-x[j])+abs(y[i]-y[j])
for i in range(n):
mi=10**18
for j in range(n):
if i==j:continue
mi=min(mi,dist(i,j))
for j in range(n):
if i==j:continue
if mi==dist(i,j):
ok[i][j]=1
tto[i]+=1
s=[]
for a in range(n):
for b in range(a+1,n):
for c in range(b+1,n):
for d in range(c+1,n):
nod=[a,b,c,d]
flag=1
for i in nod:
for j in nod:
if i==j:continue
flag&=ok[i][j]
if tto[i]!=3:# TODO: Your code here
if flag:s.append(4)
for a in range(n):
for b in range(a+1,n):
for c in range(b+1,n):
nod=[a,b,c]
flag=1
for i in nod:
for j in nod:
if i==j:continue
flag&=ok[i][j]
if tto[i]!=2:flag=0
if flag:s.append(3)
for a in range(n):
for b in range(a+1,n):
nod=[a,b]
flag=1
for i in nod:
for j in nod:
if i==j:continue
flag&=ok[i][j]
if tto[i]!=1:flag=0
if flag:s.append(2)
dp=[0]*(n+1)
dp[n-sum(s)]=1
for cnt in s:
newdp=[0]*(n+1)
for i in range(n+1):
dp[i]%=mod
if i+cnt<=n:newdp[i+cnt]+=dp[i]
if i+1<=n:newdp[i+1]+=dp[i]
dp=newdp[:]
ans=0
for k in range(n+1):
ans+=dp[k]*cnb(n,k)*kai[k]
ans%=mod
print(ans)
|
#############################
#############
cnb_max=10**5
mod=998244353
#############
kai=[1]*(cnb_max+1)
rkai=[1]*(cnb_max+1)
for i in range(cnb_max):
kai[i+1]=kai[i]*(i+1)%mod
rkai[cnb_max]=pow(kai[cnb_max],mod-2,mod)
for i in range(cnb_max):
rkai[cnb_max-1-i]=rkai[cnb_max-i]*(cnb_max-i)%mod
def cnb(x,y):
if y>x:
return 0
if x<0:return 0
if y<0:return 0
return (kai[x]*rkai[y]%mod)*rkai[x-y]%mod
def inv(n):
return kai[n-1]*rkai[n]%mod
##################################
n=int(input())
x=[]
y=[]
for i in range(n):
a,b=map(int,input().split())
x.append(a)
y.append(b)
ok=[[0]*n for i in range(n)]
tto=[0]*n
def dist(i,j):
return abs(x[i]-x[j])+abs(y[i]-y[j])
for i in range(n):
mi=10**18
for j in range(n):
if i==j:continue
mi=min(mi,dist(i,j))
for j in range(n):
if i==j:continue
if mi==dist(i,j):
ok[i][j]=1
tto[i]+=1
s=[]
for a in range(n):
for b in range(a+1,n):
for c in range(b+1,n):
for d in range(c+1,n):
nod=[a,b,c,d]
flag=1
for i in nod:
for j in nod:
if i==j:continue
flag&=ok[i][j]
if tto[i]!=3:{{completion}}
if flag:s.append(4)
for a in range(n):
for b in range(a+1,n):
for c in range(b+1,n):
nod=[a,b,c]
flag=1
for i in nod:
for j in nod:
if i==j:continue
flag&=ok[i][j]
if tto[i]!=2:flag=0
if flag:s.append(3)
for a in range(n):
for b in range(a+1,n):
nod=[a,b]
flag=1
for i in nod:
for j in nod:
if i==j:continue
flag&=ok[i][j]
if tto[i]!=1:flag=0
if flag:s.append(2)
dp=[0]*(n+1)
dp[n-sum(s)]=1
for cnt in s:
newdp=[0]*(n+1)
for i in range(n+1):
dp[i]%=mod
if i+cnt<=n:newdp[i+cnt]+=dp[i]
if i+1<=n:newdp[i+1]+=dp[i]
dp=newdp[:]
ans=0
for k in range(n+1):
ans+=dp[k]*cnb(n,k)*kai[k]
ans%=mod
print(ans)
|
flag=0
|
[{"input": "3\n1 0\n3 0\n2 1", "output": ["9"]}, {"input": "5\n1 2\n2 4\n3 4\n4 4\n1 3", "output": ["240"]}, {"input": "4\n1 0\n3 0\n2 1\n2 0", "output": ["24"]}]
|
block_completion_000548
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: Codeforces separates its users into $$$4$$$ divisions by their rating: For Division 1: $$$1900 \leq \mathrm{rating}$$$ For Division 2: $$$1600 \leq \mathrm{rating} \leq 1899$$$ For Division 3: $$$1400 \leq \mathrm{rating} \leq 1599$$$ For Division 4: $$$\mathrm{rating} \leq 1399$$$ Given a $$$\mathrm{rating}$$$, print in which division the $$$\mathrm{rating}$$$ belongs.
Input Specification: The first line of the input contains an integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of testcases. The description of each test consists of one line containing one integer $$$\mathrm{rating}$$$ ($$$-5000 \leq \mathrm{rating} \leq 5000$$$).
Output Specification: For each test case, output a single line containing the correct division in the format "Division X", where $$$X$$$ is an integer between $$$1$$$ and $$$4$$$ representing the division for the corresponding rating.
Notes: NoteFor test cases $$$1-4$$$, the corresponding ratings are $$$-789$$$, $$$1299$$$, $$$1300$$$, $$$1399$$$, so all of them are in division $$$4$$$.For the fifth test case, the corresponding rating is $$$1400$$$, so it is in division $$$3$$$.For the sixth test case, the corresponding rating is $$$1679$$$, so it is in division $$$2$$$.For the seventh test case, the corresponding rating is $$$2300$$$, so it is in division $$$1$$$.
Code:
for _ in range(int(input())):
# TODO: Your code here
|
for _ in range(int(input())):
{{completion}}
|
r = int(input())
print('Division', 4 - sum(r >= p for p in (1400, 1600, 1900)))
|
[{"input": "7\n-789\n1299\n1300\n1399\n1400\n1679\n2300", "output": ["Division 4\nDivision 4\nDivision 4\nDivision 4\nDivision 3\nDivision 2\nDivision 1"]}]
|
block_completion_000724
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: Codeforces separates its users into $$$4$$$ divisions by their rating: For Division 1: $$$1900 \leq \mathrm{rating}$$$ For Division 2: $$$1600 \leq \mathrm{rating} \leq 1899$$$ For Division 3: $$$1400 \leq \mathrm{rating} \leq 1599$$$ For Division 4: $$$\mathrm{rating} \leq 1399$$$ Given a $$$\mathrm{rating}$$$, print in which division the $$$\mathrm{rating}$$$ belongs.
Input Specification: The first line of the input contains an integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of testcases. The description of each test consists of one line containing one integer $$$\mathrm{rating}$$$ ($$$-5000 \leq \mathrm{rating} \leq 5000$$$).
Output Specification: For each test case, output a single line containing the correct division in the format "Division X", where $$$X$$$ is an integer between $$$1$$$ and $$$4$$$ representing the division for the corresponding rating.
Notes: NoteFor test cases $$$1-4$$$, the corresponding ratings are $$$-789$$$, $$$1299$$$, $$$1300$$$, $$$1399$$$, so all of them are in division $$$4$$$.For the fifth test case, the corresponding rating is $$$1400$$$, so it is in division $$$3$$$.For the sixth test case, the corresponding rating is $$$1679$$$, so it is in division $$$2$$$.For the seventh test case, the corresponding rating is $$$2300$$$, so it is in division $$$1$$$.
Code:
for n in[*open(0)][1:]:# TODO: Your code here
|
for n in[*open(0)][1:]:{{completion}}
|
r=int(n)/100;print('Division',1+(r<19)+(r<16)+(r<14))
|
[{"input": "7\n-789\n1299\n1300\n1399\n1400\n1679\n2300", "output": ["Division 4\nDivision 4\nDivision 4\nDivision 4\nDivision 3\nDivision 2\nDivision 1"]}]
|
block_completion_000725
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: Codeforces separates its users into $$$4$$$ divisions by their rating: For Division 1: $$$1900 \leq \mathrm{rating}$$$ For Division 2: $$$1600 \leq \mathrm{rating} \leq 1899$$$ For Division 3: $$$1400 \leq \mathrm{rating} \leq 1599$$$ For Division 4: $$$\mathrm{rating} \leq 1399$$$ Given a $$$\mathrm{rating}$$$, print in which division the $$$\mathrm{rating}$$$ belongs.
Input Specification: The first line of the input contains an integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of testcases. The description of each test consists of one line containing one integer $$$\mathrm{rating}$$$ ($$$-5000 \leq \mathrm{rating} \leq 5000$$$).
Output Specification: For each test case, output a single line containing the correct division in the format "Division X", where $$$X$$$ is an integer between $$$1$$$ and $$$4$$$ representing the division for the corresponding rating.
Notes: NoteFor test cases $$$1-4$$$, the corresponding ratings are $$$-789$$$, $$$1299$$$, $$$1300$$$, $$$1399$$$, so all of them are in division $$$4$$$.For the fifth test case, the corresponding rating is $$$1400$$$, so it is in division $$$3$$$.For the sixth test case, the corresponding rating is $$$1679$$$, so it is in division $$$2$$$.For the seventh test case, the corresponding rating is $$$2300$$$, so it is in division $$$1$$$.
Code:
from bisect import bisect
b = [-5001, 1400, 1600, 1900]
for i in range(int(input())): # TODO: Your code here
|
from bisect import bisect
b = [-5001, 1400, 1600, 1900]
for i in range(int(input())): {{completion}}
|
print(f'Division {-bisect(b, int(input()))+5}')
|
[{"input": "7\n-789\n1299\n1300\n1399\n1400\n1679\n2300", "output": ["Division 4\nDivision 4\nDivision 4\nDivision 4\nDivision 3\nDivision 2\nDivision 1"]}]
|
block_completion_000726
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: Codeforces separates its users into $$$4$$$ divisions by their rating: For Division 1: $$$1900 \leq \mathrm{rating}$$$ For Division 2: $$$1600 \leq \mathrm{rating} \leq 1899$$$ For Division 3: $$$1400 \leq \mathrm{rating} \leq 1599$$$ For Division 4: $$$\mathrm{rating} \leq 1399$$$ Given a $$$\mathrm{rating}$$$, print in which division the $$$\mathrm{rating}$$$ belongs.
Input Specification: The first line of the input contains an integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of testcases. The description of each test consists of one line containing one integer $$$\mathrm{rating}$$$ ($$$-5000 \leq \mathrm{rating} \leq 5000$$$).
Output Specification: For each test case, output a single line containing the correct division in the format "Division X", where $$$X$$$ is an integer between $$$1$$$ and $$$4$$$ representing the division for the corresponding rating.
Notes: NoteFor test cases $$$1-4$$$, the corresponding ratings are $$$-789$$$, $$$1299$$$, $$$1300$$$, $$$1399$$$, so all of them are in division $$$4$$$.For the fifth test case, the corresponding rating is $$$1400$$$, so it is in division $$$3$$$.For the sixth test case, the corresponding rating is $$$1679$$$, so it is in division $$$2$$$.For the seventh test case, the corresponding rating is $$$2300$$$, so it is in division $$$1$$$.
Code:
n=int(input())
while n:
num=int(input())
if num>=1900:print("Division 1")
elif num>=1600:# TODO: Your code here
elif num>=1400:print("Division 3")
else:print("Division 4")
n-=1
|
n=int(input())
while n:
num=int(input())
if num>=1900:print("Division 1")
elif num>=1600:{{completion}}
elif num>=1400:print("Division 3")
else:print("Division 4")
n-=1
|
print("Division 2")
|
[{"input": "7\n-789\n1299\n1300\n1399\n1400\n1679\n2300", "output": ["Division 4\nDivision 4\nDivision 4\nDivision 4\nDivision 3\nDivision 2\nDivision 1"]}]
|
block_completion_000727
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: Codeforces separates its users into $$$4$$$ divisions by their rating: For Division 1: $$$1900 \leq \mathrm{rating}$$$ For Division 2: $$$1600 \leq \mathrm{rating} \leq 1899$$$ For Division 3: $$$1400 \leq \mathrm{rating} \leq 1599$$$ For Division 4: $$$\mathrm{rating} \leq 1399$$$ Given a $$$\mathrm{rating}$$$, print in which division the $$$\mathrm{rating}$$$ belongs.
Input Specification: The first line of the input contains an integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of testcases. The description of each test consists of one line containing one integer $$$\mathrm{rating}$$$ ($$$-5000 \leq \mathrm{rating} \leq 5000$$$).
Output Specification: For each test case, output a single line containing the correct division in the format "Division X", where $$$X$$$ is an integer between $$$1$$$ and $$$4$$$ representing the division for the corresponding rating.
Notes: NoteFor test cases $$$1-4$$$, the corresponding ratings are $$$-789$$$, $$$1299$$$, $$$1300$$$, $$$1399$$$, so all of them are in division $$$4$$$.For the fifth test case, the corresponding rating is $$$1400$$$, so it is in division $$$3$$$.For the sixth test case, the corresponding rating is $$$1679$$$, so it is in division $$$2$$$.For the seventh test case, the corresponding rating is $$$2300$$$, so it is in division $$$1$$$.
Code:
n=int(input())
while n:
num=int(input())
if num>=1900:print("Division 1")
elif num>=1600:print("Division 2")
elif num>=1400:# TODO: Your code here
else:print("Division 4")
n-=1
|
n=int(input())
while n:
num=int(input())
if num>=1900:print("Division 1")
elif num>=1600:print("Division 2")
elif num>=1400:{{completion}}
else:print("Division 4")
n-=1
|
print("Division 3")
|
[{"input": "7\n-789\n1299\n1300\n1399\n1400\n1679\n2300", "output": ["Division 4\nDivision 4\nDivision 4\nDivision 4\nDivision 3\nDivision 2\nDivision 1"]}]
|
block_completion_000728
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: Codeforces separates its users into $$$4$$$ divisions by their rating: For Division 1: $$$1900 \leq \mathrm{rating}$$$ For Division 2: $$$1600 \leq \mathrm{rating} \leq 1899$$$ For Division 3: $$$1400 \leq \mathrm{rating} \leq 1599$$$ For Division 4: $$$\mathrm{rating} \leq 1399$$$ Given a $$$\mathrm{rating}$$$, print in which division the $$$\mathrm{rating}$$$ belongs.
Input Specification: The first line of the input contains an integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of testcases. The description of each test consists of one line containing one integer $$$\mathrm{rating}$$$ ($$$-5000 \leq \mathrm{rating} \leq 5000$$$).
Output Specification: For each test case, output a single line containing the correct division in the format "Division X", where $$$X$$$ is an integer between $$$1$$$ and $$$4$$$ representing the division for the corresponding rating.
Notes: NoteFor test cases $$$1-4$$$, the corresponding ratings are $$$-789$$$, $$$1299$$$, $$$1300$$$, $$$1399$$$, so all of them are in division $$$4$$$.For the fifth test case, the corresponding rating is $$$1400$$$, so it is in division $$$3$$$.For the sixth test case, the corresponding rating is $$$1679$$$, so it is in division $$$2$$$.For the seventh test case, the corresponding rating is $$$2300$$$, so it is in division $$$1$$$.
Code:
def div(n):
# TODO: Your code here
for _ in range(int(input())):
print(f'Division {div(int(input()))}')
|
def div(n):
{{completion}}
for _ in range(int(input())):
print(f'Division {div(int(input()))}')
|
return 1 if n >= 1900 else 2 if n >= 1600 else 3 if n >= 1400 else 4
|
[{"input": "7\n-789\n1299\n1300\n1399\n1400\n1679\n2300", "output": ["Division 4\nDivision 4\nDivision 4\nDivision 4\nDivision 3\nDivision 2\nDivision 1"]}]
|
block_completion_000729
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: Codeforces separates its users into $$$4$$$ divisions by their rating: For Division 1: $$$1900 \leq \mathrm{rating}$$$ For Division 2: $$$1600 \leq \mathrm{rating} \leq 1899$$$ For Division 3: $$$1400 \leq \mathrm{rating} \leq 1599$$$ For Division 4: $$$\mathrm{rating} \leq 1399$$$ Given a $$$\mathrm{rating}$$$, print in which division the $$$\mathrm{rating}$$$ belongs.
Input Specification: The first line of the input contains an integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of testcases. The description of each test consists of one line containing one integer $$$\mathrm{rating}$$$ ($$$-5000 \leq \mathrm{rating} \leq 5000$$$).
Output Specification: For each test case, output a single line containing the correct division in the format "Division X", where $$$X$$$ is an integer between $$$1$$$ and $$$4$$$ representing the division for the corresponding rating.
Notes: NoteFor test cases $$$1-4$$$, the corresponding ratings are $$$-789$$$, $$$1299$$$, $$$1300$$$, $$$1399$$$, so all of them are in division $$$4$$$.For the fifth test case, the corresponding rating is $$$1400$$$, so it is in division $$$3$$$.For the sixth test case, the corresponding rating is $$$1679$$$, so it is in division $$$2$$$.For the seventh test case, the corresponding rating is $$$2300$$$, so it is in division $$$1$$$.
Code:
def div(n):
return 1 if n >= 1900 else 2 if n >= 1600 else 3 if n >= 1400 else 4
for _ in range(int(input())):
# TODO: Your code here
|
def div(n):
return 1 if n >= 1900 else 2 if n >= 1600 else 3 if n >= 1400 else 4
for _ in range(int(input())):
{{completion}}
|
print(f'Division {div(int(input()))}')
|
[{"input": "7\n-789\n1299\n1300\n1399\n1400\n1679\n2300", "output": ["Division 4\nDivision 4\nDivision 4\nDivision 4\nDivision 3\nDivision 2\nDivision 1"]}]
|
block_completion_000730
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: Codeforces separates its users into $$$4$$$ divisions by their rating: For Division 1: $$$1900 \leq \mathrm{rating}$$$ For Division 2: $$$1600 \leq \mathrm{rating} \leq 1899$$$ For Division 3: $$$1400 \leq \mathrm{rating} \leq 1599$$$ For Division 4: $$$\mathrm{rating} \leq 1399$$$ Given a $$$\mathrm{rating}$$$, print in which division the $$$\mathrm{rating}$$$ belongs.
Input Specification: The first line of the input contains an integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of testcases. The description of each test consists of one line containing one integer $$$\mathrm{rating}$$$ ($$$-5000 \leq \mathrm{rating} \leq 5000$$$).
Output Specification: For each test case, output a single line containing the correct division in the format "Division X", where $$$X$$$ is an integer between $$$1$$$ and $$$4$$$ representing the division for the corresponding rating.
Notes: NoteFor test cases $$$1-4$$$, the corresponding ratings are $$$-789$$$, $$$1299$$$, $$$1300$$$, $$$1399$$$, so all of them are in division $$$4$$$.For the fifth test case, the corresponding rating is $$$1400$$$, so it is in division $$$3$$$.For the sixth test case, the corresponding rating is $$$1679$$$, so it is in division $$$2$$$.For the seventh test case, the corresponding rating is $$$2300$$$, so it is in division $$$1$$$.
Code:
x = input()
for i in range(int(x)):
z = input()
if int(z) >= 1900:
print('Division 1')
elif int(z) >= 1600:
# TODO: Your code here
elif int(z) >= 1400:
print('Division 3')
else:
print('Division 4')
|
x = input()
for i in range(int(x)):
z = input()
if int(z) >= 1900:
print('Division 1')
elif int(z) >= 1600:
{{completion}}
elif int(z) >= 1400:
print('Division 3')
else:
print('Division 4')
|
print('Division 2')
|
[{"input": "7\n-789\n1299\n1300\n1399\n1400\n1679\n2300", "output": ["Division 4\nDivision 4\nDivision 4\nDivision 4\nDivision 3\nDivision 2\nDivision 1"]}]
|
block_completion_000731
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: Codeforces separates its users into $$$4$$$ divisions by their rating: For Division 1: $$$1900 \leq \mathrm{rating}$$$ For Division 2: $$$1600 \leq \mathrm{rating} \leq 1899$$$ For Division 3: $$$1400 \leq \mathrm{rating} \leq 1599$$$ For Division 4: $$$\mathrm{rating} \leq 1399$$$ Given a $$$\mathrm{rating}$$$, print in which division the $$$\mathrm{rating}$$$ belongs.
Input Specification: The first line of the input contains an integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of testcases. The description of each test consists of one line containing one integer $$$\mathrm{rating}$$$ ($$$-5000 \leq \mathrm{rating} \leq 5000$$$).
Output Specification: For each test case, output a single line containing the correct division in the format "Division X", where $$$X$$$ is an integer between $$$1$$$ and $$$4$$$ representing the division for the corresponding rating.
Notes: NoteFor test cases $$$1-4$$$, the corresponding ratings are $$$-789$$$, $$$1299$$$, $$$1300$$$, $$$1399$$$, so all of them are in division $$$4$$$.For the fifth test case, the corresponding rating is $$$1400$$$, so it is in division $$$3$$$.For the sixth test case, the corresponding rating is $$$1679$$$, so it is in division $$$2$$$.For the seventh test case, the corresponding rating is $$$2300$$$, so it is in division $$$1$$$.
Code:
x = input()
for i in range(int(x)):
z = input()
if int(z) >= 1900:
print('Division 1')
elif int(z) >= 1600:
print('Division 2')
elif int(z) >= 1400:
# TODO: Your code here
else:
print('Division 4')
|
x = input()
for i in range(int(x)):
z = input()
if int(z) >= 1900:
print('Division 1')
elif int(z) >= 1600:
print('Division 2')
elif int(z) >= 1400:
{{completion}}
else:
print('Division 4')
|
print('Division 3')
|
[{"input": "7\n-789\n1299\n1300\n1399\n1400\n1679\n2300", "output": ["Division 4\nDivision 4\nDivision 4\nDivision 4\nDivision 3\nDivision 2\nDivision 1"]}]
|
block_completion_000732
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: Codeforces separates its users into $$$4$$$ divisions by their rating: For Division 1: $$$1900 \leq \mathrm{rating}$$$ For Division 2: $$$1600 \leq \mathrm{rating} \leq 1899$$$ For Division 3: $$$1400 \leq \mathrm{rating} \leq 1599$$$ For Division 4: $$$\mathrm{rating} \leq 1399$$$ Given a $$$\mathrm{rating}$$$, print in which division the $$$\mathrm{rating}$$$ belongs.
Input Specification: The first line of the input contains an integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of testcases. The description of each test consists of one line containing one integer $$$\mathrm{rating}$$$ ($$$-5000 \leq \mathrm{rating} \leq 5000$$$).
Output Specification: For each test case, output a single line containing the correct division in the format "Division X", where $$$X$$$ is an integer between $$$1$$$ and $$$4$$$ representing the division for the corresponding rating.
Notes: NoteFor test cases $$$1-4$$$, the corresponding ratings are $$$-789$$$, $$$1299$$$, $$$1300$$$, $$$1399$$$, so all of them are in division $$$4$$$.For the fifth test case, the corresponding rating is $$$1400$$$, so it is in division $$$3$$$.For the sixth test case, the corresponding rating is $$$1679$$$, so it is in division $$$2$$$.For the seventh test case, the corresponding rating is $$$2300$$$, so it is in division $$$1$$$.
Code:
k = 0
a = int(input())
for x in range(1, a+1):
b = int(input())
if 1900<= b:
d = 1
elif 1600 <= b <= 1899:
# TODO: Your code here
elif 1400 <= b <= 1599:
d = 3
elif b <= 1399:
d = 4
print('Division', d)
|
k = 0
a = int(input())
for x in range(1, a+1):
b = int(input())
if 1900<= b:
d = 1
elif 1600 <= b <= 1899:
{{completion}}
elif 1400 <= b <= 1599:
d = 3
elif b <= 1399:
d = 4
print('Division', d)
|
d = 2
|
[{"input": "7\n-789\n1299\n1300\n1399\n1400\n1679\n2300", "output": ["Division 4\nDivision 4\nDivision 4\nDivision 4\nDivision 3\nDivision 2\nDivision 1"]}]
|
block_completion_000733
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: Codeforces separates its users into $$$4$$$ divisions by their rating: For Division 1: $$$1900 \leq \mathrm{rating}$$$ For Division 2: $$$1600 \leq \mathrm{rating} \leq 1899$$$ For Division 3: $$$1400 \leq \mathrm{rating} \leq 1599$$$ For Division 4: $$$\mathrm{rating} \leq 1399$$$ Given a $$$\mathrm{rating}$$$, print in which division the $$$\mathrm{rating}$$$ belongs.
Input Specification: The first line of the input contains an integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of testcases. The description of each test consists of one line containing one integer $$$\mathrm{rating}$$$ ($$$-5000 \leq \mathrm{rating} \leq 5000$$$).
Output Specification: For each test case, output a single line containing the correct division in the format "Division X", where $$$X$$$ is an integer between $$$1$$$ and $$$4$$$ representing the division for the corresponding rating.
Notes: NoteFor test cases $$$1-4$$$, the corresponding ratings are $$$-789$$$, $$$1299$$$, $$$1300$$$, $$$1399$$$, so all of them are in division $$$4$$$.For the fifth test case, the corresponding rating is $$$1400$$$, so it is in division $$$3$$$.For the sixth test case, the corresponding rating is $$$1679$$$, so it is in division $$$2$$$.For the seventh test case, the corresponding rating is $$$2300$$$, so it is in division $$$1$$$.
Code:
k = 0
a = int(input())
for x in range(1, a+1):
b = int(input())
if 1900<= b:
d = 1
elif 1600 <= b <= 1899:
d = 2
elif 1400 <= b <= 1599:
# TODO: Your code here
elif b <= 1399:
d = 4
print('Division', d)
|
k = 0
a = int(input())
for x in range(1, a+1):
b = int(input())
if 1900<= b:
d = 1
elif 1600 <= b <= 1899:
d = 2
elif 1400 <= b <= 1599:
{{completion}}
elif b <= 1399:
d = 4
print('Division', d)
|
d = 3
|
[{"input": "7\n-789\n1299\n1300\n1399\n1400\n1679\n2300", "output": ["Division 4\nDivision 4\nDivision 4\nDivision 4\nDivision 3\nDivision 2\nDivision 1"]}]
|
block_completion_000734
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: Codeforces separates its users into $$$4$$$ divisions by their rating: For Division 1: $$$1900 \leq \mathrm{rating}$$$ For Division 2: $$$1600 \leq \mathrm{rating} \leq 1899$$$ For Division 3: $$$1400 \leq \mathrm{rating} \leq 1599$$$ For Division 4: $$$\mathrm{rating} \leq 1399$$$ Given a $$$\mathrm{rating}$$$, print in which division the $$$\mathrm{rating}$$$ belongs.
Input Specification: The first line of the input contains an integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of testcases. The description of each test consists of one line containing one integer $$$\mathrm{rating}$$$ ($$$-5000 \leq \mathrm{rating} \leq 5000$$$).
Output Specification: For each test case, output a single line containing the correct division in the format "Division X", where $$$X$$$ is an integer between $$$1$$$ and $$$4$$$ representing the division for the corresponding rating.
Notes: NoteFor test cases $$$1-4$$$, the corresponding ratings are $$$-789$$$, $$$1299$$$, $$$1300$$$, $$$1399$$$, so all of them are in division $$$4$$$.For the fifth test case, the corresponding rating is $$$1400$$$, so it is in division $$$3$$$.For the sixth test case, the corresponding rating is $$$1679$$$, so it is in division $$$2$$$.For the seventh test case, the corresponding rating is $$$2300$$$, so it is in division $$$1$$$.
Code:
t = int(input())
while t > 0:
n = int(input())
if n >= 1900:
print("Division",1)
elif n >= 1600 and n <1900:
# TODO: Your code here
elif n >= 1400 and n < 1600:
print("Division", 3)
else:
print("Division",4)
t -= 1
|
t = int(input())
while t > 0:
n = int(input())
if n >= 1900:
print("Division",1)
elif n >= 1600 and n <1900:
{{completion}}
elif n >= 1400 and n < 1600:
print("Division", 3)
else:
print("Division",4)
t -= 1
|
print("Division", 2)
|
[{"input": "7\n-789\n1299\n1300\n1399\n1400\n1679\n2300", "output": ["Division 4\nDivision 4\nDivision 4\nDivision 4\nDivision 3\nDivision 2\nDivision 1"]}]
|
block_completion_000735
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: Codeforces separates its users into $$$4$$$ divisions by their rating: For Division 1: $$$1900 \leq \mathrm{rating}$$$ For Division 2: $$$1600 \leq \mathrm{rating} \leq 1899$$$ For Division 3: $$$1400 \leq \mathrm{rating} \leq 1599$$$ For Division 4: $$$\mathrm{rating} \leq 1399$$$ Given a $$$\mathrm{rating}$$$, print in which division the $$$\mathrm{rating}$$$ belongs.
Input Specification: The first line of the input contains an integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of testcases. The description of each test consists of one line containing one integer $$$\mathrm{rating}$$$ ($$$-5000 \leq \mathrm{rating} \leq 5000$$$).
Output Specification: For each test case, output a single line containing the correct division in the format "Division X", where $$$X$$$ is an integer between $$$1$$$ and $$$4$$$ representing the division for the corresponding rating.
Notes: NoteFor test cases $$$1-4$$$, the corresponding ratings are $$$-789$$$, $$$1299$$$, $$$1300$$$, $$$1399$$$, so all of them are in division $$$4$$$.For the fifth test case, the corresponding rating is $$$1400$$$, so it is in division $$$3$$$.For the sixth test case, the corresponding rating is $$$1679$$$, so it is in division $$$2$$$.For the seventh test case, the corresponding rating is $$$2300$$$, so it is in division $$$1$$$.
Code:
t = int(input())
while t > 0:
n = int(input())
if n >= 1900:
print("Division",1)
elif n >= 1600 and n <1900:
print("Division", 2)
elif n >= 1400 and n < 1600:
# TODO: Your code here
else:
print("Division",4)
t -= 1
|
t = int(input())
while t > 0:
n = int(input())
if n >= 1900:
print("Division",1)
elif n >= 1600 and n <1900:
print("Division", 2)
elif n >= 1400 and n < 1600:
{{completion}}
else:
print("Division",4)
t -= 1
|
print("Division", 3)
|
[{"input": "7\n-789\n1299\n1300\n1399\n1400\n1679\n2300", "output": ["Division 4\nDivision 4\nDivision 4\nDivision 4\nDivision 3\nDivision 2\nDivision 1"]}]
|
block_completion_000736
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: Codeforces separates its users into $$$4$$$ divisions by their rating: For Division 1: $$$1900 \leq \mathrm{rating}$$$ For Division 2: $$$1600 \leq \mathrm{rating} \leq 1899$$$ For Division 3: $$$1400 \leq \mathrm{rating} \leq 1599$$$ For Division 4: $$$\mathrm{rating} \leq 1399$$$ Given a $$$\mathrm{rating}$$$, print in which division the $$$\mathrm{rating}$$$ belongs.
Input Specification: The first line of the input contains an integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of testcases. The description of each test consists of one line containing one integer $$$\mathrm{rating}$$$ ($$$-5000 \leq \mathrm{rating} \leq 5000$$$).
Output Specification: For each test case, output a single line containing the correct division in the format "Division X", where $$$X$$$ is an integer between $$$1$$$ and $$$4$$$ representing the division for the corresponding rating.
Notes: NoteFor test cases $$$1-4$$$, the corresponding ratings are $$$-789$$$, $$$1299$$$, $$$1300$$$, $$$1399$$$, so all of them are in division $$$4$$$.For the fifth test case, the corresponding rating is $$$1400$$$, so it is in division $$$3$$$.For the sixth test case, the corresponding rating is $$$1679$$$, so it is in division $$$2$$$.For the seventh test case, the corresponding rating is $$$2300$$$, so it is in division $$$1$$$.
Code:
x=int(input(""))
for i in range (x):
c=int(input(""))
if c<=1399:
print(" Division 4")
elif 1400<=c<=1599:
# TODO: Your code here
elif 1600<=c<=1899:
print(" Division 2")
else :
print(" Division 1")
|
x=int(input(""))
for i in range (x):
c=int(input(""))
if c<=1399:
print(" Division 4")
elif 1400<=c<=1599:
{{completion}}
elif 1600<=c<=1899:
print(" Division 2")
else :
print(" Division 1")
|
print(" Division 3")
|
[{"input": "7\n-789\n1299\n1300\n1399\n1400\n1679\n2300", "output": ["Division 4\nDivision 4\nDivision 4\nDivision 4\nDivision 3\nDivision 2\nDivision 1"]}]
|
block_completion_000737
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: Codeforces separates its users into $$$4$$$ divisions by their rating: For Division 1: $$$1900 \leq \mathrm{rating}$$$ For Division 2: $$$1600 \leq \mathrm{rating} \leq 1899$$$ For Division 3: $$$1400 \leq \mathrm{rating} \leq 1599$$$ For Division 4: $$$\mathrm{rating} \leq 1399$$$ Given a $$$\mathrm{rating}$$$, print in which division the $$$\mathrm{rating}$$$ belongs.
Input Specification: The first line of the input contains an integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of testcases. The description of each test consists of one line containing one integer $$$\mathrm{rating}$$$ ($$$-5000 \leq \mathrm{rating} \leq 5000$$$).
Output Specification: For each test case, output a single line containing the correct division in the format "Division X", where $$$X$$$ is an integer between $$$1$$$ and $$$4$$$ representing the division for the corresponding rating.
Notes: NoteFor test cases $$$1-4$$$, the corresponding ratings are $$$-789$$$, $$$1299$$$, $$$1300$$$, $$$1399$$$, so all of them are in division $$$4$$$.For the fifth test case, the corresponding rating is $$$1400$$$, so it is in division $$$3$$$.For the sixth test case, the corresponding rating is $$$1679$$$, so it is in division $$$2$$$.For the seventh test case, the corresponding rating is $$$2300$$$, so it is in division $$$1$$$.
Code:
x=int(input(""))
for i in range (x):
c=int(input(""))
if c<=1399:
print(" Division 4")
elif 1400<=c<=1599:
print(" Division 3")
elif 1600<=c<=1899:
# TODO: Your code here
else :
print(" Division 1")
|
x=int(input(""))
for i in range (x):
c=int(input(""))
if c<=1399:
print(" Division 4")
elif 1400<=c<=1599:
print(" Division 3")
elif 1600<=c<=1899:
{{completion}}
else :
print(" Division 1")
|
print(" Division 2")
|
[{"input": "7\n-789\n1299\n1300\n1399\n1400\n1679\n2300", "output": ["Division 4\nDivision 4\nDivision 4\nDivision 4\nDivision 3\nDivision 2\nDivision 1"]}]
|
block_completion_000738
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: Codeforces separates its users into $$$4$$$ divisions by their rating: For Division 1: $$$1900 \leq \mathrm{rating}$$$ For Division 2: $$$1600 \leq \mathrm{rating} \leq 1899$$$ For Division 3: $$$1400 \leq \mathrm{rating} \leq 1599$$$ For Division 4: $$$\mathrm{rating} \leq 1399$$$ Given a $$$\mathrm{rating}$$$, print in which division the $$$\mathrm{rating}$$$ belongs.
Input Specification: The first line of the input contains an integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of testcases. The description of each test consists of one line containing one integer $$$\mathrm{rating}$$$ ($$$-5000 \leq \mathrm{rating} \leq 5000$$$).
Output Specification: For each test case, output a single line containing the correct division in the format "Division X", where $$$X$$$ is an integer between $$$1$$$ and $$$4$$$ representing the division for the corresponding rating.
Notes: NoteFor test cases $$$1-4$$$, the corresponding ratings are $$$-789$$$, $$$1299$$$, $$$1300$$$, $$$1399$$$, so all of them are in division $$$4$$$.For the fifth test case, the corresponding rating is $$$1400$$$, so it is in division $$$3$$$.For the sixth test case, the corresponding rating is $$$1679$$$, so it is in division $$$2$$$.For the seventh test case, the corresponding rating is $$$2300$$$, so it is in division $$$1$$$.
Code:
'''
How pros write B)
'''
for i in range(int(input())):
# TODO: Your code here
|
'''
How pros write B)
'''
for i in range(int(input())):
{{completion}}
|
x = int(input())
print("Division 4" if x < 1400 else "Division 3" if x < 1600 else "Division 2" if x < 1900 else "Division 1")
|
[{"input": "7\n-789\n1299\n1300\n1399\n1400\n1679\n2300", "output": ["Division 4\nDivision 4\nDivision 4\nDivision 4\nDivision 3\nDivision 2\nDivision 1"]}]
|
block_completion_000739
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: There are $$$n$$$ candies put from left to right on a table. The candies are numbered from left to right. The $$$i$$$-th candy has weight $$$w_i$$$. Alice and Bob eat candies. Alice can eat any number of candies from the left (she can't skip candies, she eats them in a row). Bob can eat any number of candies from the right (he can't skip candies, he eats them in a row). Of course, if Alice ate a candy, Bob can't eat it (and vice versa).They want to be fair. Their goal is to eat the same total weight of candies. What is the most number of candies they can eat in total?
Input Specification: The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The first line of each test case contains an integer $$$n$$$ ($$$1 \leq n \leq 2\cdot10^5$$$) — the number of candies on the table. The second line of each test case contains $$$n$$$ integers $$$w_1, w_2, \dots, w_n$$$ ($$$1 \leq w_i \leq 10^4$$$) — the weights of candies from left to right. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2\cdot10^5$$$.
Output Specification: For each test case, print a single integer — the maximum number of candies Alice and Bob can eat in total while satisfying the condition.
Notes: NoteFor the first test case, Alice will eat one candy from the left and Bob will eat one candy from the right. There is no better way for them to eat the same total amount of weight. The answer is $$$2$$$ because they eat two candies in total.For the second test case, Alice will eat the first three candies from the left (with total weight $$$7$$$) and Bob will eat the first three candies from the right (with total weight $$$7$$$). They cannot eat more candies since all the candies have been eaten, so the answer is $$$6$$$ (because they eat six candies in total).For the third test case, there is no way Alice and Bob will eat the same non-zero weight so the answer is $$$0$$$.For the fourth test case, Alice will eat candies with weights $$$[7, 3, 20]$$$ and Bob will eat candies with weights $$$[10, 8, 11, 1]$$$, they each eat $$$30$$$ weight. There is no better partition so the answer is $$$7$$$.
Code:
from collections import deque
for _ in range(int(input())):
n = int(input())
l = deque(map(int, input().split()))
a, b = 0, 0
ans = 0
cur = 0
while l:
cur+=1
if a>=b:
b += l.pop()
else:
# TODO: Your code here
if a==b:
ans = cur
print(ans)
|
from collections import deque
for _ in range(int(input())):
n = int(input())
l = deque(map(int, input().split()))
a, b = 0, 0
ans = 0
cur = 0
while l:
cur+=1
if a>=b:
b += l.pop()
else:
{{completion}}
if a==b:
ans = cur
print(ans)
|
a += l.popleft()
|
[{"input": "4\n3\n10 20 10\n6\n2 1 4 2 4 1\n5\n1 2 4 8 16\n9\n7 3 20 5 15 1 11 8 10", "output": ["2\n6\n0\n7"]}]
|
block_completion_000796
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: There are $$$n$$$ candies put from left to right on a table. The candies are numbered from left to right. The $$$i$$$-th candy has weight $$$w_i$$$. Alice and Bob eat candies. Alice can eat any number of candies from the left (she can't skip candies, she eats them in a row). Bob can eat any number of candies from the right (he can't skip candies, he eats them in a row). Of course, if Alice ate a candy, Bob can't eat it (and vice versa).They want to be fair. Their goal is to eat the same total weight of candies. What is the most number of candies they can eat in total?
Input Specification: The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The first line of each test case contains an integer $$$n$$$ ($$$1 \leq n \leq 2\cdot10^5$$$) — the number of candies on the table. The second line of each test case contains $$$n$$$ integers $$$w_1, w_2, \dots, w_n$$$ ($$$1 \leq w_i \leq 10^4$$$) — the weights of candies from left to right. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2\cdot10^5$$$.
Output Specification: For each test case, print a single integer — the maximum number of candies Alice and Bob can eat in total while satisfying the condition.
Notes: NoteFor the first test case, Alice will eat one candy from the left and Bob will eat one candy from the right. There is no better way for them to eat the same total amount of weight. The answer is $$$2$$$ because they eat two candies in total.For the second test case, Alice will eat the first three candies from the left (with total weight $$$7$$$) and Bob will eat the first three candies from the right (with total weight $$$7$$$). They cannot eat more candies since all the candies have been eaten, so the answer is $$$6$$$ (because they eat six candies in total).For the third test case, there is no way Alice and Bob will eat the same non-zero weight so the answer is $$$0$$$.For the fourth test case, Alice will eat candies with weights $$$[7, 3, 20]$$$ and Bob will eat candies with weights $$$[10, 8, 11, 1]$$$, they each eat $$$30$$$ weight. There is no better partition so the answer is $$$7$$$.
Code:
for n in[*open(0)][2::2]:
n=[*map(int,n.split())]
a,b,l,f=[0]*4;r=len(n)-1
while l<=r:
if a<=b:
a+=n[l]
l+=1
elif b<a:
# TODO: Your code here
if a==b:
f=len(n)-r+l-1
print(f)
|
for n in[*open(0)][2::2]:
n=[*map(int,n.split())]
a,b,l,f=[0]*4;r=len(n)-1
while l<=r:
if a<=b:
a+=n[l]
l+=1
elif b<a:
{{completion}}
if a==b:
f=len(n)-r+l-1
print(f)
|
b+=n[r]
r-=1
|
[{"input": "4\n3\n10 20 10\n6\n2 1 4 2 4 1\n5\n1 2 4 8 16\n9\n7 3 20 5 15 1 11 8 10", "output": ["2\n6\n0\n7"]}]
|
block_completion_000797
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: There are $$$n$$$ candies put from left to right on a table. The candies are numbered from left to right. The $$$i$$$-th candy has weight $$$w_i$$$. Alice and Bob eat candies. Alice can eat any number of candies from the left (she can't skip candies, she eats them in a row). Bob can eat any number of candies from the right (he can't skip candies, he eats them in a row). Of course, if Alice ate a candy, Bob can't eat it (and vice versa).They want to be fair. Their goal is to eat the same total weight of candies. What is the most number of candies they can eat in total?
Input Specification: The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The first line of each test case contains an integer $$$n$$$ ($$$1 \leq n \leq 2\cdot10^5$$$) — the number of candies on the table. The second line of each test case contains $$$n$$$ integers $$$w_1, w_2, \dots, w_n$$$ ($$$1 \leq w_i \leq 10^4$$$) — the weights of candies from left to right. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2\cdot10^5$$$.
Output Specification: For each test case, print a single integer — the maximum number of candies Alice and Bob can eat in total while satisfying the condition.
Notes: NoteFor the first test case, Alice will eat one candy from the left and Bob will eat one candy from the right. There is no better way for them to eat the same total amount of weight. The answer is $$$2$$$ because they eat two candies in total.For the second test case, Alice will eat the first three candies from the left (with total weight $$$7$$$) and Bob will eat the first three candies from the right (with total weight $$$7$$$). They cannot eat more candies since all the candies have been eaten, so the answer is $$$6$$$ (because they eat six candies in total).For the third test case, there is no way Alice and Bob will eat the same non-zero weight so the answer is $$$0$$$.For the fourth test case, Alice will eat candies with weights $$$[7, 3, 20]$$$ and Bob will eat candies with weights $$$[10, 8, 11, 1]$$$, they each eat $$$30$$$ weight. There is no better partition so the answer is $$$7$$$.
Code:
from bisect import *
from itertools import *
t = int(input())
for _ in range(t):
n = int(input())
w = list(map(int, input().split()))
aw = list(accumulate(w))
bw = list(accumulate(w[::-1]))
mx = 0
for i, a in enumerate(aw):
c = bisect_left(bw, a, hi=len(bw)-i-2)
if a==bw[c] and i<(len(bw)-c-1):
# TODO: Your code here
print(mx)
|
from bisect import *
from itertools import *
t = int(input())
for _ in range(t):
n = int(input())
w = list(map(int, input().split()))
aw = list(accumulate(w))
bw = list(accumulate(w[::-1]))
mx = 0
for i, a in enumerate(aw):
c = bisect_left(bw, a, hi=len(bw)-i-2)
if a==bw[c] and i<(len(bw)-c-1):
{{completion}}
print(mx)
|
mx = max(mx, (i+1)+(c+1))
|
[{"input": "4\n3\n10 20 10\n6\n2 1 4 2 4 1\n5\n1 2 4 8 16\n9\n7 3 20 5 15 1 11 8 10", "output": ["2\n6\n0\n7"]}]
|
block_completion_000798
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: There are $$$n$$$ candies put from left to right on a table. The candies are numbered from left to right. The $$$i$$$-th candy has weight $$$w_i$$$. Alice and Bob eat candies. Alice can eat any number of candies from the left (she can't skip candies, she eats them in a row). Bob can eat any number of candies from the right (he can't skip candies, he eats them in a row). Of course, if Alice ate a candy, Bob can't eat it (and vice versa).They want to be fair. Their goal is to eat the same total weight of candies. What is the most number of candies they can eat in total?
Input Specification: The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The first line of each test case contains an integer $$$n$$$ ($$$1 \leq n \leq 2\cdot10^5$$$) — the number of candies on the table. The second line of each test case contains $$$n$$$ integers $$$w_1, w_2, \dots, w_n$$$ ($$$1 \leq w_i \leq 10^4$$$) — the weights of candies from left to right. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2\cdot10^5$$$.
Output Specification: For each test case, print a single integer — the maximum number of candies Alice and Bob can eat in total while satisfying the condition.
Notes: NoteFor the first test case, Alice will eat one candy from the left and Bob will eat one candy from the right. There is no better way for them to eat the same total amount of weight. The answer is $$$2$$$ because they eat two candies in total.For the second test case, Alice will eat the first three candies from the left (with total weight $$$7$$$) and Bob will eat the first three candies from the right (with total weight $$$7$$$). They cannot eat more candies since all the candies have been eaten, so the answer is $$$6$$$ (because they eat six candies in total).For the third test case, there is no way Alice and Bob will eat the same non-zero weight so the answer is $$$0$$$.For the fourth test case, Alice will eat candies with weights $$$[7, 3, 20]$$$ and Bob will eat candies with weights $$$[10, 8, 11, 1]$$$, they each eat $$$30$$$ weight. There is no better partition so the answer is $$$7$$$.
Code:
def solve():
n=int(input())
a=[*map(int,input().split())]
b=a[:]
for i in range(n-1):a[i+1]+=a[i]
for i in range(n-1,0,-1):b[i-1]+=b[i]
l,r=0,n-1
sol=0
while r-l>=1:
if a[l]==b[r]:sol=l+n-r+1;l+=1
if a[l]<b[r]:l+=1
else:# TODO: Your code here
return sol
for _ in [0]*int(input()):print(solve())
|
def solve():
n=int(input())
a=[*map(int,input().split())]
b=a[:]
for i in range(n-1):a[i+1]+=a[i]
for i in range(n-1,0,-1):b[i-1]+=b[i]
l,r=0,n-1
sol=0
while r-l>=1:
if a[l]==b[r]:sol=l+n-r+1;l+=1
if a[l]<b[r]:l+=1
else:{{completion}}
return sol
for _ in [0]*int(input()):print(solve())
|
r-=1
|
[{"input": "4\n3\n10 20 10\n6\n2 1 4 2 4 1\n5\n1 2 4 8 16\n9\n7 3 20 5 15 1 11 8 10", "output": ["2\n6\n0\n7"]}]
|
block_completion_000799
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: There are $$$n$$$ candies put from left to right on a table. The candies are numbered from left to right. The $$$i$$$-th candy has weight $$$w_i$$$. Alice and Bob eat candies. Alice can eat any number of candies from the left (she can't skip candies, she eats them in a row). Bob can eat any number of candies from the right (he can't skip candies, he eats them in a row). Of course, if Alice ate a candy, Bob can't eat it (and vice versa).They want to be fair. Their goal is to eat the same total weight of candies. What is the most number of candies they can eat in total?
Input Specification: The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The first line of each test case contains an integer $$$n$$$ ($$$1 \leq n \leq 2\cdot10^5$$$) — the number of candies on the table. The second line of each test case contains $$$n$$$ integers $$$w_1, w_2, \dots, w_n$$$ ($$$1 \leq w_i \leq 10^4$$$) — the weights of candies from left to right. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2\cdot10^5$$$.
Output Specification: For each test case, print a single integer — the maximum number of candies Alice and Bob can eat in total while satisfying the condition.
Notes: NoteFor the first test case, Alice will eat one candy from the left and Bob will eat one candy from the right. There is no better way for them to eat the same total amount of weight. The answer is $$$2$$$ because they eat two candies in total.For the second test case, Alice will eat the first three candies from the left (with total weight $$$7$$$) and Bob will eat the first three candies from the right (with total weight $$$7$$$). They cannot eat more candies since all the candies have been eaten, so the answer is $$$6$$$ (because they eat six candies in total).For the third test case, there is no way Alice and Bob will eat the same non-zero weight so the answer is $$$0$$$.For the fourth test case, Alice will eat candies with weights $$$[7, 3, 20]$$$ and Bob will eat candies with weights $$$[10, 8, 11, 1]$$$, they each eat $$$30$$$ weight. There is no better partition so the answer is $$$7$$$.
Code:
I=lambda:[int(i) for i in input().split()]
for _ in range(I()[0]):
n=I()[0]
l=I()
l2=[]
s1,s2=0,0
p1,p2=0,n-1
while (p1-1<=p2):
if s1 == s2: # TODO: Your code here
if s1 < s2: s1+=l[p1]; p1+=1
if s2 < s1: s2+=l[p2]; p2-=1
print(l2[-1])
|
I=lambda:[int(i) for i in input().split()]
for _ in range(I()[0]):
n=I()[0]
l=I()
l2=[]
s1,s2=0,0
p1,p2=0,n-1
while (p1-1<=p2):
if s1 == s2: {{completion}}
if s1 < s2: s1+=l[p1]; p1+=1
if s2 < s1: s2+=l[p2]; p2-=1
print(l2[-1])
|
l2.append(p1 + n-1-p2); s1+=l[p1]; p1+=1
|
[{"input": "4\n3\n10 20 10\n6\n2 1 4 2 4 1\n5\n1 2 4 8 16\n9\n7 3 20 5 15 1 11 8 10", "output": ["2\n6\n0\n7"]}]
|
block_completion_000800
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: There are $$$n$$$ candies put from left to right on a table. The candies are numbered from left to right. The $$$i$$$-th candy has weight $$$w_i$$$. Alice and Bob eat candies. Alice can eat any number of candies from the left (she can't skip candies, she eats them in a row). Bob can eat any number of candies from the right (he can't skip candies, he eats them in a row). Of course, if Alice ate a candy, Bob can't eat it (and vice versa).They want to be fair. Their goal is to eat the same total weight of candies. What is the most number of candies they can eat in total?
Input Specification: The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The first line of each test case contains an integer $$$n$$$ ($$$1 \leq n \leq 2\cdot10^5$$$) — the number of candies on the table. The second line of each test case contains $$$n$$$ integers $$$w_1, w_2, \dots, w_n$$$ ($$$1 \leq w_i \leq 10^4$$$) — the weights of candies from left to right. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2\cdot10^5$$$.
Output Specification: For each test case, print a single integer — the maximum number of candies Alice and Bob can eat in total while satisfying the condition.
Notes: NoteFor the first test case, Alice will eat one candy from the left and Bob will eat one candy from the right. There is no better way for them to eat the same total amount of weight. The answer is $$$2$$$ because they eat two candies in total.For the second test case, Alice will eat the first three candies from the left (with total weight $$$7$$$) and Bob will eat the first three candies from the right (with total weight $$$7$$$). They cannot eat more candies since all the candies have been eaten, so the answer is $$$6$$$ (because they eat six candies in total).For the third test case, there is no way Alice and Bob will eat the same non-zero weight so the answer is $$$0$$$.For the fourth test case, Alice will eat candies with weights $$$[7, 3, 20]$$$ and Bob will eat candies with weights $$$[10, 8, 11, 1]$$$, they each eat $$$30$$$ weight. There is no better partition so the answer is $$$7$$$.
Code:
I=lambda:[int(i) for i in input().split()]
for _ in range(I()[0]):
n=I()[0]
l=I()
l2=[]
s1,s2=0,0
p1,p2=0,n-1
while (p1-1<=p2):
if s1 == s2: l2.append(p1 + n-1-p2); s1+=l[p1]; p1+=1
if s1 < s2: # TODO: Your code here
if s2 < s1: s2+=l[p2]; p2-=1
print(l2[-1])
|
I=lambda:[int(i) for i in input().split()]
for _ in range(I()[0]):
n=I()[0]
l=I()
l2=[]
s1,s2=0,0
p1,p2=0,n-1
while (p1-1<=p2):
if s1 == s2: l2.append(p1 + n-1-p2); s1+=l[p1]; p1+=1
if s1 < s2: {{completion}}
if s2 < s1: s2+=l[p2]; p2-=1
print(l2[-1])
|
s1+=l[p1]; p1+=1
|
[{"input": "4\n3\n10 20 10\n6\n2 1 4 2 4 1\n5\n1 2 4 8 16\n9\n7 3 20 5 15 1 11 8 10", "output": ["2\n6\n0\n7"]}]
|
block_completion_000801
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: There are $$$n$$$ candies put from left to right on a table. The candies are numbered from left to right. The $$$i$$$-th candy has weight $$$w_i$$$. Alice and Bob eat candies. Alice can eat any number of candies from the left (she can't skip candies, she eats them in a row). Bob can eat any number of candies from the right (he can't skip candies, he eats them in a row). Of course, if Alice ate a candy, Bob can't eat it (and vice versa).They want to be fair. Their goal is to eat the same total weight of candies. What is the most number of candies they can eat in total?
Input Specification: The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The first line of each test case contains an integer $$$n$$$ ($$$1 \leq n \leq 2\cdot10^5$$$) — the number of candies on the table. The second line of each test case contains $$$n$$$ integers $$$w_1, w_2, \dots, w_n$$$ ($$$1 \leq w_i \leq 10^4$$$) — the weights of candies from left to right. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2\cdot10^5$$$.
Output Specification: For each test case, print a single integer — the maximum number of candies Alice and Bob can eat in total while satisfying the condition.
Notes: NoteFor the first test case, Alice will eat one candy from the left and Bob will eat one candy from the right. There is no better way for them to eat the same total amount of weight. The answer is $$$2$$$ because they eat two candies in total.For the second test case, Alice will eat the first three candies from the left (with total weight $$$7$$$) and Bob will eat the first three candies from the right (with total weight $$$7$$$). They cannot eat more candies since all the candies have been eaten, so the answer is $$$6$$$ (because they eat six candies in total).For the third test case, there is no way Alice and Bob will eat the same non-zero weight so the answer is $$$0$$$.For the fourth test case, Alice will eat candies with weights $$$[7, 3, 20]$$$ and Bob will eat candies with weights $$$[10, 8, 11, 1]$$$, they each eat $$$30$$$ weight. There is no better partition so the answer is $$$7$$$.
Code:
import math as m
i = input()
inp = []
for a in range(0,int(i)):
inp += [[input(), input()]]
def maxx(arr):
if arr == []:
return 0
else:
return max(arr)
for s in inp:
n = int(s[0])
arr = [int(x) for x in s[1].split()]
i = 0
j = n-1
lsum = arr[i]
rsum = arr[j]
best = 0
while i < j:
if lsum < rsum:
i += 1
lsum += arr[i]
elif rsum < lsum:
# TODO: Your code here
else:
best = i + (n-1-j) + 2
i += 1
lsum += arr[i]
print(best)
|
import math as m
i = input()
inp = []
for a in range(0,int(i)):
inp += [[input(), input()]]
def maxx(arr):
if arr == []:
return 0
else:
return max(arr)
for s in inp:
n = int(s[0])
arr = [int(x) for x in s[1].split()]
i = 0
j = n-1
lsum = arr[i]
rsum = arr[j]
best = 0
while i < j:
if lsum < rsum:
i += 1
lsum += arr[i]
elif rsum < lsum:
{{completion}}
else:
best = i + (n-1-j) + 2
i += 1
lsum += arr[i]
print(best)
|
j -= 1
rsum += arr[j]
|
[{"input": "4\n3\n10 20 10\n6\n2 1 4 2 4 1\n5\n1 2 4 8 16\n9\n7 3 20 5 15 1 11 8 10", "output": ["2\n6\n0\n7"]}]
|
block_completion_000802
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: There are $$$n$$$ candies put from left to right on a table. The candies are numbered from left to right. The $$$i$$$-th candy has weight $$$w_i$$$. Alice and Bob eat candies. Alice can eat any number of candies from the left (she can't skip candies, she eats them in a row). Bob can eat any number of candies from the right (he can't skip candies, he eats them in a row). Of course, if Alice ate a candy, Bob can't eat it (and vice versa).They want to be fair. Their goal is to eat the same total weight of candies. What is the most number of candies they can eat in total?
Input Specification: The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The first line of each test case contains an integer $$$n$$$ ($$$1 \leq n \leq 2\cdot10^5$$$) — the number of candies on the table. The second line of each test case contains $$$n$$$ integers $$$w_1, w_2, \dots, w_n$$$ ($$$1 \leq w_i \leq 10^4$$$) — the weights of candies from left to right. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2\cdot10^5$$$.
Output Specification: For each test case, print a single integer — the maximum number of candies Alice and Bob can eat in total while satisfying the condition.
Notes: NoteFor the first test case, Alice will eat one candy from the left and Bob will eat one candy from the right. There is no better way for them to eat the same total amount of weight. The answer is $$$2$$$ because they eat two candies in total.For the second test case, Alice will eat the first three candies from the left (with total weight $$$7$$$) and Bob will eat the first three candies from the right (with total weight $$$7$$$). They cannot eat more candies since all the candies have been eaten, so the answer is $$$6$$$ (because they eat six candies in total).For the third test case, there is no way Alice and Bob will eat the same non-zero weight so the answer is $$$0$$$.For the fourth test case, Alice will eat candies with weights $$$[7, 3, 20]$$$ and Bob will eat candies with weights $$$[10, 8, 11, 1]$$$, they each eat $$$30$$$ weight. There is no better partition so the answer is $$$7$$$.
Code:
import math as m
i = input()
inp = []
for a in range(0,int(i)):
inp += [[input(), input()]]
def maxx(arr):
if arr == []:
return 0
else:
return max(arr)
for s in inp:
n = int(s[0])
arr = [int(x) for x in s[1].split()]
i = 0
j = n-1
lsum = arr[i]
rsum = arr[j]
best = 0
while i < j:
if lsum < rsum:
i += 1
lsum += arr[i]
elif rsum < lsum:
j -= 1
rsum += arr[j]
else:
# TODO: Your code here
print(best)
|
import math as m
i = input()
inp = []
for a in range(0,int(i)):
inp += [[input(), input()]]
def maxx(arr):
if arr == []:
return 0
else:
return max(arr)
for s in inp:
n = int(s[0])
arr = [int(x) for x in s[1].split()]
i = 0
j = n-1
lsum = arr[i]
rsum = arr[j]
best = 0
while i < j:
if lsum < rsum:
i += 1
lsum += arr[i]
elif rsum < lsum:
j -= 1
rsum += arr[j]
else:
{{completion}}
print(best)
|
best = i + (n-1-j) + 2
i += 1
lsum += arr[i]
|
[{"input": "4\n3\n10 20 10\n6\n2 1 4 2 4 1\n5\n1 2 4 8 16\n9\n7 3 20 5 15 1 11 8 10", "output": ["2\n6\n0\n7"]}]
|
block_completion_000803
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: There are $$$n$$$ candies put from left to right on a table. The candies are numbered from left to right. The $$$i$$$-th candy has weight $$$w_i$$$. Alice and Bob eat candies. Alice can eat any number of candies from the left (she can't skip candies, she eats them in a row). Bob can eat any number of candies from the right (he can't skip candies, he eats them in a row). Of course, if Alice ate a candy, Bob can't eat it (and vice versa).They want to be fair. Their goal is to eat the same total weight of candies. What is the most number of candies they can eat in total?
Input Specification: The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The first line of each test case contains an integer $$$n$$$ ($$$1 \leq n \leq 2\cdot10^5$$$) — the number of candies on the table. The second line of each test case contains $$$n$$$ integers $$$w_1, w_2, \dots, w_n$$$ ($$$1 \leq w_i \leq 10^4$$$) — the weights of candies from left to right. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2\cdot10^5$$$.
Output Specification: For each test case, print a single integer — the maximum number of candies Alice and Bob can eat in total while satisfying the condition.
Notes: NoteFor the first test case, Alice will eat one candy from the left and Bob will eat one candy from the right. There is no better way for them to eat the same total amount of weight. The answer is $$$2$$$ because they eat two candies in total.For the second test case, Alice will eat the first three candies from the left (with total weight $$$7$$$) and Bob will eat the first three candies from the right (with total weight $$$7$$$). They cannot eat more candies since all the candies have been eaten, so the answer is $$$6$$$ (because they eat six candies in total).For the third test case, there is no way Alice and Bob will eat the same non-zero weight so the answer is $$$0$$$.For the fourth test case, Alice will eat candies with weights $$$[7, 3, 20]$$$ and Bob will eat candies with weights $$$[10, 8, 11, 1]$$$, they each eat $$$30$$$ weight. There is no better partition so the answer is $$$7$$$.
Code:
for t in range(int(input())):
n=int(input())
l1=list(map(int,input().split()))
l=0
h=n-1
ans=0
a=0
b=0
while(l<=h):
if(a>b):
b=b+l1[h]
h=h-1
else:
# TODO: Your code here
if(a==b):
ans=l+n-h-1
print(ans)
|
for t in range(int(input())):
n=int(input())
l1=list(map(int,input().split()))
l=0
h=n-1
ans=0
a=0
b=0
while(l<=h):
if(a>b):
b=b+l1[h]
h=h-1
else:
{{completion}}
if(a==b):
ans=l+n-h-1
print(ans)
|
a=a+l1[l]
l=l+1
|
[{"input": "4\n3\n10 20 10\n6\n2 1 4 2 4 1\n5\n1 2 4 8 16\n9\n7 3 20 5 15 1 11 8 10", "output": ["2\n6\n0\n7"]}]
|
block_completion_000804
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: There are $$$n$$$ candies put from left to right on a table. The candies are numbered from left to right. The $$$i$$$-th candy has weight $$$w_i$$$. Alice and Bob eat candies. Alice can eat any number of candies from the left (she can't skip candies, she eats them in a row). Bob can eat any number of candies from the right (he can't skip candies, he eats them in a row). Of course, if Alice ate a candy, Bob can't eat it (and vice versa).They want to be fair. Their goal is to eat the same total weight of candies. What is the most number of candies they can eat in total?
Input Specification: The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The first line of each test case contains an integer $$$n$$$ ($$$1 \leq n \leq 2\cdot10^5$$$) — the number of candies on the table. The second line of each test case contains $$$n$$$ integers $$$w_1, w_2, \dots, w_n$$$ ($$$1 \leq w_i \leq 10^4$$$) — the weights of candies from left to right. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2\cdot10^5$$$.
Output Specification: For each test case, print a single integer — the maximum number of candies Alice and Bob can eat in total while satisfying the condition.
Notes: NoteFor the first test case, Alice will eat one candy from the left and Bob will eat one candy from the right. There is no better way for them to eat the same total amount of weight. The answer is $$$2$$$ because they eat two candies in total.For the second test case, Alice will eat the first three candies from the left (with total weight $$$7$$$) and Bob will eat the first three candies from the right (with total weight $$$7$$$). They cannot eat more candies since all the candies have been eaten, so the answer is $$$6$$$ (because they eat six candies in total).For the third test case, there is no way Alice and Bob will eat the same non-zero weight so the answer is $$$0$$$.For the fourth test case, Alice will eat candies with weights $$$[7, 3, 20]$$$ and Bob will eat candies with weights $$$[10, 8, 11, 1]$$$, they each eat $$$30$$$ weight. There is no better partition so the answer is $$$7$$$.
Code:
for _ in range(int(input())):
n = int(input())
a = [*map(int, input().split())]
x = sum(a) // 2
s, d = 0, {}
for idx, i in enumerate(a):
s += i
if s > x: # TODO: Your code here
d[s] = idx + 1
s, r = 0, 0
for idx, i in enumerate(a[::-1]):
s += i
if s in d: r = idx + 1 + d[s]
print(r)
|
for _ in range(int(input())):
n = int(input())
a = [*map(int, input().split())]
x = sum(a) // 2
s, d = 0, {}
for idx, i in enumerate(a):
s += i
if s > x: {{completion}}
d[s] = idx + 1
s, r = 0, 0
for idx, i in enumerate(a[::-1]):
s += i
if s in d: r = idx + 1 + d[s]
print(r)
|
break
|
[{"input": "4\n3\n10 20 10\n6\n2 1 4 2 4 1\n5\n1 2 4 8 16\n9\n7 3 20 5 15 1 11 8 10", "output": ["2\n6\n0\n7"]}]
|
block_completion_000805
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: There are $$$n$$$ candies put from left to right on a table. The candies are numbered from left to right. The $$$i$$$-th candy has weight $$$w_i$$$. Alice and Bob eat candies. Alice can eat any number of candies from the left (she can't skip candies, she eats them in a row). Bob can eat any number of candies from the right (he can't skip candies, he eats them in a row). Of course, if Alice ate a candy, Bob can't eat it (and vice versa).They want to be fair. Their goal is to eat the same total weight of candies. What is the most number of candies they can eat in total?
Input Specification: The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The first line of each test case contains an integer $$$n$$$ ($$$1 \leq n \leq 2\cdot10^5$$$) — the number of candies on the table. The second line of each test case contains $$$n$$$ integers $$$w_1, w_2, \dots, w_n$$$ ($$$1 \leq w_i \leq 10^4$$$) — the weights of candies from left to right. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2\cdot10^5$$$.
Output Specification: For each test case, print a single integer — the maximum number of candies Alice and Bob can eat in total while satisfying the condition.
Notes: NoteFor the first test case, Alice will eat one candy from the left and Bob will eat one candy from the right. There is no better way for them to eat the same total amount of weight. The answer is $$$2$$$ because they eat two candies in total.For the second test case, Alice will eat the first three candies from the left (with total weight $$$7$$$) and Bob will eat the first three candies from the right (with total weight $$$7$$$). They cannot eat more candies since all the candies have been eaten, so the answer is $$$6$$$ (because they eat six candies in total).For the third test case, there is no way Alice and Bob will eat the same non-zero weight so the answer is $$$0$$$.For the fourth test case, Alice will eat candies with weights $$$[7, 3, 20]$$$ and Bob will eat candies with weights $$$[10, 8, 11, 1]$$$, they each eat $$$30$$$ weight. There is no better partition so the answer is $$$7$$$.
Code:
for _ in range(int(input())):
n = int(input())
a = [*map(int, input().split())]
x = sum(a) // 2
s, d = 0, {}
for idx, i in enumerate(a):
s += i
if s > x: break
d[s] = idx + 1
s, r = 0, 0
for idx, i in enumerate(a[::-1]):
s += i
if s in d: # TODO: Your code here
print(r)
|
for _ in range(int(input())):
n = int(input())
a = [*map(int, input().split())]
x = sum(a) // 2
s, d = 0, {}
for idx, i in enumerate(a):
s += i
if s > x: break
d[s] = idx + 1
s, r = 0, 0
for idx, i in enumerate(a[::-1]):
s += i
if s in d: {{completion}}
print(r)
|
r = idx + 1 + d[s]
|
[{"input": "4\n3\n10 20 10\n6\n2 1 4 2 4 1\n5\n1 2 4 8 16\n9\n7 3 20 5 15 1 11 8 10", "output": ["2\n6\n0\n7"]}]
|
block_completion_000806
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: There are $$$n$$$ candies put from left to right on a table. The candies are numbered from left to right. The $$$i$$$-th candy has weight $$$w_i$$$. Alice and Bob eat candies. Alice can eat any number of candies from the left (she can't skip candies, she eats them in a row). Bob can eat any number of candies from the right (he can't skip candies, he eats them in a row). Of course, if Alice ate a candy, Bob can't eat it (and vice versa).They want to be fair. Their goal is to eat the same total weight of candies. What is the most number of candies they can eat in total?
Input Specification: The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The first line of each test case contains an integer $$$n$$$ ($$$1 \leq n \leq 2\cdot10^5$$$) — the number of candies on the table. The second line of each test case contains $$$n$$$ integers $$$w_1, w_2, \dots, w_n$$$ ($$$1 \leq w_i \leq 10^4$$$) — the weights of candies from left to right. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2\cdot10^5$$$.
Output Specification: For each test case, print a single integer — the maximum number of candies Alice and Bob can eat in total while satisfying the condition.
Notes: NoteFor the first test case, Alice will eat one candy from the left and Bob will eat one candy from the right. There is no better way for them to eat the same total amount of weight. The answer is $$$2$$$ because they eat two candies in total.For the second test case, Alice will eat the first three candies from the left (with total weight $$$7$$$) and Bob will eat the first three candies from the right (with total weight $$$7$$$). They cannot eat more candies since all the candies have been eaten, so the answer is $$$6$$$ (because they eat six candies in total).For the third test case, there is no way Alice and Bob will eat the same non-zero weight so the answer is $$$0$$$.For the fourth test case, Alice will eat candies with weights $$$[7, 3, 20]$$$ and Bob will eat candies with weights $$$[10, 8, 11, 1]$$$, they each eat $$$30$$$ weight. There is no better partition so the answer is $$$7$$$.
Code:
n = int(input())
for i in range(n):
h = 1
g = 1
counter = 0
m = int(input())
lst = list(map(int, input().split()))
index =0
e= 0
r =m-1
highest = lst[m-1]
lowest = lst[0]
while h+g <= m:
if lowest<highest:
lowest+=lst[e+1]
e+=1
g +=1
elif highest<lowest:
# TODO: Your code here
elif highest == lowest:
lowest+=lst[e+1]
e+=1
g+=1
index = e + (m - r)
print(index)
|
n = int(input())
for i in range(n):
h = 1
g = 1
counter = 0
m = int(input())
lst = list(map(int, input().split()))
index =0
e= 0
r =m-1
highest = lst[m-1]
lowest = lst[0]
while h+g <= m:
if lowest<highest:
lowest+=lst[e+1]
e+=1
g +=1
elif highest<lowest:
{{completion}}
elif highest == lowest:
lowest+=lst[e+1]
e+=1
g+=1
index = e + (m - r)
print(index)
|
highest+=lst[r-1]
r-=1
h+=1
|
[{"input": "4\n3\n10 20 10\n6\n2 1 4 2 4 1\n5\n1 2 4 8 16\n9\n7 3 20 5 15 1 11 8 10", "output": ["2\n6\n0\n7"]}]
|
block_completion_000807
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: There are $$$n$$$ candies put from left to right on a table. The candies are numbered from left to right. The $$$i$$$-th candy has weight $$$w_i$$$. Alice and Bob eat candies. Alice can eat any number of candies from the left (she can't skip candies, she eats them in a row). Bob can eat any number of candies from the right (he can't skip candies, he eats them in a row). Of course, if Alice ate a candy, Bob can't eat it (and vice versa).They want to be fair. Their goal is to eat the same total weight of candies. What is the most number of candies they can eat in total?
Input Specification: The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The first line of each test case contains an integer $$$n$$$ ($$$1 \leq n \leq 2\cdot10^5$$$) — the number of candies on the table. The second line of each test case contains $$$n$$$ integers $$$w_1, w_2, \dots, w_n$$$ ($$$1 \leq w_i \leq 10^4$$$) — the weights of candies from left to right. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2\cdot10^5$$$.
Output Specification: For each test case, print a single integer — the maximum number of candies Alice and Bob can eat in total while satisfying the condition.
Notes: NoteFor the first test case, Alice will eat one candy from the left and Bob will eat one candy from the right. There is no better way for them to eat the same total amount of weight. The answer is $$$2$$$ because they eat two candies in total.For the second test case, Alice will eat the first three candies from the left (with total weight $$$7$$$) and Bob will eat the first three candies from the right (with total weight $$$7$$$). They cannot eat more candies since all the candies have been eaten, so the answer is $$$6$$$ (because they eat six candies in total).For the third test case, there is no way Alice and Bob will eat the same non-zero weight so the answer is $$$0$$$.For the fourth test case, Alice will eat candies with weights $$$[7, 3, 20]$$$ and Bob will eat candies with weights $$$[10, 8, 11, 1]$$$, they each eat $$$30$$$ weight. There is no better partition so the answer is $$$7$$$.
Code:
n = int(input())
for i in range(n):
h = 1
g = 1
counter = 0
m = int(input())
lst = list(map(int, input().split()))
index =0
e= 0
r =m-1
highest = lst[m-1]
lowest = lst[0]
while h+g <= m:
if lowest<highest:
lowest+=lst[e+1]
e+=1
g +=1
elif highest<lowest:
highest+=lst[r-1]
r-=1
h+=1
elif highest == lowest:
# TODO: Your code here
print(index)
|
n = int(input())
for i in range(n):
h = 1
g = 1
counter = 0
m = int(input())
lst = list(map(int, input().split()))
index =0
e= 0
r =m-1
highest = lst[m-1]
lowest = lst[0]
while h+g <= m:
if lowest<highest:
lowest+=lst[e+1]
e+=1
g +=1
elif highest<lowest:
highest+=lst[r-1]
r-=1
h+=1
elif highest == lowest:
{{completion}}
print(index)
|
lowest+=lst[e+1]
e+=1
g+=1
index = e + (m - r)
|
[{"input": "4\n3\n10 20 10\n6\n2 1 4 2 4 1\n5\n1 2 4 8 16\n9\n7 3 20 5 15 1 11 8 10", "output": ["2\n6\n0\n7"]}]
|
block_completion_000808
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: There are $$$n$$$ candies put from left to right on a table. The candies are numbered from left to right. The $$$i$$$-th candy has weight $$$w_i$$$. Alice and Bob eat candies. Alice can eat any number of candies from the left (she can't skip candies, she eats them in a row). Bob can eat any number of candies from the right (he can't skip candies, he eats them in a row). Of course, if Alice ate a candy, Bob can't eat it (and vice versa).They want to be fair. Their goal is to eat the same total weight of candies. What is the most number of candies they can eat in total?
Input Specification: The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The first line of each test case contains an integer $$$n$$$ ($$$1 \leq n \leq 2\cdot10^5$$$) — the number of candies on the table. The second line of each test case contains $$$n$$$ integers $$$w_1, w_2, \dots, w_n$$$ ($$$1 \leq w_i \leq 10^4$$$) — the weights of candies from left to right. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2\cdot10^5$$$.
Output Specification: For each test case, print a single integer — the maximum number of candies Alice and Bob can eat in total while satisfying the condition.
Notes: NoteFor the first test case, Alice will eat one candy from the left and Bob will eat one candy from the right. There is no better way for them to eat the same total amount of weight. The answer is $$$2$$$ because they eat two candies in total.For the second test case, Alice will eat the first three candies from the left (with total weight $$$7$$$) and Bob will eat the first three candies from the right (with total weight $$$7$$$). They cannot eat more candies since all the candies have been eaten, so the answer is $$$6$$$ (because they eat six candies in total).For the third test case, there is no way Alice and Bob will eat the same non-zero weight so the answer is $$$0$$$.For the fourth test case, Alice will eat candies with weights $$$[7, 3, 20]$$$ and Bob will eat candies with weights $$$[10, 8, 11, 1]$$$, they each eat $$$30$$$ weight. There is no better partition so the answer is $$$7$$$.
Code:
def read():
return int(input())
def readline():
return list(map(int,input().split()))
def solve():
n=read()
arr=readline()
ans,cur=0,0
a,suma=-1,0
b,sumb=n,0
while True:
if a>=b: break
elif suma>sumb:
# TODO: Your code here
elif suma<sumb:
a+=1
suma+=arr[a]
cur+=1
else :
ans=cur
a+=1
b-=1
suma+=arr[a]
sumb+=arr[b]
cur+=2
print(ans)
if __name__ == "__main__":
T=read()
for i in range(T):
solve()
|
def read():
return int(input())
def readline():
return list(map(int,input().split()))
def solve():
n=read()
arr=readline()
ans,cur=0,0
a,suma=-1,0
b,sumb=n,0
while True:
if a>=b: break
elif suma>sumb:
{{completion}}
elif suma<sumb:
a+=1
suma+=arr[a]
cur+=1
else :
ans=cur
a+=1
b-=1
suma+=arr[a]
sumb+=arr[b]
cur+=2
print(ans)
if __name__ == "__main__":
T=read()
for i in range(T):
solve()
|
b-=1
sumb+=arr[b]
cur+=1
|
[{"input": "4\n3\n10 20 10\n6\n2 1 4 2 4 1\n5\n1 2 4 8 16\n9\n7 3 20 5 15 1 11 8 10", "output": ["2\n6\n0\n7"]}]
|
block_completion_000809
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: There are $$$n$$$ candies put from left to right on a table. The candies are numbered from left to right. The $$$i$$$-th candy has weight $$$w_i$$$. Alice and Bob eat candies. Alice can eat any number of candies from the left (she can't skip candies, she eats them in a row). Bob can eat any number of candies from the right (he can't skip candies, he eats them in a row). Of course, if Alice ate a candy, Bob can't eat it (and vice versa).They want to be fair. Their goal is to eat the same total weight of candies. What is the most number of candies they can eat in total?
Input Specification: The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The first line of each test case contains an integer $$$n$$$ ($$$1 \leq n \leq 2\cdot10^5$$$) — the number of candies on the table. The second line of each test case contains $$$n$$$ integers $$$w_1, w_2, \dots, w_n$$$ ($$$1 \leq w_i \leq 10^4$$$) — the weights of candies from left to right. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2\cdot10^5$$$.
Output Specification: For each test case, print a single integer — the maximum number of candies Alice and Bob can eat in total while satisfying the condition.
Notes: NoteFor the first test case, Alice will eat one candy from the left and Bob will eat one candy from the right. There is no better way for them to eat the same total amount of weight. The answer is $$$2$$$ because they eat two candies in total.For the second test case, Alice will eat the first three candies from the left (with total weight $$$7$$$) and Bob will eat the first three candies from the right (with total weight $$$7$$$). They cannot eat more candies since all the candies have been eaten, so the answer is $$$6$$$ (because they eat six candies in total).For the third test case, there is no way Alice and Bob will eat the same non-zero weight so the answer is $$$0$$$.For the fourth test case, Alice will eat candies with weights $$$[7, 3, 20]$$$ and Bob will eat candies with weights $$$[10, 8, 11, 1]$$$, they each eat $$$30$$$ weight. There is no better partition so the answer is $$$7$$$.
Code:
def read():
return int(input())
def readline():
return list(map(int,input().split()))
def solve():
n=read()
arr=readline()
ans,cur=0,0
a,suma=-1,0
b,sumb=n,0
while True:
if a>=b: break
elif suma>sumb:
b-=1
sumb+=arr[b]
cur+=1
elif suma<sumb:
# TODO: Your code here
else :
ans=cur
a+=1
b-=1
suma+=arr[a]
sumb+=arr[b]
cur+=2
print(ans)
if __name__ == "__main__":
T=read()
for i in range(T):
solve()
|
def read():
return int(input())
def readline():
return list(map(int,input().split()))
def solve():
n=read()
arr=readline()
ans,cur=0,0
a,suma=-1,0
b,sumb=n,0
while True:
if a>=b: break
elif suma>sumb:
b-=1
sumb+=arr[b]
cur+=1
elif suma<sumb:
{{completion}}
else :
ans=cur
a+=1
b-=1
suma+=arr[a]
sumb+=arr[b]
cur+=2
print(ans)
if __name__ == "__main__":
T=read()
for i in range(T):
solve()
|
a+=1
suma+=arr[a]
cur+=1
|
[{"input": "4\n3\n10 20 10\n6\n2 1 4 2 4 1\n5\n1 2 4 8 16\n9\n7 3 20 5 15 1 11 8 10", "output": ["2\n6\n0\n7"]}]
|
block_completion_000810
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: There is a grid with $$$n$$$ rows and $$$m$$$ columns, and three types of cells: An empty cell, denoted with '.'. A stone, denoted with '*'. An obstacle, denoted with the lowercase Latin letter 'o'. All stones fall down until they meet the floor (the bottom row), an obstacle, or other stone which is already immovable. (In other words, all the stones just fall down as long as they can fall.)Simulate the process. What does the resulting grid look like?
Input Specification: The input consists of multiple test cases. The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 100$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$m$$$ ($$$1 \leq n, m \leq 50$$$) — the number of rows and the number of columns in the grid, respectively. Then $$$n$$$ lines follow, each containing $$$m$$$ characters. Each of these characters is either '.', '*', or 'o' — an empty cell, a stone, or an obstacle, respectively.
Output Specification: For each test case, output a grid with $$$n$$$ rows and $$$m$$$ columns, showing the result of the process. You don't need to output a new line after each test, it is in the samples just for clarity.
Code:
for _ in range(int(input())):
n, _ = map(int, input().split())
a = map("".join, zip(*(input() for _ in range(n))))
a = ("o".join("".join(sorted(y, reverse=True)) for y in x.split("o")) for x in a)
for x in zip(*a):
# TODO: Your code here
|
for _ in range(int(input())):
n, _ = map(int, input().split())
a = map("".join, zip(*(input() for _ in range(n))))
a = ("o".join("".join(sorted(y, reverse=True)) for y in x.split("o")) for x in a)
for x in zip(*a):
{{completion}}
|
print("".join(x))
|
[{"input": "3\n6 10\n.*.*....*.\n.*.......*\n...o....o.\n.*.*....*.\n..........\n.o......o*\n2 9\n...***ooo\n.*o.*o.*o\n5 5\n*****\n*....\n*****\n....*\n*****", "output": ["..........\n...*....*.\n.*.o....o.\n.*........\n.*......**\n.o.*....o*\n\n....**ooo\n.*o**o.*o\n\n.....\n*...*\n*****\n*****\n*****"]}]
|
block_completion_000845
|
block
|
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