message stringlengths 2 65.1k | message_type stringclasses 2
values | message_id int64 0 1 | conversation_id int64 0 108k | cluster float64 14 14 | __index_level_0__ int64 0 217k |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
The only difference between easy and hard versions is the constraints.
Vova likes pictures with kittens. The news feed in the social network he uses can be represented as an array of n consecutive pictures (with kittens, of course). Vova likes all these pictures, but some are more beautiful than the others: the i-th picture has beauty a_i.
Vova wants to repost exactly x pictures in such a way that:
* each segment of the news feed of at least k consecutive pictures has at least one picture reposted by Vova;
* the sum of beauty values of reposted pictures is maximum possible.
For example, if k=1 then Vova has to repost all the pictures in the news feed. If k=2 then Vova can skip some pictures, but between every pair of consecutive pictures Vova has to repost at least one of them.
Your task is to calculate the maximum possible sum of values of reposted pictures if Vova follows conditions described above, or say that there is no way to satisfy all conditions.
Input
The first line of the input contains three integers n, k and x (1 β€ k, x β€ n β€ 200) β the number of pictures in the news feed, the minimum length of segment with at least one repost in it and the number of pictures Vova is ready to repost.
The second line of the input contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ 10^9), where a_i is the beauty of the i-th picture.
Output
Print -1 if there is no way to repost some pictures to satisfy all the conditions in the problem statement.
Otherwise print one integer β the maximum sum of values of reposted pictures if Vova follows conditions described in the problem statement.
Examples
Input
5 2 3
5 1 3 10 1
Output
18
Input
6 1 5
10 30 30 70 10 10
Output
-1
Input
4 3 1
1 100 1 1
Output
100 | instruction | 0 | 44,633 | 14 | 89,266 |
Tags: dp
Correct Solution:
```
# -*- coding: utf-8 -*-
import sys
def input(): return sys.stdin.readline().strip()
def list2d(a, b, c): return [[c] * b for i in range(a)]
def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
def list4d(a, b, c, d, e): return [[[[e] * d for j in range(c)] for j in range(b)] for i in range(a)]
def ceil(x, y=1): return int(-(-x // y))
def INT(): return int(input())
def MAP(): return map(int, input().split())
def LIST(N=None): return list(MAP()) if N is None else [INT() for i in range(N)]
def Yes(): print('Yes')
def No(): print('No')
def YES(): print('YES')
def NO(): print('NO')
INF = 10 ** 18
MOD = 10 ** 9 + 7
N, K, X = MAP()
A = [0] + LIST()
dp = list2d(X+1, N+1, -1)
dp[0][0] = 0
for i in range(X):
for j in range(N):
if dp[i][j] == -1:
continue
for k in range(j+1, min(j+K+1, N+1)):
dp[i+1][k] = max(dp[i+1][k], dp[i][j] + A[k])
ans = max(dp[X][-K:])
print(ans)
``` | output | 1 | 44,633 | 14 | 89,267 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The only difference between easy and hard versions is the constraints.
Vova likes pictures with kittens. The news feed in the social network he uses can be represented as an array of n consecutive pictures (with kittens, of course). Vova likes all these pictures, but some are more beautiful than the others: the i-th picture has beauty a_i.
Vova wants to repost exactly x pictures in such a way that:
* each segment of the news feed of at least k consecutive pictures has at least one picture reposted by Vova;
* the sum of beauty values of reposted pictures is maximum possible.
For example, if k=1 then Vova has to repost all the pictures in the news feed. If k=2 then Vova can skip some pictures, but between every pair of consecutive pictures Vova has to repost at least one of them.
Your task is to calculate the maximum possible sum of values of reposted pictures if Vova follows conditions described above, or say that there is no way to satisfy all conditions.
Input
The first line of the input contains three integers n, k and x (1 β€ k, x β€ n β€ 200) β the number of pictures in the news feed, the minimum length of segment with at least one repost in it and the number of pictures Vova is ready to repost.
The second line of the input contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ 10^9), where a_i is the beauty of the i-th picture.
Output
Print -1 if there is no way to repost some pictures to satisfy all the conditions in the problem statement.
Otherwise print one integer β the maximum sum of values of reposted pictures if Vova follows conditions described in the problem statement.
Examples
Input
5 2 3
5 1 3 10 1
Output
18
Input
6 1 5
10 30 30 70 10 10
Output
-1
Input
4 3 1
1 100 1 1
Output
100 | instruction | 0 | 44,634 | 14 | 89,268 |
Tags: dp
Correct Solution:
```
N,K,X = map(int,input().split())
arr = [0] + [int(i) for i in input().split()]
dp = [[float('-inf') for _ in range(X+1)] for _ in range(N+1)]
dp[0][X] = 0
for i in range(1,N+1):
for x in range(X):
for p in range(1,K+1):
if i - p < 0:
break
if dp[i-p][x+1] < 0:
continue
dp[i][x] = max(dp[i][x], dp[i-p][x+1] + arr[i])
ans = max([max(v) for v in dp[N-K+1:]])
if ans < 0:
print(-1)
else:
print(ans)
``` | output | 1 | 44,634 | 14 | 89,269 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The only difference between easy and hard versions is the constraints.
Vova likes pictures with kittens. The news feed in the social network he uses can be represented as an array of n consecutive pictures (with kittens, of course). Vova likes all these pictures, but some are more beautiful than the others: the i-th picture has beauty a_i.
Vova wants to repost exactly x pictures in such a way that:
* each segment of the news feed of at least k consecutive pictures has at least one picture reposted by Vova;
* the sum of beauty values of reposted pictures is maximum possible.
For example, if k=1 then Vova has to repost all the pictures in the news feed. If k=2 then Vova can skip some pictures, but between every pair of consecutive pictures Vova has to repost at least one of them.
Your task is to calculate the maximum possible sum of values of reposted pictures if Vova follows conditions described above, or say that there is no way to satisfy all conditions.
Input
The first line of the input contains three integers n, k and x (1 β€ k, x β€ n β€ 200) β the number of pictures in the news feed, the minimum length of segment with at least one repost in it and the number of pictures Vova is ready to repost.
The second line of the input contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ 10^9), where a_i is the beauty of the i-th picture.
Output
Print -1 if there is no way to repost some pictures to satisfy all the conditions in the problem statement.
Otherwise print one integer β the maximum sum of values of reposted pictures if Vova follows conditions described in the problem statement.
Examples
Input
5 2 3
5 1 3 10 1
Output
18
Input
6 1 5
10 30 30 70 10 10
Output
-1
Input
4 3 1
1 100 1 1
Output
100 | instruction | 0 | 44,635 | 14 | 89,270 |
Tags: dp
Correct Solution:
```
# -*- coding: utf-8 -*-
import sys
def input(): return sys.stdin.readline().strip()
def list2d(a, b, c): return [[c] * b for i in range(a)]
def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
def list4d(a, b, c, d, e): return [[[[e] * d for j in range(c)] for j in range(b)] for i in range(a)]
def ceil(x, y=1): return int(-(-x // y))
def INT(): return int(input())
def MAP(): return map(int, input().split())
def LIST(N=None): return list(MAP()) if N is None else [INT() for i in range(N)]
def Yes(): print('Yes')
def No(): print('No')
def YES(): print('YES')
def NO(): print('NO')
INF = 10 ** 18
MOD = 10 ** 9 + 7
N, K, X = MAP()
A = LIST()
dp = list2d(X+1, N+1, -1)
dp[0][0] = 0
for i in range(X):
for j in range(N):
if dp[i][j] == -1:
continue
for k in range(j, min(j+K, N)):
dp[i+1][k+1] = max(dp[i+1][k+1], dp[i][j] + A[k])
ans = max(dp[X][-K:])
print(ans)
``` | output | 1 | 44,635 | 14 | 89,271 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The only difference between easy and hard versions is the constraints.
Vova likes pictures with kittens. The news feed in the social network he uses can be represented as an array of n consecutive pictures (with kittens, of course). Vova likes all these pictures, but some are more beautiful than the others: the i-th picture has beauty a_i.
Vova wants to repost exactly x pictures in such a way that:
* each segment of the news feed of at least k consecutive pictures has at least one picture reposted by Vova;
* the sum of beauty values of reposted pictures is maximum possible.
For example, if k=1 then Vova has to repost all the pictures in the news feed. If k=2 then Vova can skip some pictures, but between every pair of consecutive pictures Vova has to repost at least one of them.
Your task is to calculate the maximum possible sum of values of reposted pictures if Vova follows conditions described above, or say that there is no way to satisfy all conditions.
Input
The first line of the input contains three integers n, k and x (1 β€ k, x β€ n β€ 200) β the number of pictures in the news feed, the minimum length of segment with at least one repost in it and the number of pictures Vova is ready to repost.
The second line of the input contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ 10^9), where a_i is the beauty of the i-th picture.
Output
Print -1 if there is no way to repost some pictures to satisfy all the conditions in the problem statement.
Otherwise print one integer β the maximum sum of values of reposted pictures if Vova follows conditions described in the problem statement.
Examples
Input
5 2 3
5 1 3 10 1
Output
18
Input
6 1 5
10 30 30 70 10 10
Output
-1
Input
4 3 1
1 100 1 1
Output
100 | instruction | 0 | 44,636 | 14 | 89,272 |
Tags: dp
Correct Solution:
```
from functools import lru_cache
n, k, x = map(int, input().split())
arr = list(map(int, input().split()))
@lru_cache(maxsize=50000)
def dp(i,y):
if i>=0 and i<k and y==0:
return 0
elif i < 0 or y<= 0:
return -1000000000001
return max(dp(i-delta-1, y-1) + arr[i-delta-1] for delta in range(min(k,i+1)))
print(max(dp(n,x),-1))
``` | output | 1 | 44,636 | 14 | 89,273 |
Evaluate the correctness of the submitted Python 2 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between easy and hard versions is the constraints.
Vova likes pictures with kittens. The news feed in the social network he uses can be represented as an array of n consecutive pictures (with kittens, of course). Vova likes all these pictures, but some are more beautiful than the others: the i-th picture has beauty a_i.
Vova wants to repost exactly x pictures in such a way that:
* each segment of the news feed of at least k consecutive pictures has at least one picture reposted by Vova;
* the sum of beauty values of reposted pictures is maximum possible.
For example, if k=1 then Vova has to repost all the pictures in the news feed. If k=2 then Vova can skip some pictures, but between every pair of consecutive pictures Vova has to repost at least one of them.
Your task is to calculate the maximum possible sum of values of reposted pictures if Vova follows conditions described above, or say that there is no way to satisfy all conditions.
Input
The first line of the input contains three integers n, k and x (1 β€ k, x β€ n β€ 200) β the number of pictures in the news feed, the minimum length of segment with at least one repost in it and the number of pictures Vova is ready to repost.
The second line of the input contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ 10^9), where a_i is the beauty of the i-th picture.
Output
Print -1 if there is no way to repost some pictures to satisfy all the conditions in the problem statement.
Otherwise print one integer β the maximum sum of values of reposted pictures if Vova follows conditions described in the problem statement.
Examples
Input
5 2 3
5 1 3 10 1
Output
18
Input
6 1 5
10 30 30 70 10 10
Output
-1
Input
4 3 1
1 100 1 1
Output
100
Submitted Solution:
```
from sys import stdin, stdout
from collections import Counter, defaultdict
from itertools import permutations, combinations
raw_input = stdin.readline
pr = stdout.write
mod=10**9+7
def ni():
return int(raw_input())
def li():
return map(int,raw_input().split())
def pn(n):
stdout.write(str(n)+'\n')
def pa(arr):
pr(' '.join(map(str,arr))+'\n')
# fast read function for total integer input
def inp():
# this function returns whole input of
# space/line seperated integers
# Use Ctrl+D to flush stdin.
return map(int,stdin.read().split())
range = xrange # not for python 3.0+
n,k,x=li()
l=li()
mn=-10**9
dp=[[mn for i in range(x+1)] for j in range(n+1)]
ans=mn
dp[0][x]=0
for i in range(1,n+1):
for j in range(x):
for k1 in range(1,k+1):
if i-k1<0:
break
if dp[i-k1][j+1]==mn:
continue
else:
dp[i][j]=max(dp[i][j],dp[i-k1][j+1]+l[i-1])
for i in range(n-k+1,n+1):
ans=max(ans,max(dp[i]))
if ans==mn:
ans=-1
pn(ans)
``` | instruction | 0 | 44,637 | 14 | 89,274 |
Yes | output | 1 | 44,637 | 14 | 89,275 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between easy and hard versions is the constraints.
Vova likes pictures with kittens. The news feed in the social network he uses can be represented as an array of n consecutive pictures (with kittens, of course). Vova likes all these pictures, but some are more beautiful than the others: the i-th picture has beauty a_i.
Vova wants to repost exactly x pictures in such a way that:
* each segment of the news feed of at least k consecutive pictures has at least one picture reposted by Vova;
* the sum of beauty values of reposted pictures is maximum possible.
For example, if k=1 then Vova has to repost all the pictures in the news feed. If k=2 then Vova can skip some pictures, but between every pair of consecutive pictures Vova has to repost at least one of them.
Your task is to calculate the maximum possible sum of values of reposted pictures if Vova follows conditions described above, or say that there is no way to satisfy all conditions.
Input
The first line of the input contains three integers n, k and x (1 β€ k, x β€ n β€ 200) β the number of pictures in the news feed, the minimum length of segment with at least one repost in it and the number of pictures Vova is ready to repost.
The second line of the input contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ 10^9), where a_i is the beauty of the i-th picture.
Output
Print -1 if there is no way to repost some pictures to satisfy all the conditions in the problem statement.
Otherwise print one integer β the maximum sum of values of reposted pictures if Vova follows conditions described in the problem statement.
Examples
Input
5 2 3
5 1 3 10 1
Output
18
Input
6 1 5
10 30 30 70 10 10
Output
-1
Input
4 3 1
1 100 1 1
Output
100
Submitted Solution:
```
# by the authority of GOD author: manhar singh sachdev #
import os,sys
from io import BytesIO, IOBase
def main():
inf = -10**20
n,K,x = map(int,input().split())
a = list(map(int,input().split()))
dp = [[[inf]*K for _ in range(x+1)] for _ in range(n)]
# distance ; pictures reposted ; index
for i in range(K):
dp[i][1][0] = a[i]
for i in range(n-1):
for j in range(x+1):
for k in range(K):
if j != x:
dp[i+1][j+1][0] = max(dp[i+1][j+1][0],dp[i][j][k]+a[i+1])
if k != K-1:
dp[i+1][j][k+1] = max(dp[i+1][j][k+1],dp[i][j][k])
ans = max(dp[n-1][x])
print(ans if ans > 0 else -1)
# Fast IO Region
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
if __name__ == "__main__":
main()
``` | instruction | 0 | 44,638 | 14 | 89,276 |
Yes | output | 1 | 44,638 | 14 | 89,277 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between easy and hard versions is the constraints.
Vova likes pictures with kittens. The news feed in the social network he uses can be represented as an array of n consecutive pictures (with kittens, of course). Vova likes all these pictures, but some are more beautiful than the others: the i-th picture has beauty a_i.
Vova wants to repost exactly x pictures in such a way that:
* each segment of the news feed of at least k consecutive pictures has at least one picture reposted by Vova;
* the sum of beauty values of reposted pictures is maximum possible.
For example, if k=1 then Vova has to repost all the pictures in the news feed. If k=2 then Vova can skip some pictures, but between every pair of consecutive pictures Vova has to repost at least one of them.
Your task is to calculate the maximum possible sum of values of reposted pictures if Vova follows conditions described above, or say that there is no way to satisfy all conditions.
Input
The first line of the input contains three integers n, k and x (1 β€ k, x β€ n β€ 200) β the number of pictures in the news feed, the minimum length of segment with at least one repost in it and the number of pictures Vova is ready to repost.
The second line of the input contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ 10^9), where a_i is the beauty of the i-th picture.
Output
Print -1 if there is no way to repost some pictures to satisfy all the conditions in the problem statement.
Otherwise print one integer β the maximum sum of values of reposted pictures if Vova follows conditions described in the problem statement.
Examples
Input
5 2 3
5 1 3 10 1
Output
18
Input
6 1 5
10 30 30 70 10 10
Output
-1
Input
4 3 1
1 100 1 1
Output
100
Submitted Solution:
```
import sys
input = sys.stdin.readline
n, k, x = map(int, input().split())
a = list(map(int, input().split()))
dp = [-1] * (n + 1)
dp[0] = 0
for c in range(1, x + 1):
dp0 = [-1] * (n + 1)
for i in range(c, n + 1):
ma = -1
for j in range(1, min(i, k) + 1):
ma = max(ma, dp[i - j])
if ma == -1:
break
dp0[i] = ma + a[i - 1]
dp = dp0
ans = -1
for i in range(k):
ans = max(ans, dp[n - i])
print(ans)
``` | instruction | 0 | 44,639 | 14 | 89,278 |
Yes | output | 1 | 44,639 | 14 | 89,279 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between easy and hard versions is the constraints.
Vova likes pictures with kittens. The news feed in the social network he uses can be represented as an array of n consecutive pictures (with kittens, of course). Vova likes all these pictures, but some are more beautiful than the others: the i-th picture has beauty a_i.
Vova wants to repost exactly x pictures in such a way that:
* each segment of the news feed of at least k consecutive pictures has at least one picture reposted by Vova;
* the sum of beauty values of reposted pictures is maximum possible.
For example, if k=1 then Vova has to repost all the pictures in the news feed. If k=2 then Vova can skip some pictures, but between every pair of consecutive pictures Vova has to repost at least one of them.
Your task is to calculate the maximum possible sum of values of reposted pictures if Vova follows conditions described above, or say that there is no way to satisfy all conditions.
Input
The first line of the input contains three integers n, k and x (1 β€ k, x β€ n β€ 200) β the number of pictures in the news feed, the minimum length of segment with at least one repost in it and the number of pictures Vova is ready to repost.
The second line of the input contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ 10^9), where a_i is the beauty of the i-th picture.
Output
Print -1 if there is no way to repost some pictures to satisfy all the conditions in the problem statement.
Otherwise print one integer β the maximum sum of values of reposted pictures if Vova follows conditions described in the problem statement.
Examples
Input
5 2 3
5 1 3 10 1
Output
18
Input
6 1 5
10 30 30 70 10 10
Output
-1
Input
4 3 1
1 100 1 1
Output
100
Submitted Solution:
```
n, k, x = map(int, input().split())
a = [None] + list(map(int, input().split()))
lo, hi = 0, 10 ** 9 * 5000
q = [None] * (n + 1)
def get(mid):
f, r = 0, 0
q[0] = 0, 0, 0
for i in range(1, n + 1):
if q[r][2] == i - k - 1: r += 1
cur = q[r][0] + a[i] - mid, q[r][1] + 1, i
while r <= f and q[f] <= cur: f -= 1
f += 1
q[f] = cur
if q[r][2] == n - k: r += 1
return q[r]
while lo < hi:
mid = (lo + hi + 1) >> 1
_, cnt, _ = get(mid)
if cnt >= x:
lo = mid
else:
hi = mid - 1
sm, _, _ = get(lo)
ans = max(-1, sm + x * lo)
print(ans)
``` | instruction | 0 | 44,640 | 14 | 89,280 |
Yes | output | 1 | 44,640 | 14 | 89,281 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between easy and hard versions is the constraints.
Vova likes pictures with kittens. The news feed in the social network he uses can be represented as an array of n consecutive pictures (with kittens, of course). Vova likes all these pictures, but some are more beautiful than the others: the i-th picture has beauty a_i.
Vova wants to repost exactly x pictures in such a way that:
* each segment of the news feed of at least k consecutive pictures has at least one picture reposted by Vova;
* the sum of beauty values of reposted pictures is maximum possible.
For example, if k=1 then Vova has to repost all the pictures in the news feed. If k=2 then Vova can skip some pictures, but between every pair of consecutive pictures Vova has to repost at least one of them.
Your task is to calculate the maximum possible sum of values of reposted pictures if Vova follows conditions described above, or say that there is no way to satisfy all conditions.
Input
The first line of the input contains three integers n, k and x (1 β€ k, x β€ n β€ 200) β the number of pictures in the news feed, the minimum length of segment with at least one repost in it and the number of pictures Vova is ready to repost.
The second line of the input contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ 10^9), where a_i is the beauty of the i-th picture.
Output
Print -1 if there is no way to repost some pictures to satisfy all the conditions in the problem statement.
Otherwise print one integer β the maximum sum of values of reposted pictures if Vova follows conditions described in the problem statement.
Examples
Input
5 2 3
5 1 3 10 1
Output
18
Input
6 1 5
10 30 30 70 10 10
Output
-1
Input
4 3 1
1 100 1 1
Output
100
Submitted Solution:
```
import os
import sys
from io import BytesIO, IOBase
from types import GeneratorType
from collections import defaultdict
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
n,k,m=map(int,input().split())
a=[0]+list(map(int,input().split()))
dp=[[[-float("inf"),-float("inf")] for num in range(m+1)] for i in range(n+1)]
dp[0][0]=[0,0]
for i in range(1,n+1):
for num in range(m+1):
for j in range(i-k,i):
if (j>=0 and num>=1):
dp[i][num][0] = max(dp[i][num][0], dp[j][num][1])
dp[i][num][1]=max(dp[i][num][1],dp[j][num-1][1]+a[i])
ans=-float("inf")
j=n
while(j>(max(0,n-k))):
ans=max(ans,dp[j][m][1])
j+=-1
if (ans==-float("inf")):
print(-1)
else:
print(ans)
``` | instruction | 0 | 44,641 | 14 | 89,282 |
Yes | output | 1 | 44,641 | 14 | 89,283 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between easy and hard versions is the constraints.
Vova likes pictures with kittens. The news feed in the social network he uses can be represented as an array of n consecutive pictures (with kittens, of course). Vova likes all these pictures, but some are more beautiful than the others: the i-th picture has beauty a_i.
Vova wants to repost exactly x pictures in such a way that:
* each segment of the news feed of at least k consecutive pictures has at least one picture reposted by Vova;
* the sum of beauty values of reposted pictures is maximum possible.
For example, if k=1 then Vova has to repost all the pictures in the news feed. If k=2 then Vova can skip some pictures, but between every pair of consecutive pictures Vova has to repost at least one of them.
Your task is to calculate the maximum possible sum of values of reposted pictures if Vova follows conditions described above, or say that there is no way to satisfy all conditions.
Input
The first line of the input contains three integers n, k and x (1 β€ k, x β€ n β€ 200) β the number of pictures in the news feed, the minimum length of segment with at least one repost in it and the number of pictures Vova is ready to repost.
The second line of the input contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ 10^9), where a_i is the beauty of the i-th picture.
Output
Print -1 if there is no way to repost some pictures to satisfy all the conditions in the problem statement.
Otherwise print one integer β the maximum sum of values of reposted pictures if Vova follows conditions described in the problem statement.
Examples
Input
5 2 3
5 1 3 10 1
Output
18
Input
6 1 5
10 30 30 70 10 10
Output
-1
Input
4 3 1
1 100 1 1
Output
100
Submitted Solution:
```
import sys
n, k, x = (int(x) for x in sys.stdin.readline().split(' '))
a = [int(x) for x in sys.stdin.readline().split(' ')]
gm = 0
for i in range(1, k+1):
m = max(a[0:i])
for j in range(i, n, k):
m += max(a[j:j+k])
if gm < m:
gm = m
print(gm)
``` | instruction | 0 | 44,642 | 14 | 89,284 |
No | output | 1 | 44,642 | 14 | 89,285 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between easy and hard versions is the constraints.
Vova likes pictures with kittens. The news feed in the social network he uses can be represented as an array of n consecutive pictures (with kittens, of course). Vova likes all these pictures, but some are more beautiful than the others: the i-th picture has beauty a_i.
Vova wants to repost exactly x pictures in such a way that:
* each segment of the news feed of at least k consecutive pictures has at least one picture reposted by Vova;
* the sum of beauty values of reposted pictures is maximum possible.
For example, if k=1 then Vova has to repost all the pictures in the news feed. If k=2 then Vova can skip some pictures, but between every pair of consecutive pictures Vova has to repost at least one of them.
Your task is to calculate the maximum possible sum of values of reposted pictures if Vova follows conditions described above, or say that there is no way to satisfy all conditions.
Input
The first line of the input contains three integers n, k and x (1 β€ k, x β€ n β€ 200) β the number of pictures in the news feed, the minimum length of segment with at least one repost in it and the number of pictures Vova is ready to repost.
The second line of the input contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ 10^9), where a_i is the beauty of the i-th picture.
Output
Print -1 if there is no way to repost some pictures to satisfy all the conditions in the problem statement.
Otherwise print one integer β the maximum sum of values of reposted pictures if Vova follows conditions described in the problem statement.
Examples
Input
5 2 3
5 1 3 10 1
Output
18
Input
6 1 5
10 30 30 70 10 10
Output
-1
Input
4 3 1
1 100 1 1
Output
100
Submitted Solution:
```
def maxi(x, y):
ap = -1
ai = -1
while x < y + 1:
if ap < a[x] and a[x] != -1 or a[x] < 0:
if a[x] < 0:
ap = -a[x]
ai = x
else:
ap = a[x]
ai = x
x += 1
if a[ai] < 0:
return -1
return ai
n, k, x = map(int, input().split())
a = [int(x) for x in input().split()]
if n == 55:
print(12379557167)
exit()
if n // k > x:
print(-1)
exit()
ri = k - 1
su = 0
co = 0
while ri < n:
z = maxi(ri - k + 1, ri)
if z != -1:
# zx = max(a[ri - k + 1:ri])
co += 1
su += a[z]
a[z] *= -1
ri += 1
print(su)
``` | instruction | 0 | 44,643 | 14 | 89,286 |
No | output | 1 | 44,643 | 14 | 89,287 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between easy and hard versions is the constraints.
Vova likes pictures with kittens. The news feed in the social network he uses can be represented as an array of n consecutive pictures (with kittens, of course). Vova likes all these pictures, but some are more beautiful than the others: the i-th picture has beauty a_i.
Vova wants to repost exactly x pictures in such a way that:
* each segment of the news feed of at least k consecutive pictures has at least one picture reposted by Vova;
* the sum of beauty values of reposted pictures is maximum possible.
For example, if k=1 then Vova has to repost all the pictures in the news feed. If k=2 then Vova can skip some pictures, but between every pair of consecutive pictures Vova has to repost at least one of them.
Your task is to calculate the maximum possible sum of values of reposted pictures if Vova follows conditions described above, or say that there is no way to satisfy all conditions.
Input
The first line of the input contains three integers n, k and x (1 β€ k, x β€ n β€ 200) β the number of pictures in the news feed, the minimum length of segment with at least one repost in it and the number of pictures Vova is ready to repost.
The second line of the input contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ 10^9), where a_i is the beauty of the i-th picture.
Output
Print -1 if there is no way to repost some pictures to satisfy all the conditions in the problem statement.
Otherwise print one integer β the maximum sum of values of reposted pictures if Vova follows conditions described in the problem statement.
Examples
Input
5 2 3
5 1 3 10 1
Output
18
Input
6 1 5
10 30 30 70 10 10
Output
-1
Input
4 3 1
1 100 1 1
Output
100
Submitted Solution:
```
# -*- coding: utf-8 -*-
import math
import collections
import bisect
import heapq
import time
import itertools
import sys
"""
created by shhuan at 2018/11/17 00:27
"""
N, K, X = map(int, input().split())
A = [int(x) for x in input().split()]
if N // K > X:
print(-1)
exit(0)
dp = [[0 for _ in range(X+1)] for _ in range(N+1)]
dp[0][1] = A[0]
for i in range(1, N):
for k in range(1, min(i, X)+1):
dpik = 0
for j in range(i-K, i):
dpik = max(dpik, dp[j][k-1] + A[i])
dp[i][k] = dpik
print(max([x[X] for x in dp]))
``` | instruction | 0 | 44,644 | 14 | 89,288 |
No | output | 1 | 44,644 | 14 | 89,289 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between easy and hard versions is the constraints.
Vova likes pictures with kittens. The news feed in the social network he uses can be represented as an array of n consecutive pictures (with kittens, of course). Vova likes all these pictures, but some are more beautiful than the others: the i-th picture has beauty a_i.
Vova wants to repost exactly x pictures in such a way that:
* each segment of the news feed of at least k consecutive pictures has at least one picture reposted by Vova;
* the sum of beauty values of reposted pictures is maximum possible.
For example, if k=1 then Vova has to repost all the pictures in the news feed. If k=2 then Vova can skip some pictures, but between every pair of consecutive pictures Vova has to repost at least one of them.
Your task is to calculate the maximum possible sum of values of reposted pictures if Vova follows conditions described above, or say that there is no way to satisfy all conditions.
Input
The first line of the input contains three integers n, k and x (1 β€ k, x β€ n β€ 200) β the number of pictures in the news feed, the minimum length of segment with at least one repost in it and the number of pictures Vova is ready to repost.
The second line of the input contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ 10^9), where a_i is the beauty of the i-th picture.
Output
Print -1 if there is no way to repost some pictures to satisfy all the conditions in the problem statement.
Otherwise print one integer β the maximum sum of values of reposted pictures if Vova follows conditions described in the problem statement.
Examples
Input
5 2 3
5 1 3 10 1
Output
18
Input
6 1 5
10 30 30 70 10 10
Output
-1
Input
4 3 1
1 100 1 1
Output
100
Submitted Solution:
```
import sys
from math import *
def minp():
return sys.stdin.readline().strip()
def mint():
return int(minp())
def mints():
return map(int, minp().split())
n, x, k = mints()
a = list(mints())
dp = [[None]*(n+1) for i in range(k+1)]
for i in range(0,x):
dp[1][i] = a[i]
for kk in range(2,k+1):
d = dp[kk]
p = dp[kk-1]
for nn in range(n):
m = None
for i in range(max(0,nn-2*x+1),nn):
if p[i] != None:
if m == None:
m = p[i]
else:
m = max(m, p[i])
if m != None:
d[nn] = m + a[nn]
m = -1
d = dp[k]
for i in range(n-x,n):
if d[i] != None:
m = max(m, d[i])
print(m)
``` | instruction | 0 | 44,645 | 14 | 89,290 |
No | output | 1 | 44,645 | 14 | 89,291 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In some social network, there are n users communicating with each other in m groups of friends. Let's analyze the process of distributing some news between users.
Initially, some user x receives the news from some source. Then he or she sends the news to his or her friends (two users are friends if there is at least one group such that both of them belong to this group). Friends continue sending the news to their friends, and so on. The process ends when there is no pair of friends such that one of them knows the news, and another one doesn't know.
For each user x you have to determine what is the number of users that will know the news if initially only user x starts distributing it.
Input
The first line contains two integers n and m (1 β€ n, m β€ 5 β
10^5) β the number of users and the number of groups of friends, respectively.
Then m lines follow, each describing a group of friends. The i-th line begins with integer k_i (0 β€ k_i β€ n) β the number of users in the i-th group. Then k_i distinct integers follow, denoting the users belonging to the i-th group.
It is guaranteed that β _{i = 1}^{m} k_i β€ 5 β
10^5.
Output
Print n integers. The i-th integer should be equal to the number of users that will know the news if user i starts distributing it.
Example
Input
7 5
3 2 5 4
0
2 1 2
1 1
2 6 7
Output
4 4 1 4 4 2 2 | instruction | 0 | 44,652 | 14 | 89,304 |
Tags: dfs and similar, dsu, graphs
Correct Solution:
```
import sys; input = sys.stdin.readline
class Node:
def __init__(self, val):
self.parent = self
self.size = 1
def union(x, y):
xRoot, yRoot = find(x), find(y)
if xRoot == yRoot:
return
if xRoot.size >= yRoot.size:
yRoot.parent = xRoot
xRoot.size += yRoot.size
else:
xRoot.parent = yRoot
yRoot.size += xRoot.size
def find(x):
while x.parent != x:
x = x.parent
return x
class Problem:
def __init__(self, In=sys.stdin, Out=sys.stdout):
pass
def solve(self):
print(self._solve())
def _solve(self):
n, m = [int(item) for item in sys.stdin.readline().strip().split()]
node = [Node(x) for x in range(n + 1)]
groups = [[int(item) for item in sys.stdin.readline().strip().split()]
for _ in range(m)]
for g in range(m):
group = groups[g]
k = group[0]
if k == 0:
continue
for i in range(2, k + 1):
union(node[group[i]], node[group[i - 1]])
ans = [find(node[x]).size for x in range(1, n + 1)]
return ' '.join(str(x) for x in ans)
def main():
problem = Problem()
problem.solve()
if __name__ == "__main__":
main()
``` | output | 1 | 44,652 | 14 | 89,305 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In some social network, there are n users communicating with each other in m groups of friends. Let's analyze the process of distributing some news between users.
Initially, some user x receives the news from some source. Then he or she sends the news to his or her friends (two users are friends if there is at least one group such that both of them belong to this group). Friends continue sending the news to their friends, and so on. The process ends when there is no pair of friends such that one of them knows the news, and another one doesn't know.
For each user x you have to determine what is the number of users that will know the news if initially only user x starts distributing it.
Input
The first line contains two integers n and m (1 β€ n, m β€ 5 β
10^5) β the number of users and the number of groups of friends, respectively.
Then m lines follow, each describing a group of friends. The i-th line begins with integer k_i (0 β€ k_i β€ n) β the number of users in the i-th group. Then k_i distinct integers follow, denoting the users belonging to the i-th group.
It is guaranteed that β _{i = 1}^{m} k_i β€ 5 β
10^5.
Output
Print n integers. The i-th integer should be equal to the number of users that will know the news if user i starts distributing it.
Example
Input
7 5
3 2 5 4
0
2 1 2
1 1
2 6 7
Output
4 4 1 4 4 2 2 | instruction | 0 | 44,653 | 14 | 89,306 |
Tags: dfs and similar, dsu, graphs
Correct Solution:
```
import sys
input = sys.stdin.readline
n,m = map(int,input().split())
A = list(range(n+1))
def find(x):
while A[x] != x:
A[x] = A[A[x]]
x = A[x]
return x
for _ in range(m):
line = input().split()[1:]
if line:
V = [int(j) for j in line]
k = find(V[0])
for v in V[1:]:
A[find(v)] = k
count = [0] * (n + 1)
for i in range(1, n + 1):
count[find(i)] += 1
print(' '.join(map(str, [count[find(i)] for i in range(1,n+1)])))
``` | output | 1 | 44,653 | 14 | 89,307 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In some social network, there are n users communicating with each other in m groups of friends. Let's analyze the process of distributing some news between users.
Initially, some user x receives the news from some source. Then he or she sends the news to his or her friends (two users are friends if there is at least one group such that both of them belong to this group). Friends continue sending the news to their friends, and so on. The process ends when there is no pair of friends such that one of them knows the news, and another one doesn't know.
For each user x you have to determine what is the number of users that will know the news if initially only user x starts distributing it.
Input
The first line contains two integers n and m (1 β€ n, m β€ 5 β
10^5) β the number of users and the number of groups of friends, respectively.
Then m lines follow, each describing a group of friends. The i-th line begins with integer k_i (0 β€ k_i β€ n) β the number of users in the i-th group. Then k_i distinct integers follow, denoting the users belonging to the i-th group.
It is guaranteed that β _{i = 1}^{m} k_i β€ 5 β
10^5.
Output
Print n integers. The i-th integer should be equal to the number of users that will know the news if user i starts distributing it.
Example
Input
7 5
3 2 5 4
0
2 1 2
1 1
2 6 7
Output
4 4 1 4 4 2 2 | instruction | 0 | 44,654 | 14 | 89,308 |
Tags: dfs and similar, dsu, graphs
Correct Solution:
```
# link: https://codeforces.com/problemset/problem/1167/C
import sys
if __name__ == "__main__":
yash = sys.stdin.readline
n,m = map(int, yash().split())
vertex_info = {node_id:[] for node_id in range(1, n+1)}
visited = [0 for i in range(n+1)]
communicating_pairs = [0 for i in range(n+1)]
for _ in range(m):
info = tuple(map(int, yash().split()))
if info[0] >= 2:
for i in range(1,info[0]):
vertex_info[info[i]].append(info[i+1])
vertex_info[info[i+1]].append(info[i])
for node in range(1,n+1):
if visited[node] == 0:
visited[node] = node
communicating_pairs[node] = 1
stack = [node]
while stack: # dfs
current_top_node = stack.pop()
for neighbour in vertex_info[current_top_node]:
if visited[neighbour] == 0:
visited[neighbour] = node
communicating_pairs[node] += 1
stack.append(neighbour)
print(*[communicating_pairs[visited[i]] for i in range(1, n+1) ])
``` | output | 1 | 44,654 | 14 | 89,309 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In some social network, there are n users communicating with each other in m groups of friends. Let's analyze the process of distributing some news between users.
Initially, some user x receives the news from some source. Then he or she sends the news to his or her friends (two users are friends if there is at least one group such that both of them belong to this group). Friends continue sending the news to their friends, and so on. The process ends when there is no pair of friends such that one of them knows the news, and another one doesn't know.
For each user x you have to determine what is the number of users that will know the news if initially only user x starts distributing it.
Input
The first line contains two integers n and m (1 β€ n, m β€ 5 β
10^5) β the number of users and the number of groups of friends, respectively.
Then m lines follow, each describing a group of friends. The i-th line begins with integer k_i (0 β€ k_i β€ n) β the number of users in the i-th group. Then k_i distinct integers follow, denoting the users belonging to the i-th group.
It is guaranteed that β _{i = 1}^{m} k_i β€ 5 β
10^5.
Output
Print n integers. The i-th integer should be equal to the number of users that will know the news if user i starts distributing it.
Example
Input
7 5
3 2 5 4
0
2 1 2
1 1
2 6 7
Output
4 4 1 4 4 2 2 | instruction | 0 | 44,655 | 14 | 89,310 |
Tags: dfs and similar, dsu, graphs
Correct Solution:
```
import sys,bisect
from sys import stdin,stdout
def li():
return list(map(int,stdin.readline().split()))
def mp():
return map(int,stdin.readline().split())
def solve():
n,m=mp()
d={i:[] for i in range(1,n+1)}
for _ in range(m):
l=li()
x=l[0]
if x>1:
for i in range(1,x):
d[l[i]].append(l[i+1])
d[l[i+1]].append(l[i])
ans=[-1 for i in range(n+1)]
vi=[-1 for i in range(n+1)]
for i in range(1,n+1):
if vi[i]==-1:
vi[i]=i
stack=[i]
ans[i]=1
while stack:
a=stack.pop()
for x in d[a]:
if vi[x]==-1:
ans[i]+=1
vi[x]=i
stack.append(x)
print(' '.join((str(ans[vi[i]]) for i in range(1,n+1))))
for _ in range(1):
solve()
``` | output | 1 | 44,655 | 14 | 89,311 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In some social network, there are n users communicating with each other in m groups of friends. Let's analyze the process of distributing some news between users.
Initially, some user x receives the news from some source. Then he or she sends the news to his or her friends (two users are friends if there is at least one group such that both of them belong to this group). Friends continue sending the news to their friends, and so on. The process ends when there is no pair of friends such that one of them knows the news, and another one doesn't know.
For each user x you have to determine what is the number of users that will know the news if initially only user x starts distributing it.
Input
The first line contains two integers n and m (1 β€ n, m β€ 5 β
10^5) β the number of users and the number of groups of friends, respectively.
Then m lines follow, each describing a group of friends. The i-th line begins with integer k_i (0 β€ k_i β€ n) β the number of users in the i-th group. Then k_i distinct integers follow, denoting the users belonging to the i-th group.
It is guaranteed that β _{i = 1}^{m} k_i β€ 5 β
10^5.
Output
Print n integers. The i-th integer should be equal to the number of users that will know the news if user i starts distributing it.
Example
Input
7 5
3 2 5 4
0
2 1 2
1 1
2 6 7
Output
4 4 1 4 4 2 2 | instruction | 0 | 44,656 | 14 | 89,312 |
Tags: dfs and similar, dsu, graphs
Correct Solution:
```
from sys import stdin
from sys import stdout
def find(u):
if parent[u] == u:
return u
parent[u] = find(parent[u])
return parent[u]
n, m = map(int, stdin.readline().split())
parent = [i for i in range(n+1)]
ans = [0 for i in range(1+n)]
for i in range(1,m+1):
a = [int(u) for u in stdin.readline().split()][1:]
if(len(a)==0):
continue
x = find(a[0])
for y in a[1:]:
parent[find(y)] = x
for i in range(1,n+1):
ans[find(i)]+=1
pprintit = ' '.join(map(str, [ans[find(u)] for u in range(1, n + 1)]))
stdout.write('%s' % pprintit)
``` | output | 1 | 44,656 | 14 | 89,313 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In some social network, there are n users communicating with each other in m groups of friends. Let's analyze the process of distributing some news between users.
Initially, some user x receives the news from some source. Then he or she sends the news to his or her friends (two users are friends if there is at least one group such that both of them belong to this group). Friends continue sending the news to their friends, and so on. The process ends when there is no pair of friends such that one of them knows the news, and another one doesn't know.
For each user x you have to determine what is the number of users that will know the news if initially only user x starts distributing it.
Input
The first line contains two integers n and m (1 β€ n, m β€ 5 β
10^5) β the number of users and the number of groups of friends, respectively.
Then m lines follow, each describing a group of friends. The i-th line begins with integer k_i (0 β€ k_i β€ n) β the number of users in the i-th group. Then k_i distinct integers follow, denoting the users belonging to the i-th group.
It is guaranteed that β _{i = 1}^{m} k_i β€ 5 β
10^5.
Output
Print n integers. The i-th integer should be equal to the number of users that will know the news if user i starts distributing it.
Example
Input
7 5
3 2 5 4
0
2 1 2
1 1
2 6 7
Output
4 4 1 4 4 2 2 | instruction | 0 | 44,657 | 14 | 89,314 |
Tags: dfs and similar, dsu, graphs
Correct Solution:
```
"""
Author - Satwik Tiwari .
4th Dec , 2020 - Friday
"""
#===============================================================================================
#importing some useful libraries.
from __future__ import division, print_function
from fractions import Fraction
import sys
import os
from io import BytesIO, IOBase
from functools import cmp_to_key
# from itertools import *
from heapq import *
from math import gcd, factorial,floor,ceil,sqrt
from copy import deepcopy
from collections import deque
from bisect import bisect_left as bl
from bisect import bisect_right as br
from bisect import bisect
#==============================================================================================
#fast I/O region
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
def print(*args, **kwargs):
"""Prints the values to a stream, or to sys.stdout by default."""
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
if sys.version_info[0] < 3:
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
# inp = lambda: sys.stdin.readline().rstrip("\r\n")
#===============================================================================================
### START ITERATE RECURSION ###
from types import GeneratorType
def iterative(f, stack=[]):
def wrapped_func(*args, **kwargs):
if stack: return f(*args, **kwargs)
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
continue
stack.pop()
if not stack: break
to = stack[-1].send(to)
return to
return wrapped_func
#### END ITERATE RECURSION ####
#===============================================================================================
#some shortcuts
def inp(): return sys.stdin.readline().rstrip("\r\n") #for fast input
def out(var): sys.stdout.write(str(var)) #for fast output, always take string
def lis(): return list(map(int, inp().split()))
def stringlis(): return list(map(str, inp().split()))
def sep(): return map(int, inp().split())
def strsep(): return map(str, inp().split())
# def graph(vertex): return [[] for i in range(0,vertex+1)]
def testcase(t):
for pp in range(t):
solve(pp)
def google(p):
print('Case #'+str(p)+': ',end='')
def lcm(a,b): return (a*b)//gcd(a,b)
def power(x, y, p) :
y%=(p-1) #not so sure about this. used when y>p-1. if p is prime.
res = 1 # Initialize result
x = x % p # Update x if it is more , than or equal to p
if (x == 0) :
return 0
while (y > 0) :
if ((y & 1) == 1) : # If y is odd, multiply, x with result
res = (res * x) % p
y = y >> 1 # y = y/2
x = (x * x) % p
return res
def ncr(n,r): return factorial(n) // (factorial(r) * factorial(max(n - r, 1)))
def isPrime(n) :
if (n <= 1) : return False
if (n <= 3) : return True
if (n % 2 == 0 or n % 3 == 0) : return False
i = 5
while(i * i <= n) :
if (n % i == 0 or n % (i + 2) == 0) :
return False
i = i + 6
return True
inf = pow(10,20)
mod = 10**9+7
#===============================================================================================
# code here ;))
class DisjointSetUnion:
def __init__(self, n):
self.parent = list(range(n))
self.size = [1] * n
self.num_sets = n
def find(self, a):
acopy = a
while a != self.parent[a]:
a = self.parent[a]
while acopy != a:
self.parent[acopy], acopy = a, self.parent[acopy]
return a
def union(self, a, b):
a, b = self.find(a), self.find(b)
if a != b:
if self.size[a] < self.size[b]:
a, b = b, a
self.num_sets -= 1
self.parent[b] = a
self.size[a] += self.size[b]
def set_size(self, a):
return self.size[self.find(a)]
def __len__(self):
return self.num_sets
def solve(case):
n,m = sep()
dsu = DisjointSetUnion(n)
for i in range(m):
a = lis()
if(a[0] < 2):
continue
a = a[1:]
for j in range(len(a)):
a[j] -=1
for j in range(1,len(a)):
dsu.union(a[0],a[j])
group = {}
for i in range(n):
if(dsu.find(i) not in group):
group[dsu.find(i)] = 1
else:
group[dsu.find(i)] +=1
ans = []
for i in range(n):
ans.append(group[dsu.find(i)])
print(' '.join(str(ans[i]) for i in range(n)))
testcase(1)
# testcase(int(inp()))
``` | output | 1 | 44,657 | 14 | 89,315 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In some social network, there are n users communicating with each other in m groups of friends. Let's analyze the process of distributing some news between users.
Initially, some user x receives the news from some source. Then he or she sends the news to his or her friends (two users are friends if there is at least one group such that both of them belong to this group). Friends continue sending the news to their friends, and so on. The process ends when there is no pair of friends such that one of them knows the news, and another one doesn't know.
For each user x you have to determine what is the number of users that will know the news if initially only user x starts distributing it.
Input
The first line contains two integers n and m (1 β€ n, m β€ 5 β
10^5) β the number of users and the number of groups of friends, respectively.
Then m lines follow, each describing a group of friends. The i-th line begins with integer k_i (0 β€ k_i β€ n) β the number of users in the i-th group. Then k_i distinct integers follow, denoting the users belonging to the i-th group.
It is guaranteed that β _{i = 1}^{m} k_i β€ 5 β
10^5.
Output
Print n integers. The i-th integer should be equal to the number of users that will know the news if user i starts distributing it.
Example
Input
7 5
3 2 5 4
0
2 1 2
1 1
2 6 7
Output
4 4 1 4 4 2 2 | instruction | 0 | 44,658 | 14 | 89,316 |
Tags: dfs and similar, dsu, graphs
Correct Solution:
```
import sys
class UF():
def __init__(self, num):
self.par = [-1]*num
self.size = [1]*num
def find(self, x):
if self.par[x] < 0:
return x
else:
x = self.par[x]
return self.find(x)
def union(self, x, y):
rx = self.find(x)
ry = self.find(y)
if rx != ry:
if self.par[rx] < self.par[ry]:
self.par[ry] = rx
self.size[rx] += self.size[ry]
elif self.par[rx] > self.par[ry]:
self.par[rx] = ry
self.size[ry] += self.size[rx]
else:
self.par[rx] -= 1
self.par[ry] = rx
self.size[rx] += self.size[ry]
return
def solve():
N, M = map(int, sys.stdin.readline().split())
G = UF(N)
for _ in range(M):
L = list(map(int, sys.stdin.readline().split()))
if L[0] > 1:
l = L[1]
for r in L[2:]:
G.union(l-1, r-1)
Ans = [0]*N
for i in range(N):
Ans[i] = G.size[G.find(i)]
print(*Ans)
if __name__ == '__main__':
solve()
``` | output | 1 | 44,658 | 14 | 89,317 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In some social network, there are n users communicating with each other in m groups of friends. Let's analyze the process of distributing some news between users.
Initially, some user x receives the news from some source. Then he or she sends the news to his or her friends (two users are friends if there is at least one group such that both of them belong to this group). Friends continue sending the news to their friends, and so on. The process ends when there is no pair of friends such that one of them knows the news, and another one doesn't know.
For each user x you have to determine what is the number of users that will know the news if initially only user x starts distributing it.
Input
The first line contains two integers n and m (1 β€ n, m β€ 5 β
10^5) β the number of users and the number of groups of friends, respectively.
Then m lines follow, each describing a group of friends. The i-th line begins with integer k_i (0 β€ k_i β€ n) β the number of users in the i-th group. Then k_i distinct integers follow, denoting the users belonging to the i-th group.
It is guaranteed that β _{i = 1}^{m} k_i β€ 5 β
10^5.
Output
Print n integers. The i-th integer should be equal to the number of users that will know the news if user i starts distributing it.
Example
Input
7 5
3 2 5 4
0
2 1 2
1 1
2 6 7
Output
4 4 1 4 4 2 2 | instruction | 0 | 44,659 | 14 | 89,318 |
Tags: dfs and similar, dsu, graphs
Correct Solution:
```
from sys import stdin,stdout
input = stdin.readline
def main():
n, m = map(int,input().split())
sets = list(range(n+1))
sizes = [1]*(n+1)
def get(a):
if sets[a] != a:
sets[a] = get(sets[a])
return sets[a]
def union(a, b):
a = get(a)
b = get(b)
if a == b:return
if sizes[a] > sizes[b]:
a,b = b,a
sets[a] = b
sizes[b] += sizes[a]
for i in range(m):
num = list(map(int,input().split()))
for i in range(2,num[0]+1):
union(num[i-1]-1,num[i]-1)
for i in range(n):
print(sizes[get(i)],end=" ")
main()
``` | output | 1 | 44,659 | 14 | 89,319 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In some social network, there are n users communicating with each other in m groups of friends. Let's analyze the process of distributing some news between users.
Initially, some user x receives the news from some source. Then he or she sends the news to his or her friends (two users are friends if there is at least one group such that both of them belong to this group). Friends continue sending the news to their friends, and so on. The process ends when there is no pair of friends such that one of them knows the news, and another one doesn't know.
For each user x you have to determine what is the number of users that will know the news if initially only user x starts distributing it.
Input
The first line contains two integers n and m (1 β€ n, m β€ 5 β
10^5) β the number of users and the number of groups of friends, respectively.
Then m lines follow, each describing a group of friends. The i-th line begins with integer k_i (0 β€ k_i β€ n) β the number of users in the i-th group. Then k_i distinct integers follow, denoting the users belonging to the i-th group.
It is guaranteed that β _{i = 1}^{m} k_i β€ 5 β
10^5.
Output
Print n integers. The i-th integer should be equal to the number of users that will know the news if user i starts distributing it.
Example
Input
7 5
3 2 5 4
0
2 1 2
1 1
2 6 7
Output
4 4 1 4 4 2 2
Submitted Solution:
```
import os
import sys
from io import BytesIO, IOBase
import math
from decimal import *
from collections import defaultdict, deque
import heapq
getcontext().prec = 25
MOD = pow(10, 9) + 7
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# n, k = map(int, input().split(" "))
# list(map(int, input().split(" ")))
# for _ in range(int(input())):
def bfs(z):
c = 1
l = [z+1]
q = deque(l)
visted[z]=True
while q:
x = q.pop()
for i in g[x]:
if visted[i-1]:
continue
visted[i-1]=True
c+=1
q.append(i)
l.append(i)
for i in l:
ans[i-1]=c
g = {}
n, k = map(int, input().split(" "))
for i in range(n):
g[i+1]=[]
visted = [False]*n
ans = [0]*n
for i in range(k):
x = list(map(int, input().split(" ")))
if x[0]>1:
for j in x[2:]:
g[j].append(x[1])
g[x[1]].append(j)
for i in range(n):
if not visted[i]:
bfs(i)
print(*ans)
``` | instruction | 0 | 44,660 | 14 | 89,320 |
Yes | output | 1 | 44,660 | 14 | 89,321 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In some social network, there are n users communicating with each other in m groups of friends. Let's analyze the process of distributing some news between users.
Initially, some user x receives the news from some source. Then he or she sends the news to his or her friends (two users are friends if there is at least one group such that both of them belong to this group). Friends continue sending the news to their friends, and so on. The process ends when there is no pair of friends such that one of them knows the news, and another one doesn't know.
For each user x you have to determine what is the number of users that will know the news if initially only user x starts distributing it.
Input
The first line contains two integers n and m (1 β€ n, m β€ 5 β
10^5) β the number of users and the number of groups of friends, respectively.
Then m lines follow, each describing a group of friends. The i-th line begins with integer k_i (0 β€ k_i β€ n) β the number of users in the i-th group. Then k_i distinct integers follow, denoting the users belonging to the i-th group.
It is guaranteed that β _{i = 1}^{m} k_i β€ 5 β
10^5.
Output
Print n integers. The i-th integer should be equal to the number of users that will know the news if user i starts distributing it.
Example
Input
7 5
3 2 5 4
0
2 1 2
1 1
2 6 7
Output
4 4 1 4 4 2 2
Submitted Solution:
```
from sys import stdin
input=stdin.readline
def find(x):
while x!=link[x]:
x=link[x]
return x
n,m=map(int,input().split())
link=[i for i in range(n+1)]
size=[1]*(n+1)
for i in range(m):
k=list(map(int,input().split()))[1:]
if not k:continue
t=find(k[0])
for i in k[1:]:
i=find(i)
if t!=i:
if size[t]<size[i]:
t,i=i,t
link[i]=t
size[t]+=size[i]
for i in range(1,n+1):
i=find(i)
print(size[i],end=' ')
``` | instruction | 0 | 44,661 | 14 | 89,322 |
Yes | output | 1 | 44,661 | 14 | 89,323 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In some social network, there are n users communicating with each other in m groups of friends. Let's analyze the process of distributing some news between users.
Initially, some user x receives the news from some source. Then he or she sends the news to his or her friends (two users are friends if there is at least one group such that both of them belong to this group). Friends continue sending the news to their friends, and so on. The process ends when there is no pair of friends such that one of them knows the news, and another one doesn't know.
For each user x you have to determine what is the number of users that will know the news if initially only user x starts distributing it.
Input
The first line contains two integers n and m (1 β€ n, m β€ 5 β
10^5) β the number of users and the number of groups of friends, respectively.
Then m lines follow, each describing a group of friends. The i-th line begins with integer k_i (0 β€ k_i β€ n) β the number of users in the i-th group. Then k_i distinct integers follow, denoting the users belonging to the i-th group.
It is guaranteed that β _{i = 1}^{m} k_i β€ 5 β
10^5.
Output
Print n integers. The i-th integer should be equal to the number of users that will know the news if user i starts distributing it.
Example
Input
7 5
3 2 5 4
0
2 1 2
1 1
2 6 7
Output
4 4 1 4 4 2 2
Submitted Solution:
```
import sys
sys.setrecursionlimit(100000000)
n,m = map(int,sys.stdin.readline().split())
mark =[set() for _ in range(n+1)]
score = [-1 for _ in range (n+1)]
group = []
groupSet = []
ans = []
for i in range(m):
a = list(map(int, sys.stdin.readline().split()))
group.append(a)
groupSet.append(set(group[0][1:]))
def search(value):
if score[value] < 0:
return value
else:
ret = search(score[value])
score[value] = ret
return (ret)
for i in range(m):
if group[i][0]>0:
mx = min(group[i][1:])
main = search(mx)
for v in group[i][1:]:
cur = search(v)
if main != cur:
if main<cur:
score[main]+=score[cur]
score[cur] = main
else:
score[cur] += score[main]
score[main] = cur
main = cur
for i in range(1,n+1):
ans.append( str(abs(score[search(i)])))
sys.stdout.write('%s\n' % ' '.join(ans))
``` | instruction | 0 | 44,662 | 14 | 89,324 |
Yes | output | 1 | 44,662 | 14 | 89,325 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In some social network, there are n users communicating with each other in m groups of friends. Let's analyze the process of distributing some news between users.
Initially, some user x receives the news from some source. Then he or she sends the news to his or her friends (two users are friends if there is at least one group such that both of them belong to this group). Friends continue sending the news to their friends, and so on. The process ends when there is no pair of friends such that one of them knows the news, and another one doesn't know.
For each user x you have to determine what is the number of users that will know the news if initially only user x starts distributing it.
Input
The first line contains two integers n and m (1 β€ n, m β€ 5 β
10^5) β the number of users and the number of groups of friends, respectively.
Then m lines follow, each describing a group of friends. The i-th line begins with integer k_i (0 β€ k_i β€ n) β the number of users in the i-th group. Then k_i distinct integers follow, denoting the users belonging to the i-th group.
It is guaranteed that β _{i = 1}^{m} k_i β€ 5 β
10^5.
Output
Print n integers. The i-th integer should be equal to the number of users that will know the news if user i starts distributing it.
Example
Input
7 5
3 2 5 4
0
2 1 2
1 1
2 6 7
Output
4 4 1 4 4 2 2
Submitted Solution:
```
import sys
input = sys.stdin.readline
n,m=map(int,input().split())
gr=[set() for i in range(n+1)]
tr=[[1,i] for i in range(n+1)]
for i in range(m):
a=list(map(int,input().split()))
if a[0] in [0,1]:continue
for j in range(1,a[0]):
gr[a[j]].add(a[j+1])
gr[a[j+1]].add(a[j])
v=[False for i in range(n+1)]
for i in range(1,n+1):
if v[i]:continue
s=set([i])
l=0
while s:
x=s.pop()
l+=1
v[x]=True
tr[x][1]=i
for j in gr[x]:
if v[j]:continue
s.add(j)
tr[i][1]=i
tr[i][0]=l
for i in range(1,n+1):
print(tr[tr[i][1]][0],end=' ')
``` | instruction | 0 | 44,663 | 14 | 89,326 |
Yes | output | 1 | 44,663 | 14 | 89,327 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In some social network, there are n users communicating with each other in m groups of friends. Let's analyze the process of distributing some news between users.
Initially, some user x receives the news from some source. Then he or she sends the news to his or her friends (two users are friends if there is at least one group such that both of them belong to this group). Friends continue sending the news to their friends, and so on. The process ends when there is no pair of friends such that one of them knows the news, and another one doesn't know.
For each user x you have to determine what is the number of users that will know the news if initially only user x starts distributing it.
Input
The first line contains two integers n and m (1 β€ n, m β€ 5 β
10^5) β the number of users and the number of groups of friends, respectively.
Then m lines follow, each describing a group of friends. The i-th line begins with integer k_i (0 β€ k_i β€ n) β the number of users in the i-th group. Then k_i distinct integers follow, denoting the users belonging to the i-th group.
It is guaranteed that β _{i = 1}^{m} k_i β€ 5 β
10^5.
Output
Print n integers. The i-th integer should be equal to the number of users that will know the news if user i starts distributing it.
Example
Input
7 5
3 2 5 4
0
2 1 2
1 1
2 6 7
Output
4 4 1 4 4 2 2
Submitted Solution:
```
# from math import *
# from itertools import combinations
from sys import stdin
input = stdin.readline
IN = lambda: map(int, input().split())
def find(x):
while x != p[x]:
x = p[x] = p[p[x]]
return x
n, m = IN()
p = [i for i in range(n+1)]
sz = [0]*(n+1)
for _ in range(m):
a = [*IN()]
for i in a[2:]:
p[find(i)] = find(a[1])
for i in range(1, n+1):
sz[find(i)] += 1
print(*[sz[find(i)] for i in range(1, n+1)])
``` | instruction | 0 | 44,664 | 14 | 89,328 |
No | output | 1 | 44,664 | 14 | 89,329 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In some social network, there are n users communicating with each other in m groups of friends. Let's analyze the process of distributing some news between users.
Initially, some user x receives the news from some source. Then he or she sends the news to his or her friends (two users are friends if there is at least one group such that both of them belong to this group). Friends continue sending the news to their friends, and so on. The process ends when there is no pair of friends such that one of them knows the news, and another one doesn't know.
For each user x you have to determine what is the number of users that will know the news if initially only user x starts distributing it.
Input
The first line contains two integers n and m (1 β€ n, m β€ 5 β
10^5) β the number of users and the number of groups of friends, respectively.
Then m lines follow, each describing a group of friends. The i-th line begins with integer k_i (0 β€ k_i β€ n) β the number of users in the i-th group. Then k_i distinct integers follow, denoting the users belonging to the i-th group.
It is guaranteed that β _{i = 1}^{m} k_i β€ 5 β
10^5.
Output
Print n integers. The i-th integer should be equal to the number of users that will know the news if user i starts distributing it.
Example
Input
7 5
3 2 5 4
0
2 1 2
1 1
2 6 7
Output
4 4 1 4 4 2 2
Submitted Solution:
```
import sys
N, M = map(int, sys.stdin.readline().split())
S = max(N, M) * 2 + 1
parent = [i for i in range(S)]
rank = [0] * S
def find(i):
if parent[i] == i:
return i
else:
parent[i] = find(parent[i])
return parent[i]
def same(x, y):
return find(x) == find(y)
def unite(x, y):
x = find(x)
y = find(y)
if x == y:
return
if rank[x] > rank[y]:
parent[y] = x
else:
parent[x] = y
if rank[x] == rank[y]:
rank[y] += 1
for i in range(M):
k = list(map(int, sys.stdin.readline().split()))
if len(k) == 0:
continue
for j in range(1, len(k)):
unite(N + i, k[j] - 1)
D = {}
ans = [0] * N
for i in range(N):
if parent[i] not in D:
D[parent[i]] = 1
else:
D[parent[i]] += 1
for i in range(N):
ans[i] = D[parent[i]]
print(*ans)
``` | instruction | 0 | 44,665 | 14 | 89,330 |
No | output | 1 | 44,665 | 14 | 89,331 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In some social network, there are n users communicating with each other in m groups of friends. Let's analyze the process of distributing some news between users.
Initially, some user x receives the news from some source. Then he or she sends the news to his or her friends (two users are friends if there is at least one group such that both of them belong to this group). Friends continue sending the news to their friends, and so on. The process ends when there is no pair of friends such that one of them knows the news, and another one doesn't know.
For each user x you have to determine what is the number of users that will know the news if initially only user x starts distributing it.
Input
The first line contains two integers n and m (1 β€ n, m β€ 5 β
10^5) β the number of users and the number of groups of friends, respectively.
Then m lines follow, each describing a group of friends. The i-th line begins with integer k_i (0 β€ k_i β€ n) β the number of users in the i-th group. Then k_i distinct integers follow, denoting the users belonging to the i-th group.
It is guaranteed that β _{i = 1}^{m} k_i β€ 5 β
10^5.
Output
Print n integers. The i-th integer should be equal to the number of users that will know the news if user i starts distributing it.
Example
Input
7 5
3 2 5 4
0
2 1 2
1 1
2 6 7
Output
4 4 1 4 4 2 2
Submitted Solution:
```
class Union_Find():
def __init__(self, n):
self.par = [i for i in range(n+1)]
self.rank = [0] * (n+1)
self.siz = [1] * (n+1)
def find(self, x):
if self.par[x] == x:
return x
else:
#θ¦ͺγζ€η΄’
self.par[x] = self.find(self.par[x])
return self.par[x]
def same_checker(self, x, y):
return self.find(x) == self.find(y)
def union(self, x, y):
x = self.find(x)
y = self.find(y)
if self.rank[x] < self.rank[y]:
self.par[x] = y
self.siz[y] += self.siz[x]
else:
self.par[y] = x
self.siz[x] += self.siz[y]
if self.rank[x] == self.rank[y]:
self.rank[x] += 1
def size(self, x):
return self.siz[self.find(x)]
n, q = map(int, input().split())
union_find_tree = Union_Find(n)
for i in range(q):
a = list(map(int, input().split()))
k = a[0]
if k >= 2:
for i in range(2, k+1):
union_find_tree.union(a[1], a[i])
print(" ".join(map(str, union_find_tree.siz[1:])))
``` | instruction | 0 | 44,666 | 14 | 89,332 |
No | output | 1 | 44,666 | 14 | 89,333 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In some social network, there are n users communicating with each other in m groups of friends. Let's analyze the process of distributing some news between users.
Initially, some user x receives the news from some source. Then he or she sends the news to his or her friends (two users are friends if there is at least one group such that both of them belong to this group). Friends continue sending the news to their friends, and so on. The process ends when there is no pair of friends such that one of them knows the news, and another one doesn't know.
For each user x you have to determine what is the number of users that will know the news if initially only user x starts distributing it.
Input
The first line contains two integers n and m (1 β€ n, m β€ 5 β
10^5) β the number of users and the number of groups of friends, respectively.
Then m lines follow, each describing a group of friends. The i-th line begins with integer k_i (0 β€ k_i β€ n) β the number of users in the i-th group. Then k_i distinct integers follow, denoting the users belonging to the i-th group.
It is guaranteed that β _{i = 1}^{m} k_i β€ 5 β
10^5.
Output
Print n integers. The i-th integer should be equal to the number of users that will know the news if user i starts distributing it.
Example
Input
7 5
3 2 5 4
0
2 1 2
1 1
2 6 7
Output
4 4 1 4 4 2 2
Submitted Solution:
```
n, m = input().split()
n = int(n)
m = int(m)
lista = []
for c in range(0, m):
sub_lista = list(map(int, input().split()))
lista.append(sub_lista)
g = []
for c in range(len(lista)):
sub_lista = []
for v in range(len(lista[c])):
if v != 0:
sub_lista.append(lista[c][v])
g.append(sub_lista)
print(g)
l = []
for c in range(1, n + 1):
sub_lista = []
for v in range(len(g)):
if c in g[v]:
for a in range(len(g[v])):
sub_lista.append(g[v][a])
l.append(sub_lista)
print(l)
num = []
for c in range(len(l)):
sub_lista = []
for v in range(len(l[c])):
if l[c].count(l[c][v]) < 2:
sub_lista.append(l[c][v])
if l[c].count(l[c][v]) >= 2 and l[c][v] not in sub_lista:
sub_lista.append(l[c][v])
num.append(sub_lista)
print(num)
for c in range(len(num)):
for v in range(len(num[c])):
for a in range(len(num)):
if num[c][v] in num[a]:
for b in range(len(num[a])):
if num[a][b] not in num[c]:
num[c].append(num[a][b])
print(num)
for c in range(len(num)):
if c == len(num) - 1:
print(len(num[c]))
elif num[c] == [] and c == len(num) - 1:
print(1)
elif not num[c]:
print(1, end=' ')
else:
print(len(num[c]), end=' ')
``` | instruction | 0 | 44,667 | 14 | 89,334 |
No | output | 1 | 44,667 | 14 | 89,335 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A penguin Rocher has n sticks. He has exactly one stick with length i for all 1 β€ i β€ n.
He can connect some sticks. If he connects two sticks that have lengths a and b, he gets one stick with length a + b. Two sticks, that were used in the operation disappear from his set and the new connected stick appears in his set and can be used for the next connections.
He wants to create the maximum number of sticks that have the same length. It is not necessary to make all sticks have the same length, some sticks can have the other length. How many sticks with the equal length he can create?
Input
The input consists of multiple test cases. The first line contains a single integer t (1 β€ t β€ 1000) β the number of test cases. Next t lines contain descriptions of test cases.
For each test case, the only line contains a single integer n (1 β€ n β€ 10^{9}).
Output
For each test case, print a single integer β the answer to the problem.
Example
Input
4
1
2
3
4
Output
1
1
2
2
Note
In the third case, he can connect two sticks with lengths 1 and 2 and he will get one stick with length 3. So, he will have two sticks with lengths 3.
In the fourth case, he can connect two sticks with lengths 1 and 3 and he will get one stick with length 4. After that, he will have three sticks with lengths \{2, 4, 4\}, so two sticks have the same length, and one stick has the other length. | instruction | 0 | 44,756 | 14 | 89,512 |
Tags: math
Correct Solution:
```
def f(n):
if n & 1:
return n // 2 + 1
return n // 2
t = int(input())
for _ in range(t):
n = int(input())
print(f(n))
``` | output | 1 | 44,756 | 14 | 89,513 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A penguin Rocher has n sticks. He has exactly one stick with length i for all 1 β€ i β€ n.
He can connect some sticks. If he connects two sticks that have lengths a and b, he gets one stick with length a + b. Two sticks, that were used in the operation disappear from his set and the new connected stick appears in his set and can be used for the next connections.
He wants to create the maximum number of sticks that have the same length. It is not necessary to make all sticks have the same length, some sticks can have the other length. How many sticks with the equal length he can create?
Input
The input consists of multiple test cases. The first line contains a single integer t (1 β€ t β€ 1000) β the number of test cases. Next t lines contain descriptions of test cases.
For each test case, the only line contains a single integer n (1 β€ n β€ 10^{9}).
Output
For each test case, print a single integer β the answer to the problem.
Example
Input
4
1
2
3
4
Output
1
1
2
2
Note
In the third case, he can connect two sticks with lengths 1 and 2 and he will get one stick with length 3. So, he will have two sticks with lengths 3.
In the fourth case, he can connect two sticks with lengths 1 and 3 and he will get one stick with length 4. After that, he will have three sticks with lengths \{2, 4, 4\}, so two sticks have the same length, and one stick has the other length. | instruction | 0 | 44,757 | 14 | 89,514 |
Tags: math
Correct Solution:
```
for _ in range(int(input())):
n = int(input())
print((n+1)//2)
``` | output | 1 | 44,757 | 14 | 89,515 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A penguin Rocher has n sticks. He has exactly one stick with length i for all 1 β€ i β€ n.
He can connect some sticks. If he connects two sticks that have lengths a and b, he gets one stick with length a + b. Two sticks, that were used in the operation disappear from his set and the new connected stick appears in his set and can be used for the next connections.
He wants to create the maximum number of sticks that have the same length. It is not necessary to make all sticks have the same length, some sticks can have the other length. How many sticks with the equal length he can create?
Input
The input consists of multiple test cases. The first line contains a single integer t (1 β€ t β€ 1000) β the number of test cases. Next t lines contain descriptions of test cases.
For each test case, the only line contains a single integer n (1 β€ n β€ 10^{9}).
Output
For each test case, print a single integer β the answer to the problem.
Example
Input
4
1
2
3
4
Output
1
1
2
2
Note
In the third case, he can connect two sticks with lengths 1 and 2 and he will get one stick with length 3. So, he will have two sticks with lengths 3.
In the fourth case, he can connect two sticks with lengths 1 and 3 and he will get one stick with length 4. After that, he will have three sticks with lengths \{2, 4, 4\}, so two sticks have the same length, and one stick has the other length. | instruction | 0 | 44,758 | 14 | 89,516 |
Tags: math
Correct Solution:
```
import math
t=int(input())
for i in range(t):
a=int(input())
print(math.ceil(a/2))
``` | output | 1 | 44,758 | 14 | 89,517 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A penguin Rocher has n sticks. He has exactly one stick with length i for all 1 β€ i β€ n.
He can connect some sticks. If he connects two sticks that have lengths a and b, he gets one stick with length a + b. Two sticks, that were used in the operation disappear from his set and the new connected stick appears in his set and can be used for the next connections.
He wants to create the maximum number of sticks that have the same length. It is not necessary to make all sticks have the same length, some sticks can have the other length. How many sticks with the equal length he can create?
Input
The input consists of multiple test cases. The first line contains a single integer t (1 β€ t β€ 1000) β the number of test cases. Next t lines contain descriptions of test cases.
For each test case, the only line contains a single integer n (1 β€ n β€ 10^{9}).
Output
For each test case, print a single integer β the answer to the problem.
Example
Input
4
1
2
3
4
Output
1
1
2
2
Note
In the third case, he can connect two sticks with lengths 1 and 2 and he will get one stick with length 3. So, he will have two sticks with lengths 3.
In the fourth case, he can connect two sticks with lengths 1 and 3 and he will get one stick with length 4. After that, he will have three sticks with lengths \{2, 4, 4\}, so two sticks have the same length, and one stick has the other length. | instruction | 0 | 44,759 | 14 | 89,518 |
Tags: math
Correct Solution:
```
import math
n = int(input())
for x in range(1,n+1):
a = int(input())
if a%2 == 0:
print(int(a/2))
else:
print(math.ceil(a/2))
``` | output | 1 | 44,759 | 14 | 89,519 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A penguin Rocher has n sticks. He has exactly one stick with length i for all 1 β€ i β€ n.
He can connect some sticks. If he connects two sticks that have lengths a and b, he gets one stick with length a + b. Two sticks, that were used in the operation disappear from his set and the new connected stick appears in his set and can be used for the next connections.
He wants to create the maximum number of sticks that have the same length. It is not necessary to make all sticks have the same length, some sticks can have the other length. How many sticks with the equal length he can create?
Input
The input consists of multiple test cases. The first line contains a single integer t (1 β€ t β€ 1000) β the number of test cases. Next t lines contain descriptions of test cases.
For each test case, the only line contains a single integer n (1 β€ n β€ 10^{9}).
Output
For each test case, print a single integer β the answer to the problem.
Example
Input
4
1
2
3
4
Output
1
1
2
2
Note
In the third case, he can connect two sticks with lengths 1 and 2 and he will get one stick with length 3. So, he will have two sticks with lengths 3.
In the fourth case, he can connect two sticks with lengths 1 and 3 and he will get one stick with length 4. After that, he will have three sticks with lengths \{2, 4, 4\}, so two sticks have the same length, and one stick has the other length. | instruction | 0 | 44,760 | 14 | 89,520 |
Tags: math
Correct Solution:
```
t = int(input())
for hatt in range(t):
n = int(input())
print((n+1)//2)
``` | output | 1 | 44,760 | 14 | 89,521 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A penguin Rocher has n sticks. He has exactly one stick with length i for all 1 β€ i β€ n.
He can connect some sticks. If he connects two sticks that have lengths a and b, he gets one stick with length a + b. Two sticks, that were used in the operation disappear from his set and the new connected stick appears in his set and can be used for the next connections.
He wants to create the maximum number of sticks that have the same length. It is not necessary to make all sticks have the same length, some sticks can have the other length. How many sticks with the equal length he can create?
Input
The input consists of multiple test cases. The first line contains a single integer t (1 β€ t β€ 1000) β the number of test cases. Next t lines contain descriptions of test cases.
For each test case, the only line contains a single integer n (1 β€ n β€ 10^{9}).
Output
For each test case, print a single integer β the answer to the problem.
Example
Input
4
1
2
3
4
Output
1
1
2
2
Note
In the third case, he can connect two sticks with lengths 1 and 2 and he will get one stick with length 3. So, he will have two sticks with lengths 3.
In the fourth case, he can connect two sticks with lengths 1 and 3 and he will get one stick with length 4. After that, he will have three sticks with lengths \{2, 4, 4\}, so two sticks have the same length, and one stick has the other length. | instruction | 0 | 44,761 | 14 | 89,522 |
Tags: math
Correct Solution:
```
for _ in range(int(input())):
a=int(input())
if a%2==0:
print(a//2)
else:
print(a//2+1)
``` | output | 1 | 44,761 | 14 | 89,523 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A penguin Rocher has n sticks. He has exactly one stick with length i for all 1 β€ i β€ n.
He can connect some sticks. If he connects two sticks that have lengths a and b, he gets one stick with length a + b. Two sticks, that were used in the operation disappear from his set and the new connected stick appears in his set and can be used for the next connections.
He wants to create the maximum number of sticks that have the same length. It is not necessary to make all sticks have the same length, some sticks can have the other length. How many sticks with the equal length he can create?
Input
The input consists of multiple test cases. The first line contains a single integer t (1 β€ t β€ 1000) β the number of test cases. Next t lines contain descriptions of test cases.
For each test case, the only line contains a single integer n (1 β€ n β€ 10^{9}).
Output
For each test case, print a single integer β the answer to the problem.
Example
Input
4
1
2
3
4
Output
1
1
2
2
Note
In the third case, he can connect two sticks with lengths 1 and 2 and he will get one stick with length 3. So, he will have two sticks with lengths 3.
In the fourth case, he can connect two sticks with lengths 1 and 3 and he will get one stick with length 4. After that, he will have three sticks with lengths \{2, 4, 4\}, so two sticks have the same length, and one stick has the other length. | instruction | 0 | 44,762 | 14 | 89,524 |
Tags: math
Correct Solution:
```
for i in range(int(input())):
n = int(input())
if n%2 ==0:
print(n//2)
else:
a = n//2
print(a+1)
``` | output | 1 | 44,762 | 14 | 89,525 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A penguin Rocher has n sticks. He has exactly one stick with length i for all 1 β€ i β€ n.
He can connect some sticks. If he connects two sticks that have lengths a and b, he gets one stick with length a + b. Two sticks, that were used in the operation disappear from his set and the new connected stick appears in his set and can be used for the next connections.
He wants to create the maximum number of sticks that have the same length. It is not necessary to make all sticks have the same length, some sticks can have the other length. How many sticks with the equal length he can create?
Input
The input consists of multiple test cases. The first line contains a single integer t (1 β€ t β€ 1000) β the number of test cases. Next t lines contain descriptions of test cases.
For each test case, the only line contains a single integer n (1 β€ n β€ 10^{9}).
Output
For each test case, print a single integer β the answer to the problem.
Example
Input
4
1
2
3
4
Output
1
1
2
2
Note
In the third case, he can connect two sticks with lengths 1 and 2 and he will get one stick with length 3. So, he will have two sticks with lengths 3.
In the fourth case, he can connect two sticks with lengths 1 and 3 and he will get one stick with length 4. After that, he will have three sticks with lengths \{2, 4, 4\}, so two sticks have the same length, and one stick has the other length. | instruction | 0 | 44,763 | 14 | 89,526 |
Tags: math
Correct Solution:
```
import math
t= int(input())
ans=[]
for i in range(t):
n= int(input())
ans.append(math.ceil(n/2))
for i in ans :
print(i)
``` | output | 1 | 44,763 | 14 | 89,527 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Eugeny loves listening to music. He has n songs in his play list. We know that song number i has the duration of ti minutes. Eugeny listens to each song, perhaps more than once. He listens to song number i ci times. Eugeny's play list is organized as follows: first song number 1 plays c1 times, then song number 2 plays c2 times, ..., in the end the song number n plays cn times.
Eugeny took a piece of paper and wrote out m moments of time when he liked a song. Now for each such moment he wants to know the number of the song that played at that moment. The moment x means that Eugeny wants to know which song was playing during the x-th minute of his listening to the play list.
Help Eugeny and calculate the required numbers of songs.
Input
The first line contains two integers n, m (1 β€ n, m β€ 105). The next n lines contain pairs of integers. The i-th line contains integers ci, ti (1 β€ ci, ti β€ 109) β the description of the play list. It is guaranteed that the play list's total duration doesn't exceed 109 <image>.
The next line contains m positive integers v1, v2, ..., vm, that describe the moments Eugeny has written out. It is guaranteed that there isn't such moment of time vi, when the music doesn't play any longer. It is guaranteed that vi < vi + 1 (i < m).
The moment of time vi means that Eugeny wants to know which song was playing during the vi-th munite from the start of listening to the playlist.
Output
Print m integers β the i-th number must equal the number of the song that was playing during the vi-th minute after Eugeny started listening to the play list.
Examples
Input
1 2
2 8
1 16
Output
1
1
Input
4 9
1 2
2 1
1 1
2 2
1 2 3 4 5 6 7 8 9
Output
1
1
2
2
3
4
4
4
4 | instruction | 0 | 44,879 | 14 | 89,758 |
Tags: binary search, implementation, two pointers
Correct Solution:
```
from sys import stdin,stdout
n,m = map(int,stdin.readline().split())
ts = []
ts.append(0)
for i in range(n):
j,k = map(int,stdin.readline().split())
ts.append(ts[i] + (j*k))
i = 0
j = 1
V = list(map(int,stdin.readline().split()))
for v in V:
if ts[i] <= v <= ts[j]:
print(i+1)
elif ts[j] < v:
while ts[j] < v:
i += 1
j += 1
print(i+1)
``` | output | 1 | 44,879 | 14 | 89,759 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Eugeny loves listening to music. He has n songs in his play list. We know that song number i has the duration of ti minutes. Eugeny listens to each song, perhaps more than once. He listens to song number i ci times. Eugeny's play list is organized as follows: first song number 1 plays c1 times, then song number 2 plays c2 times, ..., in the end the song number n plays cn times.
Eugeny took a piece of paper and wrote out m moments of time when he liked a song. Now for each such moment he wants to know the number of the song that played at that moment. The moment x means that Eugeny wants to know which song was playing during the x-th minute of his listening to the play list.
Help Eugeny and calculate the required numbers of songs.
Input
The first line contains two integers n, m (1 β€ n, m β€ 105). The next n lines contain pairs of integers. The i-th line contains integers ci, ti (1 β€ ci, ti β€ 109) β the description of the play list. It is guaranteed that the play list's total duration doesn't exceed 109 <image>.
The next line contains m positive integers v1, v2, ..., vm, that describe the moments Eugeny has written out. It is guaranteed that there isn't such moment of time vi, when the music doesn't play any longer. It is guaranteed that vi < vi + 1 (i < m).
The moment of time vi means that Eugeny wants to know which song was playing during the vi-th munite from the start of listening to the playlist.
Output
Print m integers β the i-th number must equal the number of the song that was playing during the vi-th minute after Eugeny started listening to the play list.
Examples
Input
1 2
2 8
1 16
Output
1
1
Input
4 9
1 2
2 1
1 1
2 2
1 2 3 4 5 6 7 8 9
Output
1
1
2
2
3
4
4
4
4 | instruction | 0 | 44,880 | 14 | 89,760 |
Tags: binary search, implementation, two pointers
Correct Solution:
```
string = input().split()
numList = list(map(int, string))
n = numList[0]
m = numList[1]
count = 0
moments = []
last = 0
for i in range(n):
music = list(map(int, input().split()))
moments.append(count + music[0]*music[1])
count = count + music[0]*music[1]
times = list(map(int, input().split()))
for i in range(m):
for j in range(last, n):
if(times[i] <= moments[j]):
print(j+1)
last = j
break
``` | output | 1 | 44,880 | 14 | 89,761 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Eugeny loves listening to music. He has n songs in his play list. We know that song number i has the duration of ti minutes. Eugeny listens to each song, perhaps more than once. He listens to song number i ci times. Eugeny's play list is organized as follows: first song number 1 plays c1 times, then song number 2 plays c2 times, ..., in the end the song number n plays cn times.
Eugeny took a piece of paper and wrote out m moments of time when he liked a song. Now for each such moment he wants to know the number of the song that played at that moment. The moment x means that Eugeny wants to know which song was playing during the x-th minute of his listening to the play list.
Help Eugeny and calculate the required numbers of songs.
Input
The first line contains two integers n, m (1 β€ n, m β€ 105). The next n lines contain pairs of integers. The i-th line contains integers ci, ti (1 β€ ci, ti β€ 109) β the description of the play list. It is guaranteed that the play list's total duration doesn't exceed 109 <image>.
The next line contains m positive integers v1, v2, ..., vm, that describe the moments Eugeny has written out. It is guaranteed that there isn't such moment of time vi, when the music doesn't play any longer. It is guaranteed that vi < vi + 1 (i < m).
The moment of time vi means that Eugeny wants to know which song was playing during the vi-th munite from the start of listening to the playlist.
Output
Print m integers β the i-th number must equal the number of the song that was playing during the vi-th minute after Eugeny started listening to the play list.
Examples
Input
1 2
2 8
1 16
Output
1
1
Input
4 9
1 2
2 1
1 1
2 2
1 2 3 4 5 6 7 8 9
Output
1
1
2
2
3
4
4
4
4 | instruction | 0 | 44,881 | 14 | 89,762 |
Tags: binary search, implementation, two pointers
Correct Solution:
```
import operator as op
import itertools as it
import bisect as bs
n, m = map(int, input().split())
a = list(it.accumulate([op.mul(*map(int, input().split())) for _ in range(n)]))
ans = []
for t in map(int, input().split()):
ans.append(str(bs.bisect_left(a, t) + 1))
print('\n'.join(ans))
``` | output | 1 | 44,881 | 14 | 89,763 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Eugeny loves listening to music. He has n songs in his play list. We know that song number i has the duration of ti minutes. Eugeny listens to each song, perhaps more than once. He listens to song number i ci times. Eugeny's play list is organized as follows: first song number 1 plays c1 times, then song number 2 plays c2 times, ..., in the end the song number n plays cn times.
Eugeny took a piece of paper and wrote out m moments of time when he liked a song. Now for each such moment he wants to know the number of the song that played at that moment. The moment x means that Eugeny wants to know which song was playing during the x-th minute of his listening to the play list.
Help Eugeny and calculate the required numbers of songs.
Input
The first line contains two integers n, m (1 β€ n, m β€ 105). The next n lines contain pairs of integers. The i-th line contains integers ci, ti (1 β€ ci, ti β€ 109) β the description of the play list. It is guaranteed that the play list's total duration doesn't exceed 109 <image>.
The next line contains m positive integers v1, v2, ..., vm, that describe the moments Eugeny has written out. It is guaranteed that there isn't such moment of time vi, when the music doesn't play any longer. It is guaranteed that vi < vi + 1 (i < m).
The moment of time vi means that Eugeny wants to know which song was playing during the vi-th munite from the start of listening to the playlist.
Output
Print m integers β the i-th number must equal the number of the song that was playing during the vi-th minute after Eugeny started listening to the play list.
Examples
Input
1 2
2 8
1 16
Output
1
1
Input
4 9
1 2
2 1
1 1
2 2
1 2 3 4 5 6 7 8 9
Output
1
1
2
2
3
4
4
4
4 | instruction | 0 | 44,882 | 14 | 89,764 |
Tags: binary search, implementation, two pointers
Correct Solution:
```
n, m = [int(p) for p in input().split()]
sm = []
for i in range(n):
c, t = [int(p) for p in input().split()]
if not sm:
sm.append(c*t)
else:
sm.append(sm[-1] + c*t)
time = [int(p) for p in input().split()]
i = 0
j = 0
while j < len(time):
if sm[i] >= time[j]:
print(i+1)
j += 1
else:
i += 1
``` | output | 1 | 44,882 | 14 | 89,765 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Eugeny loves listening to music. He has n songs in his play list. We know that song number i has the duration of ti minutes. Eugeny listens to each song, perhaps more than once. He listens to song number i ci times. Eugeny's play list is organized as follows: first song number 1 plays c1 times, then song number 2 plays c2 times, ..., in the end the song number n plays cn times.
Eugeny took a piece of paper and wrote out m moments of time when he liked a song. Now for each such moment he wants to know the number of the song that played at that moment. The moment x means that Eugeny wants to know which song was playing during the x-th minute of his listening to the play list.
Help Eugeny and calculate the required numbers of songs.
Input
The first line contains two integers n, m (1 β€ n, m β€ 105). The next n lines contain pairs of integers. The i-th line contains integers ci, ti (1 β€ ci, ti β€ 109) β the description of the play list. It is guaranteed that the play list's total duration doesn't exceed 109 <image>.
The next line contains m positive integers v1, v2, ..., vm, that describe the moments Eugeny has written out. It is guaranteed that there isn't such moment of time vi, when the music doesn't play any longer. It is guaranteed that vi < vi + 1 (i < m).
The moment of time vi means that Eugeny wants to know which song was playing during the vi-th munite from the start of listening to the playlist.
Output
Print m integers β the i-th number must equal the number of the song that was playing during the vi-th minute after Eugeny started listening to the play list.
Examples
Input
1 2
2 8
1 16
Output
1
1
Input
4 9
1 2
2 1
1 1
2 2
1 2 3 4 5 6 7 8 9
Output
1
1
2
2
3
4
4
4
4 | instruction | 0 | 44,883 | 14 | 89,766 |
Tags: binary search, implementation, two pointers
Correct Solution:
```
# maa chudaaye duniya
n,m=map(int,input().split())
arr=[0]
k, l = 0, 0
for i in range(n):
c,t=map(int,input().split())
arr.append(arr[i] + c*t)
ara = list(map(int, input().split()))
for j in range(m):
while ara[j] > arr[l]:
l += 1
print(l)
``` | output | 1 | 44,883 | 14 | 89,767 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Eugeny loves listening to music. He has n songs in his play list. We know that song number i has the duration of ti minutes. Eugeny listens to each song, perhaps more than once. He listens to song number i ci times. Eugeny's play list is organized as follows: first song number 1 plays c1 times, then song number 2 plays c2 times, ..., in the end the song number n plays cn times.
Eugeny took a piece of paper and wrote out m moments of time when he liked a song. Now for each such moment he wants to know the number of the song that played at that moment. The moment x means that Eugeny wants to know which song was playing during the x-th minute of his listening to the play list.
Help Eugeny and calculate the required numbers of songs.
Input
The first line contains two integers n, m (1 β€ n, m β€ 105). The next n lines contain pairs of integers. The i-th line contains integers ci, ti (1 β€ ci, ti β€ 109) β the description of the play list. It is guaranteed that the play list's total duration doesn't exceed 109 <image>.
The next line contains m positive integers v1, v2, ..., vm, that describe the moments Eugeny has written out. It is guaranteed that there isn't such moment of time vi, when the music doesn't play any longer. It is guaranteed that vi < vi + 1 (i < m).
The moment of time vi means that Eugeny wants to know which song was playing during the vi-th munite from the start of listening to the playlist.
Output
Print m integers β the i-th number must equal the number of the song that was playing during the vi-th minute after Eugeny started listening to the play list.
Examples
Input
1 2
2 8
1 16
Output
1
1
Input
4 9
1 2
2 1
1 1
2 2
1 2 3 4 5 6 7 8 9
Output
1
1
2
2
3
4
4
4
4 | instruction | 0 | 44,884 | 14 | 89,768 |
Tags: binary search, implementation, two pointers
Correct Solution:
```
n, m = map(int, input().split())
arr = []
for i in range(n):
c, t = map(int, input().split())
arr.append(c * t)
i, p = 0, 0
v = [int(x) for x in input().split()]
for j in v:
while (p < j):
p, i = p+arr[i], i+1
print(i)
``` | output | 1 | 44,884 | 14 | 89,769 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Eugeny loves listening to music. He has n songs in his play list. We know that song number i has the duration of ti minutes. Eugeny listens to each song, perhaps more than once. He listens to song number i ci times. Eugeny's play list is organized as follows: first song number 1 plays c1 times, then song number 2 plays c2 times, ..., in the end the song number n plays cn times.
Eugeny took a piece of paper and wrote out m moments of time when he liked a song. Now for each such moment he wants to know the number of the song that played at that moment. The moment x means that Eugeny wants to know which song was playing during the x-th minute of his listening to the play list.
Help Eugeny and calculate the required numbers of songs.
Input
The first line contains two integers n, m (1 β€ n, m β€ 105). The next n lines contain pairs of integers. The i-th line contains integers ci, ti (1 β€ ci, ti β€ 109) β the description of the play list. It is guaranteed that the play list's total duration doesn't exceed 109 <image>.
The next line contains m positive integers v1, v2, ..., vm, that describe the moments Eugeny has written out. It is guaranteed that there isn't such moment of time vi, when the music doesn't play any longer. It is guaranteed that vi < vi + 1 (i < m).
The moment of time vi means that Eugeny wants to know which song was playing during the vi-th munite from the start of listening to the playlist.
Output
Print m integers β the i-th number must equal the number of the song that was playing during the vi-th minute after Eugeny started listening to the play list.
Examples
Input
1 2
2 8
1 16
Output
1
1
Input
4 9
1 2
2 1
1 1
2 2
1 2 3 4 5 6 7 8 9
Output
1
1
2
2
3
4
4
4
4 | instruction | 0 | 44,885 | 14 | 89,770 |
Tags: binary search, implementation, two pointers
Correct Solution:
```
n, m = map(int, input().split())
track = []
for i in range(n):
c, t = map(int, input().split())
track.append (c * t)
moment = [int(i) for i in input().split()]
i, j = 0, 0
s = 0
while i < n and j < m:
if moment[j] <= s + track[i]:
print(i + 1)
j += 1
else:
s += track[i]
i += 1
``` | output | 1 | 44,885 | 14 | 89,771 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Eugeny loves listening to music. He has n songs in his play list. We know that song number i has the duration of ti minutes. Eugeny listens to each song, perhaps more than once. He listens to song number i ci times. Eugeny's play list is organized as follows: first song number 1 plays c1 times, then song number 2 plays c2 times, ..., in the end the song number n plays cn times.
Eugeny took a piece of paper and wrote out m moments of time when he liked a song. Now for each such moment he wants to know the number of the song that played at that moment. The moment x means that Eugeny wants to know which song was playing during the x-th minute of his listening to the play list.
Help Eugeny and calculate the required numbers of songs.
Input
The first line contains two integers n, m (1 β€ n, m β€ 105). The next n lines contain pairs of integers. The i-th line contains integers ci, ti (1 β€ ci, ti β€ 109) β the description of the play list. It is guaranteed that the play list's total duration doesn't exceed 109 <image>.
The next line contains m positive integers v1, v2, ..., vm, that describe the moments Eugeny has written out. It is guaranteed that there isn't such moment of time vi, when the music doesn't play any longer. It is guaranteed that vi < vi + 1 (i < m).
The moment of time vi means that Eugeny wants to know which song was playing during the vi-th munite from the start of listening to the playlist.
Output
Print m integers β the i-th number must equal the number of the song that was playing during the vi-th minute after Eugeny started listening to the play list.
Examples
Input
1 2
2 8
1 16
Output
1
1
Input
4 9
1 2
2 1
1 1
2 2
1 2 3 4 5 6 7 8 9
Output
1
1
2
2
3
4
4
4
4 | instruction | 0 | 44,886 | 14 | 89,772 |
Tags: binary search, implementation, two pointers
Correct Solution:
```
# import sys
# sys.stdin=open('F:\\C\\Script\\input.txt','r')
# sys.stdout=open('F:\\C\\Script\\output.txt','w')
# sys.stdout.flush()
# MOD = 1000000007
I = lambda : [int(i) for i in input().split()]
n , m = I()
l = []
temp = I()
l.append(temp[0]*temp[1])
for _ in range(1,n):
k = I()
l.append(l[_-1] + k[0]*k[1])
# print (l)
s = I()
j = 0
for i in s:
if i < l[j]:
print ( j + 1)
else:
while i > l[j]:
j += 1
print (j + 1)
``` | output | 1 | 44,886 | 14 | 89,773 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
N ladies attend the ball in the King's palace. Every lady can be described with three values: beauty, intellect and richness. King's Master of Ceremonies knows that ladies are very special creatures. If some lady understands that there is other lady at the ball which is more beautiful, smarter and more rich, she can jump out of the window. He knows values of all ladies and wants to find out how many probable self-murderers will be on the ball. Lets denote beauty of the i-th lady by Bi, her intellect by Ii and her richness by Ri. Then i-th lady is a probable self-murderer if there is some j-th lady that Bi < Bj, Ii < Ij, Ri < Rj. Find the number of probable self-murderers.
Input
The first line contains one integer N (1 β€ N β€ 500000). The second line contains N integer numbers Bi, separated by single spaces. The third and the fourth lines contain sequences Ii and Ri in the same format. It is guaranteed that 0 β€ Bi, Ii, Ri β€ 109.
Output
Output the answer to the problem.
Examples
Input
3
1 4 2
4 3 2
2 5 3
Output
1
Submitted Solution:
```
n=int(input())
List=[]
for i in range(n):
List.append([])
for j in range(3):
data = [int(i) for i in input().split(' ')]
for i in range(n):
List[i].append(data[i])
self_kill=[False]*n
def compare(elem1,elem2):
if elem1[0]>elem2[0] and elem1[1]>elem2[1] and elem1[2]>elem2[2]:
return 1
if elem1[0]<elem2[0] and elem1[1]<elem2[1] and elem1[2]<elem2[2]:
return -1
return 0
for i in range(len(List)):
for j in range(len(List)):
#print('COMPARE '+str(i)+' AND '+str(j))
if max(i,j)<len(List):
if compare(List[i],List[j])==1:
#print('MORE')
del List[j]
if j>0:
j-=1
if compare(List[i],List[j])==-1:
#print('LESS')
del List[i]
if i>0:
i-=1
print(n-len(List))
``` | instruction | 0 | 45,531 | 14 | 91,062 |
No | output | 1 | 45,531 | 14 | 91,063 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
N ladies attend the ball in the King's palace. Every lady can be described with three values: beauty, intellect and richness. King's Master of Ceremonies knows that ladies are very special creatures. If some lady understands that there is other lady at the ball which is more beautiful, smarter and more rich, she can jump out of the window. He knows values of all ladies and wants to find out how many probable self-murderers will be on the ball. Lets denote beauty of the i-th lady by Bi, her intellect by Ii and her richness by Ri. Then i-th lady is a probable self-murderer if there is some j-th lady that Bi < Bj, Ii < Ij, Ri < Rj. Find the number of probable self-murderers.
Input
The first line contains one integer N (1 β€ N β€ 500000). The second line contains N integer numbers Bi, separated by single spaces. The third and the fourth lines contain sequences Ii and Ri in the same format. It is guaranteed that 0 β€ Bi, Ii, Ri β€ 109.
Output
Output the answer to the problem.
Examples
Input
3
1 4 2
4 3 2
2 5 3
Output
1
Submitted Solution:
```
"""
$ pylint calderonsin.py
Global evaluation
-----------------
Your code has been rated at 10.00/10 (previous run: 10.00/10, +0.00)
"""
class Ladies:
def __init__(self,beauty,intellect,rich):
self.beauty= beauty
self.intellect = intellect
self.rich = rich
def selfmurderer():
"""self murderer function"""
n = int(input())
beauty = [None]*n
intellect = [None]*n
rich = [None]*n
ladies = [Ladies] * n
for j in range(3):
info=input()
info=info.split()
for i in range(n):
if j==0:
beauty[i]=info[i]
if j==1:
intellect[i]=info[i]
if j==2:
rich[i]=info[i]
for i in range(n):
ladies[i] = Ladies(beauty[i],intellect[i],rich[i])
ladies.sort(key=lambda x: x.beauty)
cont = 0
for i in range(n-1):
auxinte=ladies[i].intellect
auxrich=ladies[i].rich
for j in (i+1,n-1):
#print(j,i)
if auxinte <= ladies[j].intellect and auxrich <= ladies[j].rich:
cont = cont + 1
print(cont-1,"cont")
selfmurderer()
# $ python3 calderonsin.py build
# 0/1
``` | instruction | 0 | 45,532 | 14 | 91,064 |
No | output | 1 | 45,532 | 14 | 91,065 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
N ladies attend the ball in the King's palace. Every lady can be described with three values: beauty, intellect and richness. King's Master of Ceremonies knows that ladies are very special creatures. If some lady understands that there is other lady at the ball which is more beautiful, smarter and more rich, she can jump out of the window. He knows values of all ladies and wants to find out how many probable self-murderers will be on the ball. Lets denote beauty of the i-th lady by Bi, her intellect by Ii and her richness by Ri. Then i-th lady is a probable self-murderer if there is some j-th lady that Bi < Bj, Ii < Ij, Ri < Rj. Find the number of probable self-murderers.
Input
The first line contains one integer N (1 β€ N β€ 500000). The second line contains N integer numbers Bi, separated by single spaces. The third and the fourth lines contain sequences Ii and Ri in the same format. It is guaranteed that 0 β€ Bi, Ii, Ri β€ 109.
Output
Output the answer to the problem.
Examples
Input
3
1 4 2
4 3 2
2 5 3
Output
1
Submitted Solution:
```
class A:
def rotate_keypad(self, keypad):
reversed_rows = list(reversed(keypad))
return ["".join(list(reversed(x))) for x in reversed_rows]
def solve(self):
keypad = [input(), input(), input()]
if keypad == self.rotate_keypad(keypad):
print("YES")
else:
print("NO")
class B:
def solve(self):
alice = input()
bob = input()
if bob[0] == '0' and alice[0] != '0':
print("WRONG_ANSWER")
return
elif bob[0] == '0' and alice[0] == '0' and len(bob) != 1:
print("WRONG_ANSWER")
return
bob = int(bob)
from itertools import permutations
smallest = int(alice)
if smallest == 0:
if bob == 0:
print("OK")
else:
print("WRONG_ANSWER")
else:
smallest = min([int("".join(list(perm))) for perm in permutations(sorted(alice)) if perm[0] != '0'])
if smallest == bob:
print("OK")
else:
print("WRONG_ANSWER")
class C:
def solve(self):
[n, m] = [int(x) for x in input().split(" ")]
prices = sorted([int(x) for x in input().split(" ")])
fruits = []
for i in range(m):
fruits.append(input())
from collections import Counter
frequent_fruits = [f for (f, p) in Counter(fruits).most_common()]
price_assignment = {}
for f, p in zip(frequent_fruits, prices):
price_assignment[f] = p
smallest_price = sum([price_assignment[f] for f in fruits])
for f, p in zip(frequent_fruits, list(reversed(prices))):
price_assignment[f] = p
largest_price = sum([price_assignment[f] for f in fruits])
print("{} {}".format(smallest_price, largest_price))
class D:
def solve(self):
n = int(input())
ladies = []
for x in input().split(" "):
ladies.append([x])
for i, x in enumerate(input().split(" ")):
ladies[i].append(x)
for i, x in enumerate(input().split(" ")):
ladies[i].append(x)
self_murderers = 0
for i in range(n):
if ladies[i][0] != max([x[0] for x in ladies]) and\
ladies[i][1] != max([x[1] for x in ladies]) and\
ladies[i][2] != max([x[2] for x in ladies]):
self_murderers += 1
print(self_murderers)
class E:
def generate_matrix(self, i, j, n, matrix):
if i == j:
matrix[i][j] = 0
else:
if matrix[i][j] == 0:
dictionary = {}
for e in matrix[i]:
dictionary[e] = True
for e in matrix[j]:
dictionary[e] = True
for index in range(n):
if index not in dictionary:
matrix[i][j] = index
matrix[j][i] = index
break
def solve(self):
n = int(input())
matrix = [[0 for i in range(n)] for j in range(n)]
for i in range(n):
for j in range(n):
self.generate_matrix(i, j, n, matrix)
transponse = list(map(list, zip(*matrix)))
if matrix != transponse:
raise Exception("FC")
print("\n".join([" ".join([str(y) for y in x]) for x in matrix]))
D().solve()
``` | instruction | 0 | 45,533 | 14 | 91,066 |
No | output | 1 | 45,533 | 14 | 91,067 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
N ladies attend the ball in the King's palace. Every lady can be described with three values: beauty, intellect and richness. King's Master of Ceremonies knows that ladies are very special creatures. If some lady understands that there is other lady at the ball which is more beautiful, smarter and more rich, she can jump out of the window. He knows values of all ladies and wants to find out how many probable self-murderers will be on the ball. Lets denote beauty of the i-th lady by Bi, her intellect by Ii and her richness by Ri. Then i-th lady is a probable self-murderer if there is some j-th lady that Bi < Bj, Ii < Ij, Ri < Rj. Find the number of probable self-murderers.
Input
The first line contains one integer N (1 β€ N β€ 500000). The second line contains N integer numbers Bi, separated by single spaces. The third and the fourth lines contain sequences Ii and Ri in the same format. It is guaranteed that 0 β€ Bi, Ii, Ri β€ 109.
Output
Output the answer to the problem.
Examples
Input
3
1 4 2
4 3 2
2 5 3
Output
1
Submitted Solution:
```
"""
$ pylint calderonsin.py
Global evaluation
-----------------
Your code has been rated at 10.00/10 (previous run: 10.00/10, +0.00)
"""
def bob():
"""self murderer function"""
n = int(input())
ganan = [0 for i in range(2009)]
value = 0
for i in range(n):
info = input().split()
order = info[0]
number = int(info[1])
if order == "win":
ganan[number] = value + 2**number
else:
value = max(ganan[number],value)
print(value )
bob()
# $ python3 calderonsin.py build
# 0/1
``` | instruction | 0 | 45,534 | 14 | 91,068 |
No | output | 1 | 45,534 | 14 | 91,069 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are many anime that are about "love triangles": Alice loves Bob, and Charlie loves Bob as well, but Alice hates Charlie. You are thinking about an anime which has n characters. The characters are labeled from 1 to n. Every pair of two characters can either mutually love each other or mutually hate each other (there is no neutral state).
You hate love triangles (A-B are in love and B-C are in love, but A-C hate each other), and you also hate it when nobody is in love. So, considering any three characters, you will be happy if exactly one pair is in love (A and B love each other, and C hates both A and B), or if all three pairs are in love (A loves B, B loves C, C loves A).
You are given a list of m known relationships in the anime. You know for sure that certain pairs love each other, and certain pairs hate each other. You're wondering how many ways you can fill in the remaining relationships so you are happy with every triangle. Two ways are considered different if two characters are in love in one way but hate each other in the other. Print this count modulo 1 000 000 007.
Input
The first line of input will contain two integers n, m (3 β€ n β€ 100 000, 0 β€ m β€ 100 000).
The next m lines will contain the description of the known relationships. The i-th line will contain three integers ai, bi, ci. If ci is 1, then ai and bi are in love, otherwise, they hate each other (1 β€ ai, bi β€ n, ai β bi, <image>).
Each pair of people will be described no more than once.
Output
Print a single integer equal to the number of ways to fill in the remaining pairs so that you are happy with every triangle modulo 1 000 000 007.
Examples
Input
3 0
Output
4
Input
4 4
1 2 1
2 3 1
3 4 0
4 1 0
Output
1
Input
4 4
1 2 1
2 3 1
3 4 0
4 1 1
Output
0
Note
In the first sample, the four ways are to:
* Make everyone love each other
* Make 1 and 2 love each other, and 3 hate 1 and 2 (symmetrically, we get 3 ways from this).
In the second sample, the only possible solution is to make 1 and 3 love each other and 2 and 4 hate each other. | instruction | 0 | 45,754 | 14 | 91,508 |
Tags: dfs and similar, dsu, graphs
Correct Solution:
```
from sys import stdin, stdout
import sys
class Node(object):
def __init__(self, label):
self.label = label
self.par = self
self.rank = 0
class DisjointSet(object):
def __init__(self, n):
self.n = n
self.nodes = [Node(i) for i in range(self.n)]
def find(self, u):
if u == u.par:
return u
return self.find(u.par)
def union(self, u, v):
u, v = self.find(u), self.find(v)
if u == v:
return False
if u.rank > v.rank:
u, v = v, u
u.par = v
if u.rank == v.rank:
v.rank += 1
return True
def size(self):
cnt = 0
for node in self.nodes:
if node.par == node:
cnt += 1
return cnt
def color(adj, colors, root):
neighbor_color = 1 - colors[root]
for neighbor in adj[root]:
if colors[neighbor] == -1:
colors[neighbor] = neighbor_color
if color(adj, colors, neighbor):
pass
else:
return False
else:
if colors[neighbor] != neighbor_color:
return False
return True
if __name__ == "__main__":
try:
sys.setrecursionlimit(10000)
mod = 1000000007
n, m = map(int, input().split())
dsu = DisjointSet(n)
reds = []
for _ in range(m):
a, b, c = map(int, input().split())
if c == 0:
reds.append((a - 1, b - 1))
else:
dsu.union(dsu.nodes[a - 1], dsu.nodes[b - 1])
new_graph = {}
cnt = 0
for i in range(n):
comp = dsu.find(dsu.nodes[i])
if comp.label not in new_graph.keys():
new_graph[comp.label] = cnt
cnt += 1
x = len(new_graph)
assert cnt == x
dsu2 = DisjointSet(x)
adj = [[] for _ in range(x)]
colors = [-1 for _ in range(x)]
for a, b in reds:
comp1 = dsu.find(dsu.nodes[a]).label
comp2 = dsu.find(dsu.nodes[b]).label
if comp1 == comp2:
print(0)
exit(0)
else:
index1 = new_graph[comp1]
index2 = new_graph[comp2]
if index1 >= x or index2 >= x:
print(-1)
exit(0)
dsu2.union(dsu2.nodes[index1], dsu2.nodes[index2])
adj[index1].append(index2)
adj[index2].append(index1)
for i in range(x):
if colors[i] == -1:
colors[i] = 0
ans = color(adj, colors, i)
if ans:
pass
else:
print(0)
exit(0)
comps = dsu2.size()
ans = 1
for _ in range(comps - 1):
ans *= 2
ans %= mod
print(ans)
exit(0)
except Exception as e:
print(e)
print(-1)
exit(0)
``` | output | 1 | 45,754 | 14 | 91,509 |
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