message stringlengths 2 48.6k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 318 108k | cluster float64 8 8 | __index_level_0__ int64 636 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Leha decided to move to a quiet town Vičkopolis, because he was tired by living in Bankopolis. Upon arrival he immediately began to expand his network of hacked computers. During the week Leha managed to get access to n computers throughout the town. Incidentally all the computers, which were hacked by Leha, lie on the same straight line, due to the reason that there is the only one straight street in Vičkopolis.
Let's denote the coordinate system on this street. Besides let's number all the hacked computers with integers from 1 to n. So the i-th hacked computer is located at the point xi. Moreover the coordinates of all computers are distinct.
Leha is determined to have a little rest after a hard week. Therefore he is going to invite his friend Noora to a restaurant. However the girl agrees to go on a date with the only one condition: Leha have to solve a simple task.
Leha should calculate a sum of F(a) for all a, where a is a non-empty subset of the set, that consists of all hacked computers. Formally, let's denote A the set of all integers from 1 to n. Noora asks the hacker to find value of the expression <image>. Here F(a) is calculated as the maximum among the distances between all pairs of computers from the set a. Formally, <image>. Since the required sum can be quite large Noora asks to find it modulo 109 + 7.
Though, Leha is too tired. Consequently he is not able to solve this task. Help the hacker to attend a date.
Input
The first line contains one integer n (1 ≤ n ≤ 3·105) denoting the number of hacked computers.
The second line contains n integers x1, x2, ..., xn (1 ≤ xi ≤ 109) denoting the coordinates of hacked computers. It is guaranteed that all xi are distinct.
Output
Print a single integer — the required sum modulo 109 + 7.
Examples
Input
2
4 7
Output
3
Input
3
4 3 1
Output
9
Note
There are three non-empty subsets in the first sample test:<image>, <image> and <image>. The first and the second subset increase the sum by 0 and the third subset increases the sum by 7 - 4 = 3. In total the answer is 0 + 0 + 3 = 3.
There are seven non-empty subsets in the second sample test. Among them only the following subsets increase the answer: <image>, <image>, <image>, <image>. In total the sum is (4 - 3) + (4 - 1) + (3 - 1) + (4 - 1) = 9.
Submitted Solution:
```
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
MOD = 10**9 + 7
MAX = 3 * 10**5
def y():
pass
def main():
n = int(input())
x = [int(x) for x in input().split()]
x.sort()
f = [1]
for i in range(1, n + 1):
f.append((f[-1] * i) % MOD)
s = 0
for i in range(n - 1):
m = n - 2 - i
s = (s + 2**m * (x[-1] - x[i])) % MOD
x.reverse()
for i in range(1, n - 1):
m = n - 2 - i
s = (s + 2**m * (x[i] - x[-1])) % MOD
print(s)
if __name__ == '__main__':
main()
``` | instruction | 0 | 31,478 | 8 | 62,956 |
No | output | 1 | 31,478 | 8 | 62,957 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Leha decided to move to a quiet town Vičkopolis, because he was tired by living in Bankopolis. Upon arrival he immediately began to expand his network of hacked computers. During the week Leha managed to get access to n computers throughout the town. Incidentally all the computers, which were hacked by Leha, lie on the same straight line, due to the reason that there is the only one straight street in Vičkopolis.
Let's denote the coordinate system on this street. Besides let's number all the hacked computers with integers from 1 to n. So the i-th hacked computer is located at the point xi. Moreover the coordinates of all computers are distinct.
Leha is determined to have a little rest after a hard week. Therefore he is going to invite his friend Noora to a restaurant. However the girl agrees to go on a date with the only one condition: Leha have to solve a simple task.
Leha should calculate a sum of F(a) for all a, where a is a non-empty subset of the set, that consists of all hacked computers. Formally, let's denote A the set of all integers from 1 to n. Noora asks the hacker to find value of the expression <image>. Here F(a) is calculated as the maximum among the distances between all pairs of computers from the set a. Formally, <image>. Since the required sum can be quite large Noora asks to find it modulo 109 + 7.
Though, Leha is too tired. Consequently he is not able to solve this task. Help the hacker to attend a date.
Input
The first line contains one integer n (1 ≤ n ≤ 3·105) denoting the number of hacked computers.
The second line contains n integers x1, x2, ..., xn (1 ≤ xi ≤ 109) denoting the coordinates of hacked computers. It is guaranteed that all xi are distinct.
Output
Print a single integer — the required sum modulo 109 + 7.
Examples
Input
2
4 7
Output
3
Input
3
4 3 1
Output
9
Note
There are three non-empty subsets in the first sample test:<image>, <image> and <image>. The first and the second subset increase the sum by 0 and the third subset increases the sum by 7 - 4 = 3. In total the answer is 0 + 0 + 3 = 3.
There are seven non-empty subsets in the second sample test. Among them only the following subsets increase the answer: <image>, <image>, <image>, <image>. In total the sum is (4 - 3) + (4 - 1) + (3 - 1) + (4 - 1) = 9.
Submitted Solution:
```
n = int(input())
arr = list(map(int,input().split()))
arr.sort()
mod = 10**9 + 7
print(mod)
s = 0
for i in range(1,n):
s+= arr[i]*(1 << (i-1)) # arr[i]*(2**(i-1))
ans = 0
for i in range(n-1):
# Мы находимся в индексе i, для каждого индекса g (g>i)
# мы можем составить 2^(g-i-1) (g-i-1 - кол-во элементов между i и g, каждый элемент можем
# включить так и не включить, от этого степень двойки)
# На этом шаге s - сумма максимальных элементах во всех подмножествах с минимальным
# элементом arr[i], а (arr[i])*((1<<(n-i-1)) - 1), кол-во таких подмножеств*arr[i]
ans+=s-(arr[i])*((1<<(n-i-1)) - 1)
ans%=mod
# Удаляем из s слудеющий элемент
s-=arr[i+1]
# кол-во подмножест оканчивающихся на g-ый элемент сокращается в двое для каждого g, а значит сумма
# тоже сокращается в двое
s//=2
print(ans%mod)
``` | instruction | 0 | 31,479 | 8 | 62,958 |
No | output | 1 | 31,479 | 8 | 62,959 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Maxim wants to buy an apartment in a new house at Line Avenue of Metropolis. The house has n apartments that are numbered from 1 to n and are arranged in a row. Two apartments are adjacent if their indices differ by 1. Some of the apartments can already be inhabited, others are available for sale.
Maxim often visits his neighbors, so apartment is good for him if it is available for sale and there is at least one already inhabited apartment adjacent to it. Maxim knows that there are exactly k already inhabited apartments, but he doesn't know their indices yet.
Find out what could be the minimum possible and the maximum possible number of apartments that are good for Maxim.
Input
The only line of the input contains two integers: n and k (1 ≤ n ≤ 109, 0 ≤ k ≤ n).
Output
Print the minimum possible and the maximum possible number of apartments good for Maxim.
Example
Input
6 3
Output
1 3
Note
In the sample test, the number of good apartments could be minimum possible if, for example, apartments with indices 1, 2 and 3 were inhabited. In this case only apartment 4 is good. The maximum possible number could be, for example, if apartments with indices 1, 3 and 5 were inhabited. In this case all other apartments: 2, 4 and 6 are good. | instruction | 0 | 31,480 | 8 | 62,960 |
Tags: constructive algorithms, math
Correct Solution:
```
n, k = map(int, input().split())
s = 0
if k == n:
minkv = 0
maxkv = 0
elif k == 0:
minkv = 0
maxkv = 0
else:
minkv = 1
s = n - k
if n % 3 != 0:
t = n // 3 + 1
else:
t = n // 3
if k == t:
maxkv = n - t
elif k < t:
maxkv = k * 2
else:
maxkv = n - t - (k - t)
print(minkv,maxkv)
``` | output | 1 | 31,480 | 8 | 62,961 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Maxim wants to buy an apartment in a new house at Line Avenue of Metropolis. The house has n apartments that are numbered from 1 to n and are arranged in a row. Two apartments are adjacent if their indices differ by 1. Some of the apartments can already be inhabited, others are available for sale.
Maxim often visits his neighbors, so apartment is good for him if it is available for sale and there is at least one already inhabited apartment adjacent to it. Maxim knows that there are exactly k already inhabited apartments, but he doesn't know their indices yet.
Find out what could be the minimum possible and the maximum possible number of apartments that are good for Maxim.
Input
The only line of the input contains two integers: n and k (1 ≤ n ≤ 109, 0 ≤ k ≤ n).
Output
Print the minimum possible and the maximum possible number of apartments good for Maxim.
Example
Input
6 3
Output
1 3
Note
In the sample test, the number of good apartments could be minimum possible if, for example, apartments with indices 1, 2 and 3 were inhabited. In this case only apartment 4 is good. The maximum possible number could be, for example, if apartments with indices 1, 3 and 5 were inhabited. In this case all other apartments: 2, 4 and 6 are good. | instruction | 0 | 31,481 | 8 | 62,962 |
Tags: constructive algorithms, math
Correct Solution:
```
n, k = map(int, input().split())
if k in [0, n]:
print(0, 0)
else:
print(1, min(n - k, k * 2))
``` | output | 1 | 31,481 | 8 | 62,963 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Maxim wants to buy an apartment in a new house at Line Avenue of Metropolis. The house has n apartments that are numbered from 1 to n and are arranged in a row. Two apartments are adjacent if their indices differ by 1. Some of the apartments can already be inhabited, others are available for sale.
Maxim often visits his neighbors, so apartment is good for him if it is available for sale and there is at least one already inhabited apartment adjacent to it. Maxim knows that there are exactly k already inhabited apartments, but he doesn't know their indices yet.
Find out what could be the minimum possible and the maximum possible number of apartments that are good for Maxim.
Input
The only line of the input contains two integers: n and k (1 ≤ n ≤ 109, 0 ≤ k ≤ n).
Output
Print the minimum possible and the maximum possible number of apartments good for Maxim.
Example
Input
6 3
Output
1 3
Note
In the sample test, the number of good apartments could be minimum possible if, for example, apartments with indices 1, 2 and 3 were inhabited. In this case only apartment 4 is good. The maximum possible number could be, for example, if apartments with indices 1, 3 and 5 were inhabited. In this case all other apartments: 2, 4 and 6 are good. | instruction | 0 | 31,482 | 8 | 62,964 |
Tags: constructive algorithms, math
Correct Solution:
```
from sys import exit
n, k = map(int, input().split())
if k == 0:
print(0, 0)
else:
print(min(1, n - k), min(2 * k, n - k))
``` | output | 1 | 31,482 | 8 | 62,965 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Maxim wants to buy an apartment in a new house at Line Avenue of Metropolis. The house has n apartments that are numbered from 1 to n and are arranged in a row. Two apartments are adjacent if their indices differ by 1. Some of the apartments can already be inhabited, others are available for sale.
Maxim often visits his neighbors, so apartment is good for him if it is available for sale and there is at least one already inhabited apartment adjacent to it. Maxim knows that there are exactly k already inhabited apartments, but he doesn't know their indices yet.
Find out what could be the minimum possible and the maximum possible number of apartments that are good for Maxim.
Input
The only line of the input contains two integers: n and k (1 ≤ n ≤ 109, 0 ≤ k ≤ n).
Output
Print the minimum possible and the maximum possible number of apartments good for Maxim.
Example
Input
6 3
Output
1 3
Note
In the sample test, the number of good apartments could be minimum possible if, for example, apartments with indices 1, 2 and 3 were inhabited. In this case only apartment 4 is good. The maximum possible number could be, for example, if apartments with indices 1, 3 and 5 were inhabited. In this case all other apartments: 2, 4 and 6 are good. | instruction | 0 | 31,483 | 8 | 62,966 |
Tags: constructive algorithms, math
Correct Solution:
```
n, k = map(int, input().split())
print(0 if n == k or k == 0 else 1, min(2 * k, n - k))
``` | output | 1 | 31,483 | 8 | 62,967 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Maxim wants to buy an apartment in a new house at Line Avenue of Metropolis. The house has n apartments that are numbered from 1 to n and are arranged in a row. Two apartments are adjacent if their indices differ by 1. Some of the apartments can already be inhabited, others are available for sale.
Maxim often visits his neighbors, so apartment is good for him if it is available for sale and there is at least one already inhabited apartment adjacent to it. Maxim knows that there are exactly k already inhabited apartments, but he doesn't know their indices yet.
Find out what could be the minimum possible and the maximum possible number of apartments that are good for Maxim.
Input
The only line of the input contains two integers: n and k (1 ≤ n ≤ 109, 0 ≤ k ≤ n).
Output
Print the minimum possible and the maximum possible number of apartments good for Maxim.
Example
Input
6 3
Output
1 3
Note
In the sample test, the number of good apartments could be minimum possible if, for example, apartments with indices 1, 2 and 3 were inhabited. In this case only apartment 4 is good. The maximum possible number could be, for example, if apartments with indices 1, 3 and 5 were inhabited. In this case all other apartments: 2, 4 and 6 are good. | instruction | 0 | 31,484 | 8 | 62,968 |
Tags: constructive algorithms, math
Correct Solution:
```
n,k=map(int,input().split())
print(min(1,k,n-k),min(n-k,2*k))
``` | output | 1 | 31,484 | 8 | 62,969 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Maxim wants to buy an apartment in a new house at Line Avenue of Metropolis. The house has n apartments that are numbered from 1 to n and are arranged in a row. Two apartments are adjacent if their indices differ by 1. Some of the apartments can already be inhabited, others are available for sale.
Maxim often visits his neighbors, so apartment is good for him if it is available for sale and there is at least one already inhabited apartment adjacent to it. Maxim knows that there are exactly k already inhabited apartments, but he doesn't know their indices yet.
Find out what could be the minimum possible and the maximum possible number of apartments that are good for Maxim.
Input
The only line of the input contains two integers: n and k (1 ≤ n ≤ 109, 0 ≤ k ≤ n).
Output
Print the minimum possible and the maximum possible number of apartments good for Maxim.
Example
Input
6 3
Output
1 3
Note
In the sample test, the number of good apartments could be minimum possible if, for example, apartments with indices 1, 2 and 3 were inhabited. In this case only apartment 4 is good. The maximum possible number could be, for example, if apartments with indices 1, 3 and 5 were inhabited. In this case all other apartments: 2, 4 and 6 are good. | instruction | 0 | 31,485 | 8 | 62,970 |
Tags: constructive algorithms, math
Correct Solution:
```
n,k=map(int,input().split())
if k==0:print(0,0)
else:print(min(n-k,1),min((k-1)*2+2,n-k))
``` | output | 1 | 31,485 | 8 | 62,971 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Maxim wants to buy an apartment in a new house at Line Avenue of Metropolis. The house has n apartments that are numbered from 1 to n and are arranged in a row. Two apartments are adjacent if their indices differ by 1. Some of the apartments can already be inhabited, others are available for sale.
Maxim often visits his neighbors, so apartment is good for him if it is available for sale and there is at least one already inhabited apartment adjacent to it. Maxim knows that there are exactly k already inhabited apartments, but he doesn't know their indices yet.
Find out what could be the minimum possible and the maximum possible number of apartments that are good for Maxim.
Input
The only line of the input contains two integers: n and k (1 ≤ n ≤ 109, 0 ≤ k ≤ n).
Output
Print the minimum possible and the maximum possible number of apartments good for Maxim.
Example
Input
6 3
Output
1 3
Note
In the sample test, the number of good apartments could be minimum possible if, for example, apartments with indices 1, 2 and 3 were inhabited. In this case only apartment 4 is good. The maximum possible number could be, for example, if apartments with indices 1, 3 and 5 were inhabited. In this case all other apartments: 2, 4 and 6 are good. | instruction | 0 | 31,486 | 8 | 62,972 |
Tags: constructive algorithms, math
Correct Solution:
```
n,k = map(int,input().split())
if k==0 :
print(0,0)
elif k==n :
print(0,0)
else :
print(1,min(n-k,k*2))
``` | output | 1 | 31,486 | 8 | 62,973 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Maxim wants to buy an apartment in a new house at Line Avenue of Metropolis. The house has n apartments that are numbered from 1 to n and are arranged in a row. Two apartments are adjacent if their indices differ by 1. Some of the apartments can already be inhabited, others are available for sale.
Maxim often visits his neighbors, so apartment is good for him if it is available for sale and there is at least one already inhabited apartment adjacent to it. Maxim knows that there are exactly k already inhabited apartments, but he doesn't know their indices yet.
Find out what could be the minimum possible and the maximum possible number of apartments that are good for Maxim.
Input
The only line of the input contains two integers: n and k (1 ≤ n ≤ 109, 0 ≤ k ≤ n).
Output
Print the minimum possible and the maximum possible number of apartments good for Maxim.
Example
Input
6 3
Output
1 3
Note
In the sample test, the number of good apartments could be minimum possible if, for example, apartments with indices 1, 2 and 3 were inhabited. In this case only apartment 4 is good. The maximum possible number could be, for example, if apartments with indices 1, 3 and 5 were inhabited. In this case all other apartments: 2, 4 and 6 are good. | instruction | 0 | 31,487 | 8 | 62,974 |
Tags: constructive algorithms, math
Correct Solution:
```
n,k=map(int,input().split())
k2=k
mn=1 if k<n else 0
temp=n//3
mx=0
if temp<=k:
mx+=temp*2
k-=temp
else:
mx+=k*2
k=0
if n%3==1:
if k>0: k-=1
if n%3==2:
if k>0:
mx+=1
k-=1
mx-=k
if k2==0: print(0,0)
else: print(mn,mx)
``` | output | 1 | 31,487 | 8 | 62,975 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Maxim wants to buy an apartment in a new house at Line Avenue of Metropolis. The house has n apartments that are numbered from 1 to n and are arranged in a row. Two apartments are adjacent if their indices differ by 1. Some of the apartments can already be inhabited, others are available for sale.
Maxim often visits his neighbors, so apartment is good for him if it is available for sale and there is at least one already inhabited apartment adjacent to it. Maxim knows that there are exactly k already inhabited apartments, but he doesn't know their indices yet.
Find out what could be the minimum possible and the maximum possible number of apartments that are good for Maxim.
Input
The only line of the input contains two integers: n and k (1 ≤ n ≤ 109, 0 ≤ k ≤ n).
Output
Print the minimum possible and the maximum possible number of apartments good for Maxim.
Example
Input
6 3
Output
1 3
Note
In the sample test, the number of good apartments could be minimum possible if, for example, apartments with indices 1, 2 and 3 were inhabited. In this case only apartment 4 is good. The maximum possible number could be, for example, if apartments with indices 1, 3 and 5 were inhabited. In this case all other apartments: 2, 4 and 6 are good.
Submitted Solution:
```
n,k=map(int,input().split())
print(min(k,1,1-(k==n)),min(k*2,n-k))
``` | instruction | 0 | 31,488 | 8 | 62,976 |
Yes | output | 1 | 31,488 | 8 | 62,977 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Maxim wants to buy an apartment in a new house at Line Avenue of Metropolis. The house has n apartments that are numbered from 1 to n and are arranged in a row. Two apartments are adjacent if their indices differ by 1. Some of the apartments can already be inhabited, others are available for sale.
Maxim often visits his neighbors, so apartment is good for him if it is available for sale and there is at least one already inhabited apartment adjacent to it. Maxim knows that there are exactly k already inhabited apartments, but he doesn't know their indices yet.
Find out what could be the minimum possible and the maximum possible number of apartments that are good for Maxim.
Input
The only line of the input contains two integers: n and k (1 ≤ n ≤ 109, 0 ≤ k ≤ n).
Output
Print the minimum possible and the maximum possible number of apartments good for Maxim.
Example
Input
6 3
Output
1 3
Note
In the sample test, the number of good apartments could be minimum possible if, for example, apartments with indices 1, 2 and 3 were inhabited. In this case only apartment 4 is good. The maximum possible number could be, for example, if apartments with indices 1, 3 and 5 were inhabited. In this case all other apartments: 2, 4 and 6 are good.
Submitted Solution:
```
n, k = map(int, input().split())
print(str(min(1, n-k, k)) + " " + str(min(2*k, n-k)))
``` | instruction | 0 | 31,489 | 8 | 62,978 |
Yes | output | 1 | 31,489 | 8 | 62,979 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Maxim wants to buy an apartment in a new house at Line Avenue of Metropolis. The house has n apartments that are numbered from 1 to n and are arranged in a row. Two apartments are adjacent if their indices differ by 1. Some of the apartments can already be inhabited, others are available for sale.
Maxim often visits his neighbors, so apartment is good for him if it is available for sale and there is at least one already inhabited apartment adjacent to it. Maxim knows that there are exactly k already inhabited apartments, but he doesn't know their indices yet.
Find out what could be the minimum possible and the maximum possible number of apartments that are good for Maxim.
Input
The only line of the input contains two integers: n and k (1 ≤ n ≤ 109, 0 ≤ k ≤ n).
Output
Print the minimum possible and the maximum possible number of apartments good for Maxim.
Example
Input
6 3
Output
1 3
Note
In the sample test, the number of good apartments could be minimum possible if, for example, apartments with indices 1, 2 and 3 were inhabited. In this case only apartment 4 is good. The maximum possible number could be, for example, if apartments with indices 1, 3 and 5 were inhabited. In this case all other apartments: 2, 4 and 6 are good.
Submitted Solution:
```
n,k = map(int,input().split())
x = (n//3)
if k==0 or k==n:
print(0,end=' ')
print(0)
elif x >= k:
print(1,end=' ')
print(k*2)
elif k>x:
print(1,end=' ')
print(n-k)
``` | instruction | 0 | 31,490 | 8 | 62,980 |
Yes | output | 1 | 31,490 | 8 | 62,981 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Maxim wants to buy an apartment in a new house at Line Avenue of Metropolis. The house has n apartments that are numbered from 1 to n and are arranged in a row. Two apartments are adjacent if their indices differ by 1. Some of the apartments can already be inhabited, others are available for sale.
Maxim often visits his neighbors, so apartment is good for him if it is available for sale and there is at least one already inhabited apartment adjacent to it. Maxim knows that there are exactly k already inhabited apartments, but he doesn't know their indices yet.
Find out what could be the minimum possible and the maximum possible number of apartments that are good for Maxim.
Input
The only line of the input contains two integers: n and k (1 ≤ n ≤ 109, 0 ≤ k ≤ n).
Output
Print the minimum possible and the maximum possible number of apartments good for Maxim.
Example
Input
6 3
Output
1 3
Note
In the sample test, the number of good apartments could be minimum possible if, for example, apartments with indices 1, 2 and 3 were inhabited. In this case only apartment 4 is good. The maximum possible number could be, for example, if apartments with indices 1, 3 and 5 were inhabited. In this case all other apartments: 2, 4 and 6 are good.
Submitted Solution:
```
#! /usr/bin/env python3
n, k = map(int, input().split())
if k == n:
print(0, 0)
elif 3 *k > n:
print(1, n-k)
else:
print(0 if k == 0 else 1, 2 * k)
``` | instruction | 0 | 31,491 | 8 | 62,982 |
Yes | output | 1 | 31,491 | 8 | 62,983 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Maxim wants to buy an apartment in a new house at Line Avenue of Metropolis. The house has n apartments that are numbered from 1 to n and are arranged in a row. Two apartments are adjacent if their indices differ by 1. Some of the apartments can already be inhabited, others are available for sale.
Maxim often visits his neighbors, so apartment is good for him if it is available for sale and there is at least one already inhabited apartment adjacent to it. Maxim knows that there are exactly k already inhabited apartments, but he doesn't know their indices yet.
Find out what could be the minimum possible and the maximum possible number of apartments that are good for Maxim.
Input
The only line of the input contains two integers: n and k (1 ≤ n ≤ 109, 0 ≤ k ≤ n).
Output
Print the minimum possible and the maximum possible number of apartments good for Maxim.
Example
Input
6 3
Output
1 3
Note
In the sample test, the number of good apartments could be minimum possible if, for example, apartments with indices 1, 2 and 3 were inhabited. In this case only apartment 4 is good. The maximum possible number could be, for example, if apartments with indices 1, 3 and 5 were inhabited. In this case all other apartments: 2, 4 and 6 are good.
Submitted Solution:
```
n,k=map(int,input().split())
if k==1:
print(1,k+1)
else:
print(1,n-k)
``` | instruction | 0 | 31,492 | 8 | 62,984 |
No | output | 1 | 31,492 | 8 | 62,985 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Maxim wants to buy an apartment in a new house at Line Avenue of Metropolis. The house has n apartments that are numbered from 1 to n and are arranged in a row. Two apartments are adjacent if their indices differ by 1. Some of the apartments can already be inhabited, others are available for sale.
Maxim often visits his neighbors, so apartment is good for him if it is available for sale and there is at least one already inhabited apartment adjacent to it. Maxim knows that there are exactly k already inhabited apartments, but he doesn't know their indices yet.
Find out what could be the minimum possible and the maximum possible number of apartments that are good for Maxim.
Input
The only line of the input contains two integers: n and k (1 ≤ n ≤ 109, 0 ≤ k ≤ n).
Output
Print the minimum possible and the maximum possible number of apartments good for Maxim.
Example
Input
6 3
Output
1 3
Note
In the sample test, the number of good apartments could be minimum possible if, for example, apartments with indices 1, 2 and 3 were inhabited. In this case only apartment 4 is good. The maximum possible number could be, for example, if apartments with indices 1, 3 and 5 were inhabited. In this case all other apartments: 2, 4 and 6 are good.
Submitted Solution:
```
from sys import maxsize, stdout, stdin,stderr
mod = int(1e9 + 7)
def tup():return map(int,stdin.readline().split())
def I(): return int(stdin.readline())
def lint(): return [int(x) for x in stdin.readline().split()]
def S(): return input().strip()
def grid(r, c): return [lint() for i in range(r)]
def debug(*args, c=6): print('\033[3{}m'.format(c), *args, '\033[0m', file=stderr)
from math import log2,sqrt
from collections import Counter,defaultdict
from itertools import permutations
import re
# def bark():
# n = I(); s =S()
# a = re.split('[A-Z]+',s)
# p = max(map(lambda x: len(set(x)),a))
# return p
from math import floor,sqrt,ceil
#
# def bark():
# n = I()
# s = int(sqrt(n))
# for i in range(s,n+1):
# if i*s >=n:
# print(2*(i+s))
# #find some area greater than or equal to n and then find it's perimeter since it's a rectange use 2*(l+b)
# break
# if __name__ == '__main__':
# bark()
#print(375.80 , 30000.30000)
#print(1234567890000000/123456789)
# from math import gcd
# def kround():
# n , k =tup()
# a , b = n , pow(10,k)
# l = (a*b) // gcd(a,b)
# print(l) #look the definiton of lcm
# #ughhhhhh lcm is the lowest common multiple that divides both the numbers
#
#
# if __name__ == '__main__':
# kround()
def aps():
n , k = tup()
if n==k or k==0:mi = 0
elif k >0 : mi = 1
x =n//2
od = n- x
if od ==k:ma = k
elif k < od:
ma = 2*k
else:ma = n-k
print(mi, ma)
if __name__ == '__main__':
aps()
``` | instruction | 0 | 31,493 | 8 | 62,986 |
No | output | 1 | 31,493 | 8 | 62,987 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Maxim wants to buy an apartment in a new house at Line Avenue of Metropolis. The house has n apartments that are numbered from 1 to n and are arranged in a row. Two apartments are adjacent if their indices differ by 1. Some of the apartments can already be inhabited, others are available for sale.
Maxim often visits his neighbors, so apartment is good for him if it is available for sale and there is at least one already inhabited apartment adjacent to it. Maxim knows that there are exactly k already inhabited apartments, but he doesn't know their indices yet.
Find out what could be the minimum possible and the maximum possible number of apartments that are good for Maxim.
Input
The only line of the input contains two integers: n and k (1 ≤ n ≤ 109, 0 ≤ k ≤ n).
Output
Print the minimum possible and the maximum possible number of apartments good for Maxim.
Example
Input
6 3
Output
1 3
Note
In the sample test, the number of good apartments could be minimum possible if, for example, apartments with indices 1, 2 and 3 were inhabited. In this case only apartment 4 is good. The maximum possible number could be, for example, if apartments with indices 1, 3 and 5 were inhabited. In this case all other apartments: 2, 4 and 6 are good.
Submitted Solution:
```
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
n, k = map(int, input().split())
if n == k:
print(0, 0)
else:
a = n // 2
if k <= a:
print(1, k)
else:
print(1, n - k)
``` | instruction | 0 | 31,494 | 8 | 62,988 |
No | output | 1 | 31,494 | 8 | 62,989 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Maxim wants to buy an apartment in a new house at Line Avenue of Metropolis. The house has n apartments that are numbered from 1 to n and are arranged in a row. Two apartments are adjacent if their indices differ by 1. Some of the apartments can already be inhabited, others are available for sale.
Maxim often visits his neighbors, so apartment is good for him if it is available for sale and there is at least one already inhabited apartment adjacent to it. Maxim knows that there are exactly k already inhabited apartments, but he doesn't know their indices yet.
Find out what could be the minimum possible and the maximum possible number of apartments that are good for Maxim.
Input
The only line of the input contains two integers: n and k (1 ≤ n ≤ 109, 0 ≤ k ≤ n).
Output
Print the minimum possible and the maximum possible number of apartments good for Maxim.
Example
Input
6 3
Output
1 3
Note
In the sample test, the number of good apartments could be minimum possible if, for example, apartments with indices 1, 2 and 3 were inhabited. In this case only apartment 4 is good. The maximum possible number could be, for example, if apartments with indices 1, 3 and 5 were inhabited. In this case all other apartments: 2, 4 and 6 are good.
Submitted Solution:
```
n,k=list(map(int, input().split()))
minv = 1
maxv = 1
if (k == 0 and k == n):
minv=0
maxv=min(k*2, n - k)
print(minv, maxv)
``` | instruction | 0 | 31,495 | 8 | 62,990 |
No | output | 1 | 31,495 | 8 | 62,991 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The German University in Cairo (GUC) dorm houses are numbered from 1 to n. Underground water pipes connect these houses together. Each pipe has certain direction (water can flow only in this direction and not vice versa), and diameter (which characterizes the maximal amount of water it can handle).
For each house, there is at most one pipe going into it and at most one pipe going out of it. With the new semester starting, GUC student and dorm resident, Lulu, wants to install tanks and taps at the dorms. For every house with an outgoing water pipe and without an incoming water pipe, Lulu should install a water tank at that house. For every house with an incoming water pipe and without an outgoing water pipe, Lulu should install a water tap at that house. Each tank house will convey water to all houses that have a sequence of pipes from the tank to it. Accordingly, each tap house will receive water originating from some tank house.
In order to avoid pipes from bursting one week later (like what happened last semester), Lulu also has to consider the diameter of the pipes. The amount of water each tank conveys should not exceed the diameter of the pipes connecting a tank to its corresponding tap. Lulu wants to find the maximal amount of water that can be safely conveyed from each tank to its corresponding tap.
Input
The first line contains two space-separated integers n and p (1 ≤ n ≤ 1000, 0 ≤ p ≤ n) — the number of houses and the number of pipes correspondingly.
Then p lines follow — the description of p pipes. The i-th line contains three integers ai bi di, indicating a pipe of diameter di going from house ai to house bi (1 ≤ ai, bi ≤ n, ai ≠ bi, 1 ≤ di ≤ 106).
It is guaranteed that for each house there is at most one pipe going into it and at most one pipe going out of it.
Output
Print integer t in the first line — the number of tank-tap pairs of houses.
For the next t lines, print 3 integers per line, separated by spaces: tanki, tapi, and diameteri, where tanki ≠ tapi (1 ≤ i ≤ t). Here tanki and tapi are indexes of tank and tap houses respectively, and diameteri is the maximum amount of water that can be conveyed. All the t lines should be ordered (increasingly) by tanki.
Examples
Input
3 2
1 2 10
2 3 20
Output
1
1 3 10
Input
3 3
1 2 20
2 3 10
3 1 5
Output
0
Input
4 2
1 2 60
3 4 50
Output
2
1 2 60
3 4 50 | instruction | 0 | 31,791 | 8 | 63,582 |
Tags: dfs and similar, graphs
Correct Solution:
```
n, p = input().split(" ")
n = int(n)
p = int(p)
pipes=[]
for i in range(p):
pipes.append(input().split(" "))
cant=[]
while len(pipes)!=0:
index = 0
ciclo = False
maximo = float('inf')
pipe = []
first = 0
last = 0
while index<len(pipes):
if first==0:
first = int(pipes[index][0])
last = int(pipes[index][1])
pipe = pipes.pop(index)
else:
if int(pipes[index][0]) == last:
if int(pipes[index][1]) == first:
pipe = pipes.pop(index)
ciclo = True
index = len(pipes)
else:
last = int(pipes[index][1])
pipe = pipes.pop(index)
index = 0
elif int(pipes[index][1]) == first:
first = int(pipes[index][0])
pipe = pipes.pop(index)
index = 0
else:
index += 1
if int(pipe[2]) < float(maximo):
maximo = pipe[2]
if not ciclo and first != last:
cant.append([first,last,maximo])
cant.sort()
print(len(cant))
for i in cant:
print(i[0],i[1],i[2])
``` | output | 1 | 31,791 | 8 | 63,583 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The German University in Cairo (GUC) dorm houses are numbered from 1 to n. Underground water pipes connect these houses together. Each pipe has certain direction (water can flow only in this direction and not vice versa), and diameter (which characterizes the maximal amount of water it can handle).
For each house, there is at most one pipe going into it and at most one pipe going out of it. With the new semester starting, GUC student and dorm resident, Lulu, wants to install tanks and taps at the dorms. For every house with an outgoing water pipe and without an incoming water pipe, Lulu should install a water tank at that house. For every house with an incoming water pipe and without an outgoing water pipe, Lulu should install a water tap at that house. Each tank house will convey water to all houses that have a sequence of pipes from the tank to it. Accordingly, each tap house will receive water originating from some tank house.
In order to avoid pipes from bursting one week later (like what happened last semester), Lulu also has to consider the diameter of the pipes. The amount of water each tank conveys should not exceed the diameter of the pipes connecting a tank to its corresponding tap. Lulu wants to find the maximal amount of water that can be safely conveyed from each tank to its corresponding tap.
Input
The first line contains two space-separated integers n and p (1 ≤ n ≤ 1000, 0 ≤ p ≤ n) — the number of houses and the number of pipes correspondingly.
Then p lines follow — the description of p pipes. The i-th line contains three integers ai bi di, indicating a pipe of diameter di going from house ai to house bi (1 ≤ ai, bi ≤ n, ai ≠ bi, 1 ≤ di ≤ 106).
It is guaranteed that for each house there is at most one pipe going into it and at most one pipe going out of it.
Output
Print integer t in the first line — the number of tank-tap pairs of houses.
For the next t lines, print 3 integers per line, separated by spaces: tanki, tapi, and diameteri, where tanki ≠ tapi (1 ≤ i ≤ t). Here tanki and tapi are indexes of tank and tap houses respectively, and diameteri is the maximum amount of water that can be conveyed. All the t lines should be ordered (increasingly) by tanki.
Examples
Input
3 2
1 2 10
2 3 20
Output
1
1 3 10
Input
3 3
1 2 20
2 3 10
3 1 5
Output
0
Input
4 2
1 2 60
3 4 50
Output
2
1 2 60
3 4 50 | instruction | 0 | 31,792 | 8 | 63,584 |
Tags: dfs and similar, graphs
Correct Solution:
```
from collections import defaultdict
n,m=map(int,input().strip().split())
inc=[None]*(n)
out=[None]*(n)
for i in range(m):
x,y,z=map(int,input().strip().split())
inc[y-1]=(x-1,z)
out[x-1]=(y-1,z)
if m==0:
print(0)
exit(0)
i=0
ans=[]
for i in range(0,n):
tank, tap, weight = None, None, float("inf")
t=i
vis = [0] * (n)
# print(vis)
while vis!=[1]*(n) and vis[t]!=1:
if tank==None and inc[t]==None:
tank=t
if out[t]:
weight=min(out[t][1],weight)
vis[t] = 1
if out[t]!=None:
t=out[t][0]
else:
break
elif tank!=None and out[t]==None:
tap=t
weight=min(inc[t][1],weight)
vis[t]=1
break
else:
vis[t]=1
weight=min(inc[t][1],weight)
if out[t]!=None:
t=out[t][0]
else:
break
if tank!=None and tap!=None:
ans.append((tank+1,tap+1,weight))
if vis==[1]*(n):
break
print(len(ans))
for i in ans:
print(*i)
``` | output | 1 | 31,792 | 8 | 63,585 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The German University in Cairo (GUC) dorm houses are numbered from 1 to n. Underground water pipes connect these houses together. Each pipe has certain direction (water can flow only in this direction and not vice versa), and diameter (which characterizes the maximal amount of water it can handle).
For each house, there is at most one pipe going into it and at most one pipe going out of it. With the new semester starting, GUC student and dorm resident, Lulu, wants to install tanks and taps at the dorms. For every house with an outgoing water pipe and without an incoming water pipe, Lulu should install a water tank at that house. For every house with an incoming water pipe and without an outgoing water pipe, Lulu should install a water tap at that house. Each tank house will convey water to all houses that have a sequence of pipes from the tank to it. Accordingly, each tap house will receive water originating from some tank house.
In order to avoid pipes from bursting one week later (like what happened last semester), Lulu also has to consider the diameter of the pipes. The amount of water each tank conveys should not exceed the diameter of the pipes connecting a tank to its corresponding tap. Lulu wants to find the maximal amount of water that can be safely conveyed from each tank to its corresponding tap.
Input
The first line contains two space-separated integers n and p (1 ≤ n ≤ 1000, 0 ≤ p ≤ n) — the number of houses and the number of pipes correspondingly.
Then p lines follow — the description of p pipes. The i-th line contains three integers ai bi di, indicating a pipe of diameter di going from house ai to house bi (1 ≤ ai, bi ≤ n, ai ≠ bi, 1 ≤ di ≤ 106).
It is guaranteed that for each house there is at most one pipe going into it and at most one pipe going out of it.
Output
Print integer t in the first line — the number of tank-tap pairs of houses.
For the next t lines, print 3 integers per line, separated by spaces: tanki, tapi, and diameteri, where tanki ≠ tapi (1 ≤ i ≤ t). Here tanki and tapi are indexes of tank and tap houses respectively, and diameteri is the maximum amount of water that can be conveyed. All the t lines should be ordered (increasingly) by tanki.
Examples
Input
3 2
1 2 10
2 3 20
Output
1
1 3 10
Input
3 3
1 2 20
2 3 10
3 1 5
Output
0
Input
4 2
1 2 60
3 4 50
Output
2
1 2 60
3 4 50 | instruction | 0 | 31,793 | 8 | 63,586 |
Tags: dfs and similar, graphs
Correct Solution:
```
#!/usr/bin/env python
inf = 2*(10**9)
res = []
In = [0]*1001
out = [0]*1001
f = [0]*1001
s = input().split()
n=int(s[0])
m=int(s[1])
u, v, d = (0,)*3
for i in range(1001):
In[i],out[i],f[i]= (-1,)*3
for i in range(m):
s = input().split()
u,v,d =int(s[0]),int(s[1]),int(s[2])
f[u], In[v], out[u] = d, u, v
for i in range(1,n+1):
if ( In[i]==-1 ):
u,v,w,cnt,c = i,0,inf,0,0
while ( out[u]!= -1):
cnt+=1
if ( cnt >n-1):
c=1
break
v = out[u]
w = min( w ,f[u] )
u=v
if(c or (w == inf )):
continue
res.append([i, v, w ])
print (len(res),end='\n')
for i in range (len(res)):
print (res[i][0],res[i][1],res[i][2] ,sep=' ')
``` | output | 1 | 31,793 | 8 | 63,587 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The German University in Cairo (GUC) dorm houses are numbered from 1 to n. Underground water pipes connect these houses together. Each pipe has certain direction (water can flow only in this direction and not vice versa), and diameter (which characterizes the maximal amount of water it can handle).
For each house, there is at most one pipe going into it and at most one pipe going out of it. With the new semester starting, GUC student and dorm resident, Lulu, wants to install tanks and taps at the dorms. For every house with an outgoing water pipe and without an incoming water pipe, Lulu should install a water tank at that house. For every house with an incoming water pipe and without an outgoing water pipe, Lulu should install a water tap at that house. Each tank house will convey water to all houses that have a sequence of pipes from the tank to it. Accordingly, each tap house will receive water originating from some tank house.
In order to avoid pipes from bursting one week later (like what happened last semester), Lulu also has to consider the diameter of the pipes. The amount of water each tank conveys should not exceed the diameter of the pipes connecting a tank to its corresponding tap. Lulu wants to find the maximal amount of water that can be safely conveyed from each tank to its corresponding tap.
Input
The first line contains two space-separated integers n and p (1 ≤ n ≤ 1000, 0 ≤ p ≤ n) — the number of houses and the number of pipes correspondingly.
Then p lines follow — the description of p pipes. The i-th line contains three integers ai bi di, indicating a pipe of diameter di going from house ai to house bi (1 ≤ ai, bi ≤ n, ai ≠ bi, 1 ≤ di ≤ 106).
It is guaranteed that for each house there is at most one pipe going into it and at most one pipe going out of it.
Output
Print integer t in the first line — the number of tank-tap pairs of houses.
For the next t lines, print 3 integers per line, separated by spaces: tanki, tapi, and diameteri, where tanki ≠ tapi (1 ≤ i ≤ t). Here tanki and tapi are indexes of tank and tap houses respectively, and diameteri is the maximum amount of water that can be conveyed. All the t lines should be ordered (increasingly) by tanki.
Examples
Input
3 2
1 2 10
2 3 20
Output
1
1 3 10
Input
3 3
1 2 20
2 3 10
3 1 5
Output
0
Input
4 2
1 2 60
3 4 50
Output
2
1 2 60
3 4 50 | instruction | 0 | 31,794 | 8 | 63,588 |
Tags: dfs and similar, graphs
Correct Solution:
```
from sys import stdin, stdout
nmbr = lambda: int(stdin.readline())
lst = lambda: list(map(int, stdin.readline().split()))
def dfs(src):
global mn
if g[src]==src or g[src]==-1:return src
neigh=g[src]
wt=cost[src,neigh]
mn=min(wt,mn)
return dfs(neigh)
for _ in range(1):#nmbr()):
n,eg=lst()
ind=[0]*(1+n)
g=[-1]*(1+n)
cost={}
ans=[]
for i in range(eg):
u,v,wt=lst()
g[u]=v
cost[u,v]=wt
ind[v]+=1
for i in range(1,n+1):
if ind[i]==0:
mn=float('inf')
des=dfs(i)
if des!=i:
ans+=[[i,des,mn]]
print(len(ans))
for k,v,w in ans:
print(k,v,w)
``` | output | 1 | 31,794 | 8 | 63,589 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The German University in Cairo (GUC) dorm houses are numbered from 1 to n. Underground water pipes connect these houses together. Each pipe has certain direction (water can flow only in this direction and not vice versa), and diameter (which characterizes the maximal amount of water it can handle).
For each house, there is at most one pipe going into it and at most one pipe going out of it. With the new semester starting, GUC student and dorm resident, Lulu, wants to install tanks and taps at the dorms. For every house with an outgoing water pipe and without an incoming water pipe, Lulu should install a water tank at that house. For every house with an incoming water pipe and without an outgoing water pipe, Lulu should install a water tap at that house. Each tank house will convey water to all houses that have a sequence of pipes from the tank to it. Accordingly, each tap house will receive water originating from some tank house.
In order to avoid pipes from bursting one week later (like what happened last semester), Lulu also has to consider the diameter of the pipes. The amount of water each tank conveys should not exceed the diameter of the pipes connecting a tank to its corresponding tap. Lulu wants to find the maximal amount of water that can be safely conveyed from each tank to its corresponding tap.
Input
The first line contains two space-separated integers n and p (1 ≤ n ≤ 1000, 0 ≤ p ≤ n) — the number of houses and the number of pipes correspondingly.
Then p lines follow — the description of p pipes. The i-th line contains three integers ai bi di, indicating a pipe of diameter di going from house ai to house bi (1 ≤ ai, bi ≤ n, ai ≠ bi, 1 ≤ di ≤ 106).
It is guaranteed that for each house there is at most one pipe going into it and at most one pipe going out of it.
Output
Print integer t in the first line — the number of tank-tap pairs of houses.
For the next t lines, print 3 integers per line, separated by spaces: tanki, tapi, and diameteri, where tanki ≠ tapi (1 ≤ i ≤ t). Here tanki and tapi are indexes of tank and tap houses respectively, and diameteri is the maximum amount of water that can be conveyed. All the t lines should be ordered (increasingly) by tanki.
Examples
Input
3 2
1 2 10
2 3 20
Output
1
1 3 10
Input
3 3
1 2 20
2 3 10
3 1 5
Output
0
Input
4 2
1 2 60
3 4 50
Output
2
1 2 60
3 4 50 | instruction | 0 | 31,795 | 8 | 63,590 |
Tags: dfs and similar, graphs
Correct Solution:
```
from collections import *
import os, sys
from io import BytesIO, IOBase
def main():
n, m = rints()
g = graph(n)
for i in range(m):
u, v, w = rints()
g.addEdge(u, v, w)
g.dfs()
class graph:
def __init__(self, n):
self.n, self.gdict = n, [[] for _ in range(n + 1)]
self.indeg, self.outdeg = [False] * (n + 1), [False] * (n + 1)
def addEdge(self, node1, node2, w=None):
self.gdict[node1].append((node2, w))
self.indeg[node2] = True
self.outdeg[node1] = True
def dfsUtil(self, v):
stack, vis = [(v, float('inf'))], [False] * (self.n + 1)
vis[v] = True
while (stack):
s, w = stack.pop()
if not self.gdict[s]:
self.ans.append(f'{v} {s} {w}')
continue
for i1, j1 in self.gdict[s]:
if not vis[i1]:
stack.append((i1, min(j1, w)))
vis[i1] = True
def dfs(self):
self.ans = []
for i in range(1, self.n + 1):
if not self.indeg[i] and self.outdeg[i]:
self.dfsUtil(i)
print(len(self.ans), *self.ans, sep='\n')
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")
BUFSIZE = 8192
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
rstr = lambda: input().strip()
rstrs = lambda: [str(x) for x in input().split()]
rstr_2d = lambda n: [rstr() for _ in range(n)]
rint = lambda: int(input())
rints = lambda: [int(x) for x in input().split()]
rint_2d = lambda n: [rint() for _ in range(n)]
rints_2d = lambda n: [rints() for _ in range(n)]
ceil1 = lambda a, b: (a + b - 1) // b
if __name__ == '__main__':
main()
``` | output | 1 | 31,795 | 8 | 63,591 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The German University in Cairo (GUC) dorm houses are numbered from 1 to n. Underground water pipes connect these houses together. Each pipe has certain direction (water can flow only in this direction and not vice versa), and diameter (which characterizes the maximal amount of water it can handle).
For each house, there is at most one pipe going into it and at most one pipe going out of it. With the new semester starting, GUC student and dorm resident, Lulu, wants to install tanks and taps at the dorms. For every house with an outgoing water pipe and without an incoming water pipe, Lulu should install a water tank at that house. For every house with an incoming water pipe and without an outgoing water pipe, Lulu should install a water tap at that house. Each tank house will convey water to all houses that have a sequence of pipes from the tank to it. Accordingly, each tap house will receive water originating from some tank house.
In order to avoid pipes from bursting one week later (like what happened last semester), Lulu also has to consider the diameter of the pipes. The amount of water each tank conveys should not exceed the diameter of the pipes connecting a tank to its corresponding tap. Lulu wants to find the maximal amount of water that can be safely conveyed from each tank to its corresponding tap.
Input
The first line contains two space-separated integers n and p (1 ≤ n ≤ 1000, 0 ≤ p ≤ n) — the number of houses and the number of pipes correspondingly.
Then p lines follow — the description of p pipes. The i-th line contains three integers ai bi di, indicating a pipe of diameter di going from house ai to house bi (1 ≤ ai, bi ≤ n, ai ≠ bi, 1 ≤ di ≤ 106).
It is guaranteed that for each house there is at most one pipe going into it and at most one pipe going out of it.
Output
Print integer t in the first line — the number of tank-tap pairs of houses.
For the next t lines, print 3 integers per line, separated by spaces: tanki, tapi, and diameteri, where tanki ≠ tapi (1 ≤ i ≤ t). Here tanki and tapi are indexes of tank and tap houses respectively, and diameteri is the maximum amount of water that can be conveyed. All the t lines should be ordered (increasingly) by tanki.
Examples
Input
3 2
1 2 10
2 3 20
Output
1
1 3 10
Input
3 3
1 2 20
2 3 10
3 1 5
Output
0
Input
4 2
1 2 60
3 4 50
Output
2
1 2 60
3 4 50 | instruction | 0 | 31,796 | 8 | 63,592 |
Tags: dfs and similar, graphs
Correct Solution:
```
#!/usr/bin/env python3
n, pcnt = map(int, input().split())
pipes = [None] * n
in_cnt = [0] * n
for i in range(pcnt):
u, v, d = map(int, input().split())
pipes[u-1] = (v-1, d)
in_cnt[v-1] += 1
ans = []
for i, p in enumerate(pipes):
if in_cnt[i] == 0 and pipes[i]:
cur, mn = i, 10_000_000
while pipes[cur]:
mn = min(mn, pipes[cur][1])
cur = pipes[cur][0]
ans.append((i+1, cur+1, mn))
print(len(ans))
for a in ans:
print(a[0], a[1], a[2])
``` | output | 1 | 31,796 | 8 | 63,593 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The German University in Cairo (GUC) dorm houses are numbered from 1 to n. Underground water pipes connect these houses together. Each pipe has certain direction (water can flow only in this direction and not vice versa), and diameter (which characterizes the maximal amount of water it can handle).
For each house, there is at most one pipe going into it and at most one pipe going out of it. With the new semester starting, GUC student and dorm resident, Lulu, wants to install tanks and taps at the dorms. For every house with an outgoing water pipe and without an incoming water pipe, Lulu should install a water tank at that house. For every house with an incoming water pipe and without an outgoing water pipe, Lulu should install a water tap at that house. Each tank house will convey water to all houses that have a sequence of pipes from the tank to it. Accordingly, each tap house will receive water originating from some tank house.
In order to avoid pipes from bursting one week later (like what happened last semester), Lulu also has to consider the diameter of the pipes. The amount of water each tank conveys should not exceed the diameter of the pipes connecting a tank to its corresponding tap. Lulu wants to find the maximal amount of water that can be safely conveyed from each tank to its corresponding tap.
Input
The first line contains two space-separated integers n and p (1 ≤ n ≤ 1000, 0 ≤ p ≤ n) — the number of houses and the number of pipes correspondingly.
Then p lines follow — the description of p pipes. The i-th line contains three integers ai bi di, indicating a pipe of diameter di going from house ai to house bi (1 ≤ ai, bi ≤ n, ai ≠ bi, 1 ≤ di ≤ 106).
It is guaranteed that for each house there is at most one pipe going into it and at most one pipe going out of it.
Output
Print integer t in the first line — the number of tank-tap pairs of houses.
For the next t lines, print 3 integers per line, separated by spaces: tanki, tapi, and diameteri, where tanki ≠ tapi (1 ≤ i ≤ t). Here tanki and tapi are indexes of tank and tap houses respectively, and diameteri is the maximum amount of water that can be conveyed. All the t lines should be ordered (increasingly) by tanki.
Examples
Input
3 2
1 2 10
2 3 20
Output
1
1 3 10
Input
3 3
1 2 20
2 3 10
3 1 5
Output
0
Input
4 2
1 2 60
3 4 50
Output
2
1 2 60
3 4 50 | instruction | 0 | 31,797 | 8 | 63,594 |
Tags: dfs and similar, graphs
Correct Solution:
```
import sys,os,io
import math,bisect,operator
inf,mod = float('inf'),10**9+7
# sys.setrecursionlimit(10 ** 6)
from itertools import groupby,accumulate
from heapq import heapify,heappop,heappush
from collections import deque,Counter,defaultdict
I = lambda : int(sys.stdin.buffer.readline())
Neo = lambda : list(map(int, sys.stdin.buffer.readline().split()))
n,p = Neo()
G = defaultdict(list)
d = defaultdict(int)
parent = [0]*(n+1)
if p == 0:
print(0)
exit()
for i in range(p):
a,b,c = Neo()
G[a].append(b)
d[(a,b)] = c
parent[b] = a
vis = [0]*(n+1)
def hello(n):
vis[n] = 1
global f,t,last
last = n
for i in G[n]:
if vis[i]:
f = 0
return
t = min(t,d[(n,i)])
hello(i)
Ans = []
# print(parent)
for i in range(1,n+1):
if vis[i] or i not in G:
continue
f = 1
t = inf
last = -1
p = i
while parent[i] != 0 and parent[i] != p:
i = parent[i]
hello(i)
if f:
Ans.append((i,last,t))
Ans.sort()
print(len(Ans))
for i,j,k in Ans:
print(i,j,k)
``` | output | 1 | 31,797 | 8 | 63,595 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The German University in Cairo (GUC) dorm houses are numbered from 1 to n. Underground water pipes connect these houses together. Each pipe has certain direction (water can flow only in this direction and not vice versa), and diameter (which characterizes the maximal amount of water it can handle).
For each house, there is at most one pipe going into it and at most one pipe going out of it. With the new semester starting, GUC student and dorm resident, Lulu, wants to install tanks and taps at the dorms. For every house with an outgoing water pipe and without an incoming water pipe, Lulu should install a water tank at that house. For every house with an incoming water pipe and without an outgoing water pipe, Lulu should install a water tap at that house. Each tank house will convey water to all houses that have a sequence of pipes from the tank to it. Accordingly, each tap house will receive water originating from some tank house.
In order to avoid pipes from bursting one week later (like what happened last semester), Lulu also has to consider the diameter of the pipes. The amount of water each tank conveys should not exceed the diameter of the pipes connecting a tank to its corresponding tap. Lulu wants to find the maximal amount of water that can be safely conveyed from each tank to its corresponding tap.
Input
The first line contains two space-separated integers n and p (1 ≤ n ≤ 1000, 0 ≤ p ≤ n) — the number of houses and the number of pipes correspondingly.
Then p lines follow — the description of p pipes. The i-th line contains three integers ai bi di, indicating a pipe of diameter di going from house ai to house bi (1 ≤ ai, bi ≤ n, ai ≠ bi, 1 ≤ di ≤ 106).
It is guaranteed that for each house there is at most one pipe going into it and at most one pipe going out of it.
Output
Print integer t in the first line — the number of tank-tap pairs of houses.
For the next t lines, print 3 integers per line, separated by spaces: tanki, tapi, and diameteri, where tanki ≠ tapi (1 ≤ i ≤ t). Here tanki and tapi are indexes of tank and tap houses respectively, and diameteri is the maximum amount of water that can be conveyed. All the t lines should be ordered (increasingly) by tanki.
Examples
Input
3 2
1 2 10
2 3 20
Output
1
1 3 10
Input
3 3
1 2 20
2 3 10
3 1 5
Output
0
Input
4 2
1 2 60
3 4 50
Output
2
1 2 60
3 4 50 | instruction | 0 | 31,798 | 8 | 63,596 |
Tags: dfs and similar, graphs
Correct Solution:
```
'code by AanchalTiwari'
def dfs(node):
stack = [node]
capacity = float("inf")
while stack:
node = stack.pop()
if outdeg[node] == 0:
return node, capacity
for child, dia in g[node]:
capacity = min(capacity, dia)
stack.append(child)
n, m = map(int, input().split())
g = {i: [] for i in range(1, n + 1)}
indeg = [0 for i in range(n + 1)]
outdeg = [0 for i in range(n+1)]
for i in range(m):
u, v, d = map(int, input().split())
g[u].append([v, d])
indeg[v] += 1
outdeg[u] += 1
ans = set()
for i in range(1, n + 1):
if indeg[i] == 0 and outdeg[i] > 0:
start = i
end, capacity = dfs(i)
indeg[end] += 1
outdeg[start] += 1
ans.add((start, end, capacity))
ans = sorted(list(ans))
print(len(ans))
for i in ans:
print(*i)
``` | output | 1 | 31,798 | 8 | 63,597 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The German University in Cairo (GUC) dorm houses are numbered from 1 to n. Underground water pipes connect these houses together. Each pipe has certain direction (water can flow only in this direction and not vice versa), and diameter (which characterizes the maximal amount of water it can handle).
For each house, there is at most one pipe going into it and at most one pipe going out of it. With the new semester starting, GUC student and dorm resident, Lulu, wants to install tanks and taps at the dorms. For every house with an outgoing water pipe and without an incoming water pipe, Lulu should install a water tank at that house. For every house with an incoming water pipe and without an outgoing water pipe, Lulu should install a water tap at that house. Each tank house will convey water to all houses that have a sequence of pipes from the tank to it. Accordingly, each tap house will receive water originating from some tank house.
In order to avoid pipes from bursting one week later (like what happened last semester), Lulu also has to consider the diameter of the pipes. The amount of water each tank conveys should not exceed the diameter of the pipes connecting a tank to its corresponding tap. Lulu wants to find the maximal amount of water that can be safely conveyed from each tank to its corresponding tap.
Input
The first line contains two space-separated integers n and p (1 ≤ n ≤ 1000, 0 ≤ p ≤ n) — the number of houses and the number of pipes correspondingly.
Then p lines follow — the description of p pipes. The i-th line contains three integers ai bi di, indicating a pipe of diameter di going from house ai to house bi (1 ≤ ai, bi ≤ n, ai ≠ bi, 1 ≤ di ≤ 106).
It is guaranteed that for each house there is at most one pipe going into it and at most one pipe going out of it.
Output
Print integer t in the first line — the number of tank-tap pairs of houses.
For the next t lines, print 3 integers per line, separated by spaces: tanki, tapi, and diameteri, where tanki ≠ tapi (1 ≤ i ≤ t). Here tanki and tapi are indexes of tank and tap houses respectively, and diameteri is the maximum amount of water that can be conveyed. All the t lines should be ordered (increasingly) by tanki.
Examples
Input
3 2
1 2 10
2 3 20
Output
1
1 3 10
Input
3 3
1 2 20
2 3 10
3 1 5
Output
0
Input
4 2
1 2 60
3 4 50
Output
2
1 2 60
3 4 50
Submitted Solution:
```
from collections import defaultdict
graph = defaultdict(list)
n,m = list(map(int,input().split()))
mat = [-1 for i in range(n+1)]
visited = [False for i in range(n+1)]
d = {}
for _ in range(m):
u,v,dist = list(map(int,input().split()))
x = str(u)+':'+str(v)
graph[v].append(u)
d[x] = dist
mat[u] = v
ans = []
for i in range(1,n+1):
if mat[i]==-1 and visited[i]==False:
visited[i] = True
path = [i]
cur = i
while True:
if cur not in graph:
break
else:
cur = graph[cur][0]
if visited[cur]==True:
break
else:
path.append(cur)
ans.append(path[-1::-1])
final = []
for i in ans:
cur = i
if len(cur)>=2:
prev = cur[0]
t = 10**10
for j in range(1,len(cur)):
y = str(prev)+':'+str(cur[j])
t = min(t,d[y])
prev = cur[j]
final.append([cur[0],cur[-1],t])
print(len(final))
final.sort()
for i in final:
print(i[0],i[1],i[2])
``` | instruction | 0 | 31,799 | 8 | 63,598 |
Yes | output | 1 | 31,799 | 8 | 63,599 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The German University in Cairo (GUC) dorm houses are numbered from 1 to n. Underground water pipes connect these houses together. Each pipe has certain direction (water can flow only in this direction and not vice versa), and diameter (which characterizes the maximal amount of water it can handle).
For each house, there is at most one pipe going into it and at most one pipe going out of it. With the new semester starting, GUC student and dorm resident, Lulu, wants to install tanks and taps at the dorms. For every house with an outgoing water pipe and without an incoming water pipe, Lulu should install a water tank at that house. For every house with an incoming water pipe and without an outgoing water pipe, Lulu should install a water tap at that house. Each tank house will convey water to all houses that have a sequence of pipes from the tank to it. Accordingly, each tap house will receive water originating from some tank house.
In order to avoid pipes from bursting one week later (like what happened last semester), Lulu also has to consider the diameter of the pipes. The amount of water each tank conveys should not exceed the diameter of the pipes connecting a tank to its corresponding tap. Lulu wants to find the maximal amount of water that can be safely conveyed from each tank to its corresponding tap.
Input
The first line contains two space-separated integers n and p (1 ≤ n ≤ 1000, 0 ≤ p ≤ n) — the number of houses and the number of pipes correspondingly.
Then p lines follow — the description of p pipes. The i-th line contains three integers ai bi di, indicating a pipe of diameter di going from house ai to house bi (1 ≤ ai, bi ≤ n, ai ≠ bi, 1 ≤ di ≤ 106).
It is guaranteed that for each house there is at most one pipe going into it and at most one pipe going out of it.
Output
Print integer t in the first line — the number of tank-tap pairs of houses.
For the next t lines, print 3 integers per line, separated by spaces: tanki, tapi, and diameteri, where tanki ≠ tapi (1 ≤ i ≤ t). Here tanki and tapi are indexes of tank and tap houses respectively, and diameteri is the maximum amount of water that can be conveyed. All the t lines should be ordered (increasingly) by tanki.
Examples
Input
3 2
1 2 10
2 3 20
Output
1
1 3 10
Input
3 3
1 2 20
2 3 10
3 1 5
Output
0
Input
4 2
1 2 60
3 4 50
Output
2
1 2 60
3 4 50
Submitted Solution:
```
from collections import defaultdict
ans = []
def DFS(graph,x,visited):
stack = [x]
min1 = float("inf")
while len(stack):
temp = stack.pop()
visited[temp] = 1
for i in graph[temp]:
if visited[i[0]] == 1:
return
else:
stack.append(i[0])
min1 = min(min1,i[1])
return [x,temp,min1]
n,p = map(int,input().split())
graph = defaultdict(list)
child = [0 for i in range(n+1)]
graph_presence = [0 for i in range(n+1)]
for i in range(p):
a,b,d = map(int,input().split())
graph[a].append( (b,d) )
child[b] = 1
graph_presence[a] = 1
graph_presence[b] = 1
visited = [0 for i in range(n+1)]
r = 0
ans = []
for i in range(1,n+1):
if child[i] == 0 and graph_presence[i] == 1:
if visited[i] == 0:
t = DFS(graph,i,visited)
if t!= None:
t = list(t)
r = r + 1
ans.append(t)
print(r)
for i in ans:
print(*i)
``` | instruction | 0 | 31,800 | 8 | 63,600 |
Yes | output | 1 | 31,800 | 8 | 63,601 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The German University in Cairo (GUC) dorm houses are numbered from 1 to n. Underground water pipes connect these houses together. Each pipe has certain direction (water can flow only in this direction and not vice versa), and diameter (which characterizes the maximal amount of water it can handle).
For each house, there is at most one pipe going into it and at most one pipe going out of it. With the new semester starting, GUC student and dorm resident, Lulu, wants to install tanks and taps at the dorms. For every house with an outgoing water pipe and without an incoming water pipe, Lulu should install a water tank at that house. For every house with an incoming water pipe and without an outgoing water pipe, Lulu should install a water tap at that house. Each tank house will convey water to all houses that have a sequence of pipes from the tank to it. Accordingly, each tap house will receive water originating from some tank house.
In order to avoid pipes from bursting one week later (like what happened last semester), Lulu also has to consider the diameter of the pipes. The amount of water each tank conveys should not exceed the diameter of the pipes connecting a tank to its corresponding tap. Lulu wants to find the maximal amount of water that can be safely conveyed from each tank to its corresponding tap.
Input
The first line contains two space-separated integers n and p (1 ≤ n ≤ 1000, 0 ≤ p ≤ n) — the number of houses and the number of pipes correspondingly.
Then p lines follow — the description of p pipes. The i-th line contains three integers ai bi di, indicating a pipe of diameter di going from house ai to house bi (1 ≤ ai, bi ≤ n, ai ≠ bi, 1 ≤ di ≤ 106).
It is guaranteed that for each house there is at most one pipe going into it and at most one pipe going out of it.
Output
Print integer t in the first line — the number of tank-tap pairs of houses.
For the next t lines, print 3 integers per line, separated by spaces: tanki, tapi, and diameteri, where tanki ≠ tapi (1 ≤ i ≤ t). Here tanki and tapi are indexes of tank and tap houses respectively, and diameteri is the maximum amount of water that can be conveyed. All the t lines should be ordered (increasingly) by tanki.
Examples
Input
3 2
1 2 10
2 3 20
Output
1
1 3 10
Input
3 3
1 2 20
2 3 10
3 1 5
Output
0
Input
4 2
1 2 60
3 4 50
Output
2
1 2 60
3 4 50
Submitted Solution:
```
import math,sys,bisect,heapq
from collections import defaultdict,Counter,deque
from itertools import groupby,accumulate
#sys.setrecursionlimit(200000000)
int1 = lambda x: int(x) - 1
#def input(): return sys.stdin.readline().strip()
input = iter(sys.stdin.buffer.read().decode().splitlines()).__next__
ilele = lambda: map(int,input().split())
alele = lambda: list(map(int, input().split()))
ilelec = lambda: map(int1,input().split())
alelec = lambda: list(map(int1, input().split()))
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)]
#MOD = 1000000000 + 7
def Y(c): print(["NO","YES"][c])
def y(c): print(["no","yes"][c])
def Yy(c): print(["No","Yes"][c])
n,p = ilele()
if p == 0:
print(0)
exit(0)
G = defaultdict(list)
P= defaultdict(set)
def addEdge(a,b,c):
G[a].append(b)
G[b].append(a)
P[a].add(c)
P[b].add(c)
vis = [False]*(n+1)
def dfs(node):
Ans = []
vis[node] = True
s = deque()
mini = 1e9
s.append(node)
while s:
x = s.pop()
Ans.append(x)
if P[x]:
mini = min(mini,min(P[x]))
for i in G.get(x,[]):
if not vis[i]:
s.append(i)
vis[i] = True
return Ans,mini
tap = set()
tank = set()
for i in range(p):
a,b,c = ilele()
addEdge(a,b,c)
tap.add(b)
tank.add(a)
#print(tank,tap,G)
Ans2 = []
for i in range(1,n+1):
if not vis[i]:
Ans,mini = dfs(i)
t1 = t2 = None
for j in Ans:
x = j in tap
y = j in tank
if x^y:
if x:
t1 = j
else:
t2 = j
if t1 and t2 and mini != None:
Ans2.append((t2,t1,mini))
print(len(Ans2))
Ans2.sort()
if Ans2:
for i,j,k in Ans2:
print(i,j,k)
``` | instruction | 0 | 31,801 | 8 | 63,602 |
Yes | output | 1 | 31,801 | 8 | 63,603 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The German University in Cairo (GUC) dorm houses are numbered from 1 to n. Underground water pipes connect these houses together. Each pipe has certain direction (water can flow only in this direction and not vice versa), and diameter (which characterizes the maximal amount of water it can handle).
For each house, there is at most one pipe going into it and at most one pipe going out of it. With the new semester starting, GUC student and dorm resident, Lulu, wants to install tanks and taps at the dorms. For every house with an outgoing water pipe and without an incoming water pipe, Lulu should install a water tank at that house. For every house with an incoming water pipe and without an outgoing water pipe, Lulu should install a water tap at that house. Each tank house will convey water to all houses that have a sequence of pipes from the tank to it. Accordingly, each tap house will receive water originating from some tank house.
In order to avoid pipes from bursting one week later (like what happened last semester), Lulu also has to consider the diameter of the pipes. The amount of water each tank conveys should not exceed the diameter of the pipes connecting a tank to its corresponding tap. Lulu wants to find the maximal amount of water that can be safely conveyed from each tank to its corresponding tap.
Input
The first line contains two space-separated integers n and p (1 ≤ n ≤ 1000, 0 ≤ p ≤ n) — the number of houses and the number of pipes correspondingly.
Then p lines follow — the description of p pipes. The i-th line contains three integers ai bi di, indicating a pipe of diameter di going from house ai to house bi (1 ≤ ai, bi ≤ n, ai ≠ bi, 1 ≤ di ≤ 106).
It is guaranteed that for each house there is at most one pipe going into it and at most one pipe going out of it.
Output
Print integer t in the first line — the number of tank-tap pairs of houses.
For the next t lines, print 3 integers per line, separated by spaces: tanki, tapi, and diameteri, where tanki ≠ tapi (1 ≤ i ≤ t). Here tanki and tapi are indexes of tank and tap houses respectively, and diameteri is the maximum amount of water that can be conveyed. All the t lines should be ordered (increasingly) by tanki.
Examples
Input
3 2
1 2 10
2 3 20
Output
1
1 3 10
Input
3 3
1 2 20
2 3 10
3 1 5
Output
0
Input
4 2
1 2 60
3 4 50
Output
2
1 2 60
3 4 50
Submitted Solution:
```
class CodeforcesTask107ASolution:
def __init__(self):
self.result = ''
self.n_p = []
self.pipes = []
def read_input(self):
self.n_p = [int(x) for x in input().split(" ")]
for _ in range(self.n_p[1]):
self.pipes.append([int(x) for x in input().split(" ")])
def process_task(self):
incomings = [[] for _ in range(self.n_p[0])]
outcomings = [[] for _ in range(self.n_p[0])]
for pipe in self.pipes:
incomings[pipe[1] - 1].append(pipe[0])
outcomings[pipe[0] - 1].append((pipe[1], pipe[2]))
result = []
for x in range(self.n_p[0]):
if not incomings[x] and outcomings[x]:
start = x + 1
diam = outcomings[x][0][1]
stop = outcomings[x][0][0]
while outcomings[stop - 1]:
diam = min(diam, outcomings[stop - 1][0][1])
stop = outcomings[stop - 1][0][0]
result.append(f"{start} {stop} {diam}")
self.result = "{0}\n{1}".format(len(result), "\n".join(result))
def get_result(self):
return self.result
if __name__ == "__main__":
Solution = CodeforcesTask107ASolution()
Solution.read_input()
Solution.process_task()
print(Solution.get_result())
``` | instruction | 0 | 31,802 | 8 | 63,604 |
Yes | output | 1 | 31,802 | 8 | 63,605 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The German University in Cairo (GUC) dorm houses are numbered from 1 to n. Underground water pipes connect these houses together. Each pipe has certain direction (water can flow only in this direction and not vice versa), and diameter (which characterizes the maximal amount of water it can handle).
For each house, there is at most one pipe going into it and at most one pipe going out of it. With the new semester starting, GUC student and dorm resident, Lulu, wants to install tanks and taps at the dorms. For every house with an outgoing water pipe and without an incoming water pipe, Lulu should install a water tank at that house. For every house with an incoming water pipe and without an outgoing water pipe, Lulu should install a water tap at that house. Each tank house will convey water to all houses that have a sequence of pipes from the tank to it. Accordingly, each tap house will receive water originating from some tank house.
In order to avoid pipes from bursting one week later (like what happened last semester), Lulu also has to consider the diameter of the pipes. The amount of water each tank conveys should not exceed the diameter of the pipes connecting a tank to its corresponding tap. Lulu wants to find the maximal amount of water that can be safely conveyed from each tank to its corresponding tap.
Input
The first line contains two space-separated integers n and p (1 ≤ n ≤ 1000, 0 ≤ p ≤ n) — the number of houses and the number of pipes correspondingly.
Then p lines follow — the description of p pipes. The i-th line contains three integers ai bi di, indicating a pipe of diameter di going from house ai to house bi (1 ≤ ai, bi ≤ n, ai ≠ bi, 1 ≤ di ≤ 106).
It is guaranteed that for each house there is at most one pipe going into it and at most one pipe going out of it.
Output
Print integer t in the first line — the number of tank-tap pairs of houses.
For the next t lines, print 3 integers per line, separated by spaces: tanki, tapi, and diameteri, where tanki ≠ tapi (1 ≤ i ≤ t). Here tanki and tapi are indexes of tank and tap houses respectively, and diameteri is the maximum amount of water that can be conveyed. All the t lines should be ordered (increasingly) by tanki.
Examples
Input
3 2
1 2 10
2 3 20
Output
1
1 3 10
Input
3 3
1 2 20
2 3 10
3 1 5
Output
0
Input
4 2
1 2 60
3 4 50
Output
2
1 2 60
3 4 50
Submitted Solution:
```
f_in = input()
f_in = f_in.split(" ")
num_houses = int(f_in[0])
num_pipes = int(f_in[1])
#si son iguales
if(num_houses == num_pipes):
print(0)
else:
answer = []
pipes_dic = {}
for i in range(num_pipes):
pipe = input()
pipe = pipe.split(" ")
if(pipe[0] not in pipes_dic):
flag = False
for key in pipes_dic:
if(pipes_dic[key][0] == pipe[1]):
flag = True
pipes_dic[key][0] = pipe[0]
if(pipes_dic[key][1] > pipe[2]):
pipes_dic[key][1] = pipe[2]
break
if(flag):
continue
pipes_dic[pipe[1]] = [ pipe[0], pipe[2] ]
else:
pipes_dic[pipe[1]] = pipes_dic[pipe[0]]
if(pipes_dic[pipe[1]][1] > pipe[2]):
pipes_dic[pipe[1]][1] = pipe[2]
pipes_dic.pop(pipe[0])
for key in pipes_dic:
if(pipes_dic[key][0] == pipe[1]):
pipes_dic[key][0] = pipes_dic[pipe[1]][0]
if(pipes_dic[key][1] > pipes_dic[pipe[1]][1]):
pipes_dic[key][1] = pipes_dic[pipe[1]][1]
pipes_dic.pop(pipe[1])
break
for i in pipes_dic:
answer.append([pipes_dic[i][0], i, pipes_dic[i][1]])
answer.sort()
print(len(answer))
for tank, tap, diameter in answer:
print(tank , tap, diameter)
``` | instruction | 0 | 31,803 | 8 | 63,606 |
No | output | 1 | 31,803 | 8 | 63,607 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The German University in Cairo (GUC) dorm houses are numbered from 1 to n. Underground water pipes connect these houses together. Each pipe has certain direction (water can flow only in this direction and not vice versa), and diameter (which characterizes the maximal amount of water it can handle).
For each house, there is at most one pipe going into it and at most one pipe going out of it. With the new semester starting, GUC student and dorm resident, Lulu, wants to install tanks and taps at the dorms. For every house with an outgoing water pipe and without an incoming water pipe, Lulu should install a water tank at that house. For every house with an incoming water pipe and without an outgoing water pipe, Lulu should install a water tap at that house. Each tank house will convey water to all houses that have a sequence of pipes from the tank to it. Accordingly, each tap house will receive water originating from some tank house.
In order to avoid pipes from bursting one week later (like what happened last semester), Lulu also has to consider the diameter of the pipes. The amount of water each tank conveys should not exceed the diameter of the pipes connecting a tank to its corresponding tap. Lulu wants to find the maximal amount of water that can be safely conveyed from each tank to its corresponding tap.
Input
The first line contains two space-separated integers n and p (1 ≤ n ≤ 1000, 0 ≤ p ≤ n) — the number of houses and the number of pipes correspondingly.
Then p lines follow — the description of p pipes. The i-th line contains three integers ai bi di, indicating a pipe of diameter di going from house ai to house bi (1 ≤ ai, bi ≤ n, ai ≠ bi, 1 ≤ di ≤ 106).
It is guaranteed that for each house there is at most one pipe going into it and at most one pipe going out of it.
Output
Print integer t in the first line — the number of tank-tap pairs of houses.
For the next t lines, print 3 integers per line, separated by spaces: tanki, tapi, and diameteri, where tanki ≠ tapi (1 ≤ i ≤ t). Here tanki and tapi are indexes of tank and tap houses respectively, and diameteri is the maximum amount of water that can be conveyed. All the t lines should be ordered (increasingly) by tanki.
Examples
Input
3 2
1 2 10
2 3 20
Output
1
1 3 10
Input
3 3
1 2 20
2 3 10
3 1 5
Output
0
Input
4 2
1 2 60
3 4 50
Output
2
1 2 60
3 4 50
Submitted Solution:
```
n,m = list(map(int,input().split()))
mat = []
d = {}
for _ in range(m):
u,v,dist = list(map(int,input().split()))
x = str(u)+':'+str(v)
d[x] = dist
flag = 0
for i in range(len(mat)):
temp = mat[i]
f,l = temp[0],temp[-1]
if v==f:
temp = [u]+temp
flag = 1
mat[i] = temp
break
elif u==l:
temp.append(v)
flag = 1
mat[i] = temp
break
if flag==0:
mat.append([u,v])
ans = []
for i in mat:
if len(i)>=2 and i[0]!=i[-1]:
ans.append(i)
mat = []
for i in ans:
if mat==[]:
mat.append(i[::])
else:
af,al = i[0],i[-1]
flag = 0
for j in range(len(mat)):
f,l = mat[j][0],mat[j][-1]
if l==af:
mat[j] = mat[j]+i[1::]
flag = 1
break
elif f==al:
mat[j] = i[::]+mat[j][1::]
flag = 1
break
if flag==0:
mat.append(i)
ans = []
for i in mat:
if len(i)>=2 and i[0]!=i[-1]:
ans.append(i)
print(len(ans))
# print(ans)
final = []
for i in ans:
t = 10**10
prev = i[0]
for j in range(1,len(i)):
cur = i[j]
x = str(prev)+':'+str(cur)
t = min(t,d[x])
prev = cur
final.append([i[0],i[-1],t])
# print(i[0],i[-1],t)
final.sort()
for i in final:
print(i[0],i[1],i[2])
``` | instruction | 0 | 31,804 | 8 | 63,608 |
No | output | 1 | 31,804 | 8 | 63,609 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The German University in Cairo (GUC) dorm houses are numbered from 1 to n. Underground water pipes connect these houses together. Each pipe has certain direction (water can flow only in this direction and not vice versa), and diameter (which characterizes the maximal amount of water it can handle).
For each house, there is at most one pipe going into it and at most one pipe going out of it. With the new semester starting, GUC student and dorm resident, Lulu, wants to install tanks and taps at the dorms. For every house with an outgoing water pipe and without an incoming water pipe, Lulu should install a water tank at that house. For every house with an incoming water pipe and without an outgoing water pipe, Lulu should install a water tap at that house. Each tank house will convey water to all houses that have a sequence of pipes from the tank to it. Accordingly, each tap house will receive water originating from some tank house.
In order to avoid pipes from bursting one week later (like what happened last semester), Lulu also has to consider the diameter of the pipes. The amount of water each tank conveys should not exceed the diameter of the pipes connecting a tank to its corresponding tap. Lulu wants to find the maximal amount of water that can be safely conveyed from each tank to its corresponding tap.
Input
The first line contains two space-separated integers n and p (1 ≤ n ≤ 1000, 0 ≤ p ≤ n) — the number of houses and the number of pipes correspondingly.
Then p lines follow — the description of p pipes. The i-th line contains three integers ai bi di, indicating a pipe of diameter di going from house ai to house bi (1 ≤ ai, bi ≤ n, ai ≠ bi, 1 ≤ di ≤ 106).
It is guaranteed that for each house there is at most one pipe going into it and at most one pipe going out of it.
Output
Print integer t in the first line — the number of tank-tap pairs of houses.
For the next t lines, print 3 integers per line, separated by spaces: tanki, tapi, and diameteri, where tanki ≠ tapi (1 ≤ i ≤ t). Here tanki and tapi are indexes of tank and tap houses respectively, and diameteri is the maximum amount of water that can be conveyed. All the t lines should be ordered (increasingly) by tanki.
Examples
Input
3 2
1 2 10
2 3 20
Output
1
1 3 10
Input
3 3
1 2 20
2 3 10
3 1 5
Output
0
Input
4 2
1 2 60
3 4 50
Output
2
1 2 60
3 4 50
Submitted Solution:
```
from collections import *
import os, sys
from io import BytesIO, IOBase
def main():
n, m = rints()
g = graph(n)
for i in range(m):
u, v, w = rints()
g.addEdge(u, v, w)
g.dfs()
class graph:
def __init__(self, n):
self.n, self.gdict = n, [[] for _ in range(n + 1)]
self.indeg = [False] * (n + 1)
def addEdge(self, node1, node2, w=None):
self.gdict[node1].append((node2, w))
self.indeg[node2] = True
def dfsUtil(self, v):
stack, cur = [(v, float('inf'))], [v, float('inf')]
while (stack):
s, w = stack.pop()
cur = [s, w]
for i1, j1 in self.gdict[s]:
stack.append((i1, min(j1, w)))
self.ans.append(f'{v} {cur[0]} {cur[1]}')
def dfs(self):
self.ans = []
for i in range(1, self.n + 1):
if not self.indeg[i]:
self.dfsUtil(i)
print(len(self.ans), *self.ans, sep='\n')
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")
BUFSIZE = 8192
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
rstr = lambda: input().strip()
rstrs = lambda: [str(x) for x in input().split()]
rstr_2d = lambda n: [rstr() for _ in range(n)]
rint = lambda: int(input())
rints = lambda: [int(x) for x in input().split()]
rint_2d = lambda n: [rint() for _ in range(n)]
rints_2d = lambda n: [rints() for _ in range(n)]
ceil1 = lambda a, b: (a + b - 1) // b
if __name__ == '__main__':
main()
``` | instruction | 0 | 31,805 | 8 | 63,610 |
No | output | 1 | 31,805 | 8 | 63,611 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The German University in Cairo (GUC) dorm houses are numbered from 1 to n. Underground water pipes connect these houses together. Each pipe has certain direction (water can flow only in this direction and not vice versa), and diameter (which characterizes the maximal amount of water it can handle).
For each house, there is at most one pipe going into it and at most one pipe going out of it. With the new semester starting, GUC student and dorm resident, Lulu, wants to install tanks and taps at the dorms. For every house with an outgoing water pipe and without an incoming water pipe, Lulu should install a water tank at that house. For every house with an incoming water pipe and without an outgoing water pipe, Lulu should install a water tap at that house. Each tank house will convey water to all houses that have a sequence of pipes from the tank to it. Accordingly, each tap house will receive water originating from some tank house.
In order to avoid pipes from bursting one week later (like what happened last semester), Lulu also has to consider the diameter of the pipes. The amount of water each tank conveys should not exceed the diameter of the pipes connecting a tank to its corresponding tap. Lulu wants to find the maximal amount of water that can be safely conveyed from each tank to its corresponding tap.
Input
The first line contains two space-separated integers n and p (1 ≤ n ≤ 1000, 0 ≤ p ≤ n) — the number of houses and the number of pipes correspondingly.
Then p lines follow — the description of p pipes. The i-th line contains three integers ai bi di, indicating a pipe of diameter di going from house ai to house bi (1 ≤ ai, bi ≤ n, ai ≠ bi, 1 ≤ di ≤ 106).
It is guaranteed that for each house there is at most one pipe going into it and at most one pipe going out of it.
Output
Print integer t in the first line — the number of tank-tap pairs of houses.
For the next t lines, print 3 integers per line, separated by spaces: tanki, tapi, and diameteri, where tanki ≠ tapi (1 ≤ i ≤ t). Here tanki and tapi are indexes of tank and tap houses respectively, and diameteri is the maximum amount of water that can be conveyed. All the t lines should be ordered (increasingly) by tanki.
Examples
Input
3 2
1 2 10
2 3 20
Output
1
1 3 10
Input
3 3
1 2 20
2 3 10
3 1 5
Output
0
Input
4 2
1 2 60
3 4 50
Output
2
1 2 60
3 4 50
Submitted Solution:
```
n, p = [int(x) for x in input().split()]
i=0
if (n == p) or (p < (n // 2) + (n % 2)):
print(0)
else:
lout=[]
lin=[]
llen=[]
endlin=[]
endlout=[]
endlen=[]
remember=[x for x in range(1,n+1)]
for count in range(p):
templist=[int(x) for x in input().split()]
lout.append(templist[0])
lin.append(templist[1])
llen.append(templist[2])
while True:
if len(remember) == 0:
break
for i in range(len(remember)):
if lout.count(remember[i]) == 0: # tap location
endlin.append(remember[i])
tempindex = lin.index(remember[i])
templen = llen[tempindex]
remember.remove(remember[i])
while True:
tempnumber = lout[tempindex]
remember.remove(tempnumber)
if lin.count(tempnumber) == 0:
endlout.append(tempnumber)
endlen.append(templen)
break
lout.remove(lout[tempindex])
lin.remove(lin[tempindex])
llen.remove(llen[tempindex])
tempindex = lin.index(tempnumber)
templen = min(templen,llen[tempindex])
break
elif lin.count(remember[i]) == 0: # tank location
endlout.append(remember[i])
tempindex = lout.index(remember[i])
templen = llen[tempindex]
remember.remove(remember[i])
while True:
tempnumber = lin[tempindex]
remember.remove(tempnumber)
if lout.count(tempnumber) == 0:
endlin.append(tempnumber)
endlen.append(templen)
break
lout.remove(lout[tempindex])
lin.remove(lin[tempindex])
llen.remove(llen[tempindex])
tempindex = lout.index(tempnumber)
templen = min(templen,llen[tempindex])
break
else:pass
if len(remember) == 0 or (i == len(remember)-1 and lin.count(remember[-1])+lout.count(remember[-1]) == 2):
break
print(n-p)
while endlout !=[]:
i = endlout.index(min(endlout))
print(str(min(endlout))+' '+str(endlin[i])+' '+str(endlen[i]))
endlout.remove(endlout[i])
endlin.remove(endlin[i])
endlen.remove(endlen[i])
``` | instruction | 0 | 31,806 | 8 | 63,612 |
No | output | 1 | 31,806 | 8 | 63,613 |
Provide a correct Python 3 solution for this coding contest problem.
In the building of Jewelry Art Gallery (JAG), there is a long corridor in the east-west direction. There is a window on the north side of the corridor, and $N$ windowpanes are attached to this window. The width of each windowpane is $W$, and the height is $H$. The $i$-th windowpane from the west covers the horizontal range between $W\times(i-1)$ and $W\times i$ from the west edge of the window.
<image>
Figure A1. Illustration of the window
You received instructions from the manager of JAG about how to slide the windowpanes. These instructions consist of $N$ integers $x_1, x_2, ..., x_N$, and $x_i \leq W$ is satisfied for all $i$. For the $i$-th windowpane, if $i$ is odd, you have to slide $i$-th windowpane to the east by $x_i$, otherwise, you have to slide $i$-th windowpane to the west by $x_i$.
You can assume that the windowpanes will not collide each other even if you slide windowpanes according to the instructions. In more detail, $N$ windowpanes are alternately mounted on two rails. That is, the $i$-th windowpane is attached to the inner rail of the building if $i$ is odd, otherwise, it is attached to the outer rail of the building.
Before you execute the instructions, you decide to obtain the area where the window is open after the instructions.
Input
The input consists of a single test case in the format below.
$N$ $H$ $W$
$x_1$ ... $x_N$
The first line consists of three integers $N, H,$ and $W$ ($1 \leq N \leq 100, 1 \leq H, W \leq 100$). It is guaranteed that $N$ is even. The following line consists of $N$ integers $x_1, ..., x_N$ while represent the instructions from the manager of JAG. $x_i$ represents the distance to slide the $i$-th windowpane ($0 \leq x_i \leq W$).
Output
Print the area where the window is open after the instructions in one line.
Examples
Input
4 3 3
1 1 2 3
Output
9
Input
8 10 18
2 12 16 14 18 4 17 16
Output
370
Input
6 2 2
0 2 2 2 2 0
Output
8
Input
4 1 4
3 3 2 2
Output
6
Input
8 7 15
5 0 9 14 0 4 4 15
Output
189 | instruction | 0 | 32,458 | 8 | 64,916 |
"Correct Solution:
```
#!usr/bin/env python3
from collections import defaultdict
from collections import deque
from heapq import heappush, heappop
import sys
import math
import bisect
import random
def LI(): return list(map(int, sys.stdin.readline().split()))
def I(): return int(sys.stdin.readline())
def LS():return list(map(list, sys.stdin.readline().split()))
def S(): return list(sys.stdin.readline())[:-1]
def IR(n):
l = [None for i in range(n)]
for i in range(n):l[i] = I()
return l
def LIR(n):
l = [None for i in range(n)]
for i in range(n):l[i] = LI()
return l
def SR(n):
l = [None for i in range(n)]
for i in range(n):l[i] = S()
return l
def LSR(n):
l = [None for i in range(n)]
for i in range(n):l[i] = LS()
return l
sys.setrecursionlimit(1000000)
mod = 1000000007
#A
def A():
n,h,w = LI()
x = LI()
t = [0 for i in range(w*n+1)]
for i in range(n):
if not i%2:
t[i*w+x[i]] += 1
t[(i+1)*w+x[i]] -= 1
else:
t[i*w-x[i]] += 1
t[(i+1)*w-x[i]] -= 1
for i in range(w*n):
t[i+1] += t[i]
ans = 0
for i in t[:-1]:
if i == 0:
ans += h
print(ans)
return
#B
def B():
return
#C
def C():
return
#D
def D():
return
#E
def E():
return
#F
def F():
return
#G
def G():
return
#H
def H():
return
#I
def I_():
return
#J
def J():
return
#Solve
if __name__ == "__main__":
A()
``` | output | 1 | 32,458 | 8 | 64,917 |
Provide a correct Python 3 solution for this coding contest problem.
In the building of Jewelry Art Gallery (JAG), there is a long corridor in the east-west direction. There is a window on the north side of the corridor, and $N$ windowpanes are attached to this window. The width of each windowpane is $W$, and the height is $H$. The $i$-th windowpane from the west covers the horizontal range between $W\times(i-1)$ and $W\times i$ from the west edge of the window.
<image>
Figure A1. Illustration of the window
You received instructions from the manager of JAG about how to slide the windowpanes. These instructions consist of $N$ integers $x_1, x_2, ..., x_N$, and $x_i \leq W$ is satisfied for all $i$. For the $i$-th windowpane, if $i$ is odd, you have to slide $i$-th windowpane to the east by $x_i$, otherwise, you have to slide $i$-th windowpane to the west by $x_i$.
You can assume that the windowpanes will not collide each other even if you slide windowpanes according to the instructions. In more detail, $N$ windowpanes are alternately mounted on two rails. That is, the $i$-th windowpane is attached to the inner rail of the building if $i$ is odd, otherwise, it is attached to the outer rail of the building.
Before you execute the instructions, you decide to obtain the area where the window is open after the instructions.
Input
The input consists of a single test case in the format below.
$N$ $H$ $W$
$x_1$ ... $x_N$
The first line consists of three integers $N, H,$ and $W$ ($1 \leq N \leq 100, 1 \leq H, W \leq 100$). It is guaranteed that $N$ is even. The following line consists of $N$ integers $x_1, ..., x_N$ while represent the instructions from the manager of JAG. $x_i$ represents the distance to slide the $i$-th windowpane ($0 \leq x_i \leq W$).
Output
Print the area where the window is open after the instructions in one line.
Examples
Input
4 3 3
1 1 2 3
Output
9
Input
8 10 18
2 12 16 14 18 4 17 16
Output
370
Input
6 2 2
0 2 2 2 2 0
Output
8
Input
4 1 4
3 3 2 2
Output
6
Input
8 7 15
5 0 9 14 0 4 4 15
Output
189 | instruction | 0 | 32,459 | 8 | 64,918 |
"Correct Solution:
```
n,h,w=map(int,input().split())
x=list(map(int,input().split()))
wide_total=n*w
wide_cover=[False]*wide_total
for i in range(n):
if (i+1)%2==1:
for j in range(i*w+x[i],i*w+x[i]+w):
wide_cover[j]=True
else:
for j in range(i*w-x[i],i*w-x[i]+w):
wide_cover[j]=True
cnt=0
for c in wide_cover:
if c==False:
cnt+=1
print(cnt*h)
``` | output | 1 | 32,459 | 8 | 64,919 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As you all know, the plum harvesting season is on! Little Milutin had his plums planted in an orchard that can be represented as an n by m matrix. While he was harvesting, he wrote the heights of all trees in a matrix of dimensions n by m.
At night, when he has spare time, he likes to perform various statistics on his trees. This time, he is curious to find out the height of his lowest tree. So far, he has discovered some interesting properties of his orchard. There is one particular property that he thinks is useful for finding the tree with the smallest heigh.
Formally, let L(i) be the leftmost tree with the smallest height in the i-th row of his orchard. He knows that L(i) ≤ L(i+1) for all 1 ≤ i ≤ n - 1. Moreover, if he takes a submatrix induced by any subset of rows and any subset of columns, L(i) ≤ L(i+1) will hold for all 1 ≤ i ≤ n'-1, where n' is the number of rows in that submatrix.
Since the season is at its peak and he is short on time, he asks you to help him find the plum tree with minimal height.
Input
This problem is interactive.
The first line of input will contain two integers n and m, representing the number of rows and the number of columns in Milutin's orchard. It is guaranteed that 1 ≤ n, m ≤ 10^6.
The following lines will contain the answers to your queries.
Output
Once you know have found the minimum value r, you should print ! r to the standard output.
Interaction
Your code is allowed to query for an entry (i, j) of a matrix (i.e. get the height of the tree which is in the i-th row and j-th column). The query should be formatted as ? i j, so that 1 ≤ i ≤ n and 1 ≤ j ≤ m.
You may assume that the entries of the matrix will be integers between 1 and 10^9.
Your solution should use not more than 4 ⋅ (n + m) queries.
This is an interactive problem. You have to use a flush operation right after printing each line. For example, in C++ you should use the function fflush(stdout), in Java — System.out.flush(), in Pascal — flush(output) and in Python — sys.stdout.flush().
Example
Input
5 5
13 15 10 9 15
15 17 12 11 17
10 12 7 6 12
17 19 14 13 19
16 18 13 12 18
Output
Submitted Solution:
```
import sys
mn = int(1e18)
def ask(i, j):
print('?', i, j)
res = int(input())
global mn
mn = min(mn, res)
return [res, j]
def answer():
global mn
print('!', mn)
sys.exit(0)
def main():
n, m = [int(x) for x in input().split()]
if m == 1:
for i in range(1, n+1):
ask(i, 1)
answer()
a = []
for i in range(1, m+1):
a.append(ask(1, i))
a.sort()
print(a)
cur = 0
for i in range(2, n):
prv, nxt = a[cur][1], a[cur+1][1]
ans_prv = ask(i, prv)
ans_nxt = ask(i, nxt)
while ans_prv > ans_nxt and cur + 1 < m:
ans_prv = ans_nxt
cur += 1
ans_nxt = ask(i, a[cur][1])
for i in range(1, m+1):
ask(n, i)
answer()
main()
``` | instruction | 0 | 32,716 | 8 | 65,432 |
No | output | 1 | 32,716 | 8 | 65,433 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As you all know, the plum harvesting season is on! Little Milutin had his plums planted in an orchard that can be represented as an n by m matrix. While he was harvesting, he wrote the heights of all trees in a matrix of dimensions n by m.
At night, when he has spare time, he likes to perform various statistics on his trees. This time, he is curious to find out the height of his lowest tree. So far, he has discovered some interesting properties of his orchard. There is one particular property that he thinks is useful for finding the tree with the smallest heigh.
Formally, let L(i) be the leftmost tree with the smallest height in the i-th row of his orchard. He knows that L(i) ≤ L(i+1) for all 1 ≤ i ≤ n - 1. Moreover, if he takes a submatrix induced by any subset of rows and any subset of columns, L(i) ≤ L(i+1) will hold for all 1 ≤ i ≤ n'-1, where n' is the number of rows in that submatrix.
Since the season is at its peak and he is short on time, he asks you to help him find the plum tree with minimal height.
Input
This problem is interactive.
The first line of input will contain two integers n and m, representing the number of rows and the number of columns in Milutin's orchard. It is guaranteed that 1 ≤ n, m ≤ 10^6.
The following lines will contain the answers to your queries.
Output
Once you know have found the minimum value r, you should print ! r to the standard output.
Interaction
Your code is allowed to query for an entry (i, j) of a matrix (i.e. get the height of the tree which is in the i-th row and j-th column). The query should be formatted as ? i j, so that 1 ≤ i ≤ n and 1 ≤ j ≤ m.
You may assume that the entries of the matrix will be integers between 1 and 10^9.
Your solution should use not more than 4 ⋅ (n + m) queries.
This is an interactive problem. You have to use a flush operation right after printing each line. For example, in C++ you should use the function fflush(stdout), in Java — System.out.flush(), in Pascal — flush(output) and in Python — sys.stdout.flush().
Example
Input
5 5
13 15 10 9 15
15 17 12 11 17
10 12 7 6 12
17 19 14 13 19
16 18 13 12 18
Output
Submitted Solution:
```
import sys
mn = int(1e18)
def ask(i, j):
print('?', i, j)
res = int(input())
global mn
mn = min(mn, res)
return [res, j]
def answer():
global mn
print('!', mn)
sys.exit(0)
def main():
n, m = [int(x) for x in input().split()]
if m == 1:
for i in range(1, n+1):
ask(i, 1)
answer()
a = []
for i in range(1, m+1):
a.append(ask(1, i))
a.sort()
cur = 0
for i in range(2, n):
prv, nxt = a[cur][1], a[cur+1][1]
ans_prv = ask(i, prv)
ans_nxt = ask(i, nxt)
while ans_prv > ans_nxt and cur + 1 < m:
ans_prv = ans_nxt
cur += 1
ans_nxt = ask(i, a[cur][1])
for i in range(1, m+1):
ask(n, i)
answer()
main()
``` | instruction | 0 | 32,717 | 8 | 65,434 |
No | output | 1 | 32,717 | 8 | 65,435 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As you all know, the plum harvesting season is on! Little Milutin had his plums planted in an orchard that can be represented as an n by m matrix. While he was harvesting, he wrote the heights of all trees in a matrix of dimensions n by m.
At night, when he has spare time, he likes to perform various statistics on his trees. This time, he is curious to find out the height of his lowest tree. So far, he has discovered some interesting properties of his orchard. There is one particular property that he thinks is useful for finding the tree with the smallest heigh.
Formally, let L(i) be the leftmost tree with the smallest height in the i-th row of his orchard. He knows that L(i) ≤ L(i+1) for all 1 ≤ i ≤ n - 1. Moreover, if he takes a submatrix induced by any subset of rows and any subset of columns, L(i) ≤ L(i+1) will hold for all 1 ≤ i ≤ n'-1, where n' is the number of rows in that submatrix.
Since the season is at its peak and he is short on time, he asks you to help him find the plum tree with minimal height.
Input
This problem is interactive.
The first line of input will contain two integers n and m, representing the number of rows and the number of columns in Milutin's orchard. It is guaranteed that 1 ≤ n, m ≤ 10^6.
The following lines will contain the answers to your queries.
Output
Once you know have found the minimum value r, you should print ! r to the standard output.
Interaction
Your code is allowed to query for an entry (i, j) of a matrix (i.e. get the height of the tree which is in the i-th row and j-th column). The query should be formatted as ? i j, so that 1 ≤ i ≤ n and 1 ≤ j ≤ m.
You may assume that the entries of the matrix will be integers between 1 and 10^9.
Your solution should use not more than 4 ⋅ (n + m) queries.
This is an interactive problem. You have to use a flush operation right after printing each line. For example, in C++ you should use the function fflush(stdout), in Java — System.out.flush(), in Pascal — flush(output) and in Python — sys.stdout.flush().
Example
Input
5 5
13 15 10 9 15
15 17 12 11 17
10 12 7 6 12
17 19 14 13 19
16 18 13 12 18
Output
Submitted Solution:
```
import sys
mn = int(1e18)
def ask(i, j):
print('?', i, j)
res = int(input())
global mn
mn = min(mn, res)
return [res, j]
def answer():
global mn
print('!', mn)
sys.exit(0)
def main():
n, m = [int(x) for x in input().split()]
if m == 1:
for i in range(1, n+1):
ask(i, 1)
answer()
a = []
for i in range(1, m+1):
a.append(ask(1, i))
a.sort()
cur = 0
for i in range(2, n+1):
prv, nxt = a[cur][1], a[cur+1][1]
ans_prv = ask(i, prv)
ans_nxt = ask(i, nxt)
while ans_prv > ans_nxt and cur + 1 < m:
ans_prv = ans_nxt
cur += 1
ans_nxt = ask(i, a[cur][1])
answer()
main()
``` | instruction | 0 | 32,718 | 8 | 65,436 |
No | output | 1 | 32,718 | 8 | 65,437 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Finally it is a day when Arthur has enough money for buying an apartment. He found a great option close to the center of the city with a nice price.
Plan of the apartment found by Arthur looks like a rectangle n × m consisting of squares of size 1 × 1. Each of those squares contains either a wall (such square is denoted by a symbol "*" on the plan) or a free space (such square is denoted on the plan by a symbol ".").
Room in an apartment is a maximal connected area consisting of free squares. Squares are considered adjacent if they share a common side.
The old Arthur dream is to live in an apartment where all rooms are rectangles. He asks you to calculate minimum number of walls you need to remove in order to achieve this goal. After removing a wall from a square it becomes a free square. While removing the walls it is possible that some rooms unite into a single one.
Input
The first line of the input contains two integers n, m (1 ≤ n, m ≤ 2000) denoting the size of the Arthur apartments.
Following n lines each contain m symbols — the plan of the apartment.
If the cell is denoted by a symbol "*" then it contains a wall.
If the cell is denoted by a symbol "." then it this cell is free from walls and also this cell is contained in some of the rooms.
Output
Output n rows each consisting of m symbols that show how the Arthur apartment plan should look like after deleting the minimum number of walls in order to make each room (maximum connected area free from walls) be a rectangle.
If there are several possible answers, output any of them.
Examples
Input
5 5
.*.*.
*****
.*.*.
*****
.*.*.
Output
.*.*.
*****
.*.*.
*****
.*.*.
Input
6 7
***.*.*
..*.*.*
*.*.*.*
*.*.*.*
..*...*
*******
Output
***...*
..*...*
..*...*
..*...*
..*...*
*******
Input
4 5
.....
.....
..***
..*..
Output
.....
.....
.....
.....
Submitted Solution:
```
import sys
def dfs(room, n, m, si, sj, visited):
x0, x1 = si, si
y0, y1 = sj, sj
stack = [(si, sj)]
while stack:
i, j = stack.pop()
if visited[i][j]:
continue
x0 = min(x0, i)
x1 = max(x1, i)
y0 = min(y0, j)
y1 = max(y1, j)
for move in [(i + 1, j), (i - 1, j), (i, j + 1), (i, j - 1)]:
if (0 <= move[0] < n) and (0 <= move[1] < m) and room[move[0]][move[1]] != '*':
stack.append(move)
visited[i][j] = True
return x0, y0, x1, y1
def clear(room, x0, y0, x1, y1):
for i in range(x0, x1 + 1):
for j in range(y0, y1 + 1):
room[i][j] = '.'
def main():
n, m = map(int, sys.stdin.readline().split())
room = []
visited = []
for i in range(n):
room.append(list(sys.stdin.readline().strip()))
visited.append([False] * m)
for i in range(n):
for j in range(m):
if not visited[i][j] and room[i][j] != '*':
x0, y0, x1, y1 = dfs(room, n, m, i, j, visited)
clear(room, x0, y0, x1, y1)
for row in room:
print (''.join(row))
if __name__ == '__main__':
main()
``` | instruction | 0 | 32,938 | 8 | 65,876 |
No | output | 1 | 32,938 | 8 | 65,877 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Finally it is a day when Arthur has enough money for buying an apartment. He found a great option close to the center of the city with a nice price.
Plan of the apartment found by Arthur looks like a rectangle n × m consisting of squares of size 1 × 1. Each of those squares contains either a wall (such square is denoted by a symbol "*" on the plan) or a free space (such square is denoted on the plan by a symbol ".").
Room in an apartment is a maximal connected area consisting of free squares. Squares are considered adjacent if they share a common side.
The old Arthur dream is to live in an apartment where all rooms are rectangles. He asks you to calculate minimum number of walls you need to remove in order to achieve this goal. After removing a wall from a square it becomes a free square. While removing the walls it is possible that some rooms unite into a single one.
Input
The first line of the input contains two integers n, m (1 ≤ n, m ≤ 2000) denoting the size of the Arthur apartments.
Following n lines each contain m symbols — the plan of the apartment.
If the cell is denoted by a symbol "*" then it contains a wall.
If the cell is denoted by a symbol "." then it this cell is free from walls and also this cell is contained in some of the rooms.
Output
Output n rows each consisting of m symbols that show how the Arthur apartment plan should look like after deleting the minimum number of walls in order to make each room (maximum connected area free from walls) be a rectangle.
If there are several possible answers, output any of them.
Examples
Input
5 5
.*.*.
*****
.*.*.
*****
.*.*.
Output
.*.*.
*****
.*.*.
*****
.*.*.
Input
6 7
***.*.*
..*.*.*
*.*.*.*
*.*.*.*
..*...*
*******
Output
***...*
..*...*
..*...*
..*...*
..*...*
*******
Input
4 5
.....
.....
..***
..*..
Output
.....
.....
.....
.....
Submitted Solution:
```
global max_x
global max_y
global min_x
global min_y
dx=[0,0,-1,1]
dy=[-1,1,0,0]
def dfs(x,y):
vis[x][y]=1
global max_x
global max_y
global min_x
global min_y
max_x=max(max_x,x)
min_x=min(min_x,x)
max_y=max(max_y,y)
min_y=min(min_y,y)
for i in range(0,4):
xx=x+dx[i]
yy=y+dy[i]
if xx>=0 and xx<n and yy>=0 and yy<m and s[xx][yy]=='.' and not vis[xx][yy]:
dfs(xx,yy)
s=input()
n,m=s.split()
n,m=int(n),int(m)
s=[[0 for i in range(m+1)] for j in range(n+1)]
vis=[[0 for i in range(m+1)] for j in range(n+1)]
for i in range(0,n):
s[i]=list(input())
print(s[i])
for i in range(0,n):
for j in range(0,m):
if not vis[i][j] and s[i][j]=='.':
max_x=min_x=i
max_y=min_y=j
dfs(i,j)
for a in range(min_x,max_x+1):
for b in range(min_y,max_y+1):
s[a][b]='.'
vis[a][b]=1
for i in range(0,n):
print(''.join(s[i]))
``` | instruction | 0 | 32,939 | 8 | 65,878 |
No | output | 1 | 32,939 | 8 | 65,879 |
Provide tags and a correct Python 3 solution for this coding contest problem.
It is nighttime and Joe the Elusive got into the country's main bank's safe. The safe has n cells positioned in a row, each of them contains some amount of diamonds. Let's make the problem more comfortable to work with and mark the cells with positive numbers from 1 to n from the left to the right.
Unfortunately, Joe didn't switch the last security system off. On the plus side, he knows the way it works.
Every minute the security system calculates the total amount of diamonds for each two adjacent cells (for the cells between whose numbers difference equals 1). As a result of this check we get an n - 1 sums. If at least one of the sums differs from the corresponding sum received during the previous check, then the security system is triggered.
Joe can move the diamonds from one cell to another between the security system's checks. He manages to move them no more than m times between two checks. One of the three following operations is regarded as moving a diamond: moving a diamond from any cell to any other one, moving a diamond from any cell to Joe's pocket, moving a diamond from Joe's pocket to any cell. Initially Joe's pocket is empty, and it can carry an unlimited amount of diamonds. It is considered that before all Joe's actions the system performs at least one check.
In the morning the bank employees will come, which is why Joe has to leave the bank before that moment. Joe has only k minutes left before morning, and on each of these k minutes he can perform no more than m operations. All that remains in Joe's pocket, is considered his loot.
Calculate the largest amount of diamonds Joe can carry with him. Don't forget that the security system shouldn't be triggered (even after Joe leaves the bank) and Joe should leave before morning.
Input
The first line contains integers n, m and k (1 ≤ n ≤ 104, 1 ≤ m, k ≤ 109). The next line contains n numbers. The i-th number is equal to the amount of diamonds in the i-th cell — it is an integer from 0 to 105.
Output
Print a single number — the maximum number of diamonds Joe can steal.
Examples
Input
2 3 1
2 3
Output
0
Input
3 2 2
4 1 3
Output
2
Note
In the second sample Joe can act like this:
The diamonds' initial positions are 4 1 3.
During the first period of time Joe moves a diamond from the 1-th cell to the 2-th one and a diamond from the 3-th cell to his pocket.
By the end of the first period the diamonds' positions are 3 2 2. The check finds no difference and the security system doesn't go off.
During the second period Joe moves a diamond from the 3-rd cell to the 2-nd one and puts a diamond from the 1-st cell to his pocket.
By the end of the second period the diamonds' positions are 2 3 1. The check finds no difference again and the security system doesn't go off.
Now Joe leaves with 2 diamonds in his pocket. | instruction | 0 | 33,065 | 8 | 66,130 |
Tags: greedy
Correct Solution:
```
from sys import stdin
n,m,k = [int(x) for x in stdin.readline().split()]
d = [int(x) for x in stdin.readline().split()]
splitD = [x for x in d[::2]]
if n == 1:
print(min(d[0], k*m))
elif n == 2:
print(0)
elif d[0] == 0 or d[-1] == 0:
print(0)
elif n % 2 == 0:
print(0)
else:
if m >= n//2+1:
times = m//(n//2+1)
print(min(times*k, min(splitD)))
else:
print(0)
``` | output | 1 | 33,065 | 8 | 66,131 |
Provide tags and a correct Python 3 solution for this coding contest problem.
It is nighttime and Joe the Elusive got into the country's main bank's safe. The safe has n cells positioned in a row, each of them contains some amount of diamonds. Let's make the problem more comfortable to work with and mark the cells with positive numbers from 1 to n from the left to the right.
Unfortunately, Joe didn't switch the last security system off. On the plus side, he knows the way it works.
Every minute the security system calculates the total amount of diamonds for each two adjacent cells (for the cells between whose numbers difference equals 1). As a result of this check we get an n - 1 sums. If at least one of the sums differs from the corresponding sum received during the previous check, then the security system is triggered.
Joe can move the diamonds from one cell to another between the security system's checks. He manages to move them no more than m times between two checks. One of the three following operations is regarded as moving a diamond: moving a diamond from any cell to any other one, moving a diamond from any cell to Joe's pocket, moving a diamond from Joe's pocket to any cell. Initially Joe's pocket is empty, and it can carry an unlimited amount of diamonds. It is considered that before all Joe's actions the system performs at least one check.
In the morning the bank employees will come, which is why Joe has to leave the bank before that moment. Joe has only k minutes left before morning, and on each of these k minutes he can perform no more than m operations. All that remains in Joe's pocket, is considered his loot.
Calculate the largest amount of diamonds Joe can carry with him. Don't forget that the security system shouldn't be triggered (even after Joe leaves the bank) and Joe should leave before morning.
Input
The first line contains integers n, m and k (1 ≤ n ≤ 104, 1 ≤ m, k ≤ 109). The next line contains n numbers. The i-th number is equal to the amount of diamonds in the i-th cell — it is an integer from 0 to 105.
Output
Print a single number — the maximum number of diamonds Joe can steal.
Examples
Input
2 3 1
2 3
Output
0
Input
3 2 2
4 1 3
Output
2
Note
In the second sample Joe can act like this:
The diamonds' initial positions are 4 1 3.
During the first period of time Joe moves a diamond from the 1-th cell to the 2-th one and a diamond from the 3-th cell to his pocket.
By the end of the first period the diamonds' positions are 3 2 2. The check finds no difference and the security system doesn't go off.
During the second period Joe moves a diamond from the 3-rd cell to the 2-nd one and puts a diamond from the 1-st cell to his pocket.
By the end of the second period the diamonds' positions are 2 3 1. The check finds no difference again and the security system doesn't go off.
Now Joe leaves with 2 diamonds in his pocket. | instruction | 0 | 33,066 | 8 | 66,132 |
Tags: greedy
Correct Solution:
```
n, m, k = map(int, input().split())
a = list(map(int, input().split()))
if n % 2 == 0:
print('0')
else:
print(min(m // (n // 2 + 1) * k, min(a[::2])))
# Made By Mostafa_Khaled
``` | output | 1 | 33,066 | 8 | 66,133 |
Provide tags and a correct Python 3 solution for this coding contest problem.
It is nighttime and Joe the Elusive got into the country's main bank's safe. The safe has n cells positioned in a row, each of them contains some amount of diamonds. Let's make the problem more comfortable to work with and mark the cells with positive numbers from 1 to n from the left to the right.
Unfortunately, Joe didn't switch the last security system off. On the plus side, he knows the way it works.
Every minute the security system calculates the total amount of diamonds for each two adjacent cells (for the cells between whose numbers difference equals 1). As a result of this check we get an n - 1 sums. If at least one of the sums differs from the corresponding sum received during the previous check, then the security system is triggered.
Joe can move the diamonds from one cell to another between the security system's checks. He manages to move them no more than m times between two checks. One of the three following operations is regarded as moving a diamond: moving a diamond from any cell to any other one, moving a diamond from any cell to Joe's pocket, moving a diamond from Joe's pocket to any cell. Initially Joe's pocket is empty, and it can carry an unlimited amount of diamonds. It is considered that before all Joe's actions the system performs at least one check.
In the morning the bank employees will come, which is why Joe has to leave the bank before that moment. Joe has only k minutes left before morning, and on each of these k minutes he can perform no more than m operations. All that remains in Joe's pocket, is considered his loot.
Calculate the largest amount of diamonds Joe can carry with him. Don't forget that the security system shouldn't be triggered (even after Joe leaves the bank) and Joe should leave before morning.
Input
The first line contains integers n, m and k (1 ≤ n ≤ 104, 1 ≤ m, k ≤ 109). The next line contains n numbers. The i-th number is equal to the amount of diamonds in the i-th cell — it is an integer from 0 to 105.
Output
Print a single number — the maximum number of diamonds Joe can steal.
Examples
Input
2 3 1
2 3
Output
0
Input
3 2 2
4 1 3
Output
2
Note
In the second sample Joe can act like this:
The diamonds' initial positions are 4 1 3.
During the first period of time Joe moves a diamond from the 1-th cell to the 2-th one and a diamond from the 3-th cell to his pocket.
By the end of the first period the diamonds' positions are 3 2 2. The check finds no difference and the security system doesn't go off.
During the second period Joe moves a diamond from the 3-rd cell to the 2-nd one and puts a diamond from the 1-st cell to his pocket.
By the end of the second period the diamonds' positions are 2 3 1. The check finds no difference again and the security system doesn't go off.
Now Joe leaves with 2 diamonds in his pocket. | instruction | 0 | 33,067 | 8 | 66,134 |
Tags: greedy
Correct Solution:
```
n, m, k = map(int, input().split())
a = list(map(int, input().split()))
if n % 2 == 0:
print('0')
else:
print(min(m // (n // 2 + 1) * k, min(a[::2])))
``` | output | 1 | 33,067 | 8 | 66,135 |
Provide tags and a correct Python 3 solution for this coding contest problem.
It is nighttime and Joe the Elusive got into the country's main bank's safe. The safe has n cells positioned in a row, each of them contains some amount of diamonds. Let's make the problem more comfortable to work with and mark the cells with positive numbers from 1 to n from the left to the right.
Unfortunately, Joe didn't switch the last security system off. On the plus side, he knows the way it works.
Every minute the security system calculates the total amount of diamonds for each two adjacent cells (for the cells between whose numbers difference equals 1). As a result of this check we get an n - 1 sums. If at least one of the sums differs from the corresponding sum received during the previous check, then the security system is triggered.
Joe can move the diamonds from one cell to another between the security system's checks. He manages to move them no more than m times between two checks. One of the three following operations is regarded as moving a diamond: moving a diamond from any cell to any other one, moving a diamond from any cell to Joe's pocket, moving a diamond from Joe's pocket to any cell. Initially Joe's pocket is empty, and it can carry an unlimited amount of diamonds. It is considered that before all Joe's actions the system performs at least one check.
In the morning the bank employees will come, which is why Joe has to leave the bank before that moment. Joe has only k minutes left before morning, and on each of these k minutes he can perform no more than m operations. All that remains in Joe's pocket, is considered his loot.
Calculate the largest amount of diamonds Joe can carry with him. Don't forget that the security system shouldn't be triggered (even after Joe leaves the bank) and Joe should leave before morning.
Input
The first line contains integers n, m and k (1 ≤ n ≤ 104, 1 ≤ m, k ≤ 109). The next line contains n numbers. The i-th number is equal to the amount of diamonds in the i-th cell — it is an integer from 0 to 105.
Output
Print a single number — the maximum number of diamonds Joe can steal.
Examples
Input
2 3 1
2 3
Output
0
Input
3 2 2
4 1 3
Output
2
Note
In the second sample Joe can act like this:
The diamonds' initial positions are 4 1 3.
During the first period of time Joe moves a diamond from the 1-th cell to the 2-th one and a diamond from the 3-th cell to his pocket.
By the end of the first period the diamonds' positions are 3 2 2. The check finds no difference and the security system doesn't go off.
During the second period Joe moves a diamond from the 3-rd cell to the 2-nd one and puts a diamond from the 1-st cell to his pocket.
By the end of the second period the diamonds' positions are 2 3 1. The check finds no difference again and the security system doesn't go off.
Now Joe leaves with 2 diamonds in his pocket. | instruction | 0 | 33,068 | 8 | 66,136 |
Tags: greedy
Correct Solution:
```
n,m,k=map(int,input().split())
a=[int(x) for x in input().split()]
if n%2==0:
print(0)
else:
print(min( min(a[::2]), m//(n//2+1)*k))
``` | output | 1 | 33,068 | 8 | 66,137 |
Provide tags and a correct Python 3 solution for this coding contest problem.
It is nighttime and Joe the Elusive got into the country's main bank's safe. The safe has n cells positioned in a row, each of them contains some amount of diamonds. Let's make the problem more comfortable to work with and mark the cells with positive numbers from 1 to n from the left to the right.
Unfortunately, Joe didn't switch the last security system off. On the plus side, he knows the way it works.
Every minute the security system calculates the total amount of diamonds for each two adjacent cells (for the cells between whose numbers difference equals 1). As a result of this check we get an n - 1 sums. If at least one of the sums differs from the corresponding sum received during the previous check, then the security system is triggered.
Joe can move the diamonds from one cell to another between the security system's checks. He manages to move them no more than m times between two checks. One of the three following operations is regarded as moving a diamond: moving a diamond from any cell to any other one, moving a diamond from any cell to Joe's pocket, moving a diamond from Joe's pocket to any cell. Initially Joe's pocket is empty, and it can carry an unlimited amount of diamonds. It is considered that before all Joe's actions the system performs at least one check.
In the morning the bank employees will come, which is why Joe has to leave the bank before that moment. Joe has only k minutes left before morning, and on each of these k minutes he can perform no more than m operations. All that remains in Joe's pocket, is considered his loot.
Calculate the largest amount of diamonds Joe can carry with him. Don't forget that the security system shouldn't be triggered (even after Joe leaves the bank) and Joe should leave before morning.
Input
The first line contains integers n, m and k (1 ≤ n ≤ 104, 1 ≤ m, k ≤ 109). The next line contains n numbers. The i-th number is equal to the amount of diamonds in the i-th cell — it is an integer from 0 to 105.
Output
Print a single number — the maximum number of diamonds Joe can steal.
Examples
Input
2 3 1
2 3
Output
0
Input
3 2 2
4 1 3
Output
2
Note
In the second sample Joe can act like this:
The diamonds' initial positions are 4 1 3.
During the first period of time Joe moves a diamond from the 1-th cell to the 2-th one and a diamond from the 3-th cell to his pocket.
By the end of the first period the diamonds' positions are 3 2 2. The check finds no difference and the security system doesn't go off.
During the second period Joe moves a diamond from the 3-rd cell to the 2-nd one and puts a diamond from the 1-st cell to his pocket.
By the end of the second period the diamonds' positions are 2 3 1. The check finds no difference again and the security system doesn't go off.
Now Joe leaves with 2 diamonds in his pocket. | instruction | 0 | 33,069 | 8 | 66,138 |
Tags: greedy
Correct Solution:
```
#Code by Sounak, IIESTS
#------------------------------warmup----------------------------
import os
import sys
import math
from io import BytesIO, IOBase
from fractions import Fraction
import collections
from itertools import permutations
from collections import defaultdict
from collections import deque
import threading
threading.stack_size(10**8)
sys.setrecursionlimit(300000)
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")
#-------------------game starts now-----------------------------------------------------
class Factorial:
def __init__(self, MOD):
self.MOD = MOD
self.factorials = [1, 1]
self.invModulos = [0, 1]
self.invFactorial_ = [1, 1]
def calc(self, n):
if n <= -1:
print("Invalid argument to calculate n!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.factorials):
return self.factorials[n]
nextArr = [0] * (n + 1 - len(self.factorials))
initialI = len(self.factorials)
prev = self.factorials[-1]
m = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = prev * i % m
self.factorials += nextArr
return self.factorials[n]
def inv(self, n):
if n <= -1:
print("Invalid argument to calculate n^(-1)")
print("n must be non-negative value. But the argument was " + str(n))
exit()
p = self.MOD
pi = n % p
if pi < len(self.invModulos):
return self.invModulos[pi]
nextArr = [0] * (n + 1 - len(self.invModulos))
initialI = len(self.invModulos)
for i in range(initialI, min(p, n + 1)):
next = -self.invModulos[p % i] * (p // i) % p
self.invModulos.append(next)
return self.invModulos[pi]
def invFactorial(self, n):
if n <= -1:
print("Invalid argument to calculate (n^(-1))!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.invFactorial_):
return self.invFactorial_[n]
self.inv(n) # To make sure already calculated n^-1
nextArr = [0] * (n + 1 - len(self.invFactorial_))
initialI = len(self.invFactorial_)
prev = self.invFactorial_[-1]
p = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = (prev * self.invModulos[i % p]) % p
self.invFactorial_ += nextArr
return self.invFactorial_[n]
class Combination:
def __init__(self, MOD):
self.MOD = MOD
self.factorial = Factorial(MOD)
def ncr(self, n, k):
if k < 0 or n < k:
return 0
k = min(k, n - k)
f = self.factorial
return f.calc(n) * f.invFactorial(max(n - k, k)) * f.invFactorial(min(k, n - k)) % self.MOD
#-------------------------------------------------------------------------
mod=10**9+7
n,m,k=map(int,input().split())
a=list(map(int,input().split()))
if n%2==0 or m<n//2+1:
print(0)
else:
mn=a[0]
for i in range (2,n,2):
mn=min(mn,a[i])
m1=m//(n//2+1)
avl=m1*k
r=min(mn,avl)
print(r)
``` | output | 1 | 33,069 | 8 | 66,139 |
Provide tags and a correct Python 3 solution for this coding contest problem.
It is nighttime and Joe the Elusive got into the country's main bank's safe. The safe has n cells positioned in a row, each of them contains some amount of diamonds. Let's make the problem more comfortable to work with and mark the cells with positive numbers from 1 to n from the left to the right.
Unfortunately, Joe didn't switch the last security system off. On the plus side, he knows the way it works.
Every minute the security system calculates the total amount of diamonds for each two adjacent cells (for the cells between whose numbers difference equals 1). As a result of this check we get an n - 1 sums. If at least one of the sums differs from the corresponding sum received during the previous check, then the security system is triggered.
Joe can move the diamonds from one cell to another between the security system's checks. He manages to move them no more than m times between two checks. One of the three following operations is regarded as moving a diamond: moving a diamond from any cell to any other one, moving a diamond from any cell to Joe's pocket, moving a diamond from Joe's pocket to any cell. Initially Joe's pocket is empty, and it can carry an unlimited amount of diamonds. It is considered that before all Joe's actions the system performs at least one check.
In the morning the bank employees will come, which is why Joe has to leave the bank before that moment. Joe has only k minutes left before morning, and on each of these k minutes he can perform no more than m operations. All that remains in Joe's pocket, is considered his loot.
Calculate the largest amount of diamonds Joe can carry with him. Don't forget that the security system shouldn't be triggered (even after Joe leaves the bank) and Joe should leave before morning.
Input
The first line contains integers n, m and k (1 ≤ n ≤ 104, 1 ≤ m, k ≤ 109). The next line contains n numbers. The i-th number is equal to the amount of diamonds in the i-th cell — it is an integer from 0 to 105.
Output
Print a single number — the maximum number of diamonds Joe can steal.
Examples
Input
2 3 1
2 3
Output
0
Input
3 2 2
4 1 3
Output
2
Note
In the second sample Joe can act like this:
The diamonds' initial positions are 4 1 3.
During the first period of time Joe moves a diamond from the 1-th cell to the 2-th one and a diamond from the 3-th cell to his pocket.
By the end of the first period the diamonds' positions are 3 2 2. The check finds no difference and the security system doesn't go off.
During the second period Joe moves a diamond from the 3-rd cell to the 2-nd one and puts a diamond from the 1-st cell to his pocket.
By the end of the second period the diamonds' positions are 2 3 1. The check finds no difference again and the security system doesn't go off.
Now Joe leaves with 2 diamonds in his pocket. | instruction | 0 | 33,070 | 8 | 66,140 |
Tags: greedy
Correct Solution:
```
z=lambda: list(map(int,input().split()))
n,m,k=z()
print(n%2*min(m//(n//2+1)*k, *z()[::2]))
``` | output | 1 | 33,070 | 8 | 66,141 |
Provide tags and a correct Python 3 solution for this coding contest problem.
It is nighttime and Joe the Elusive got into the country's main bank's safe. The safe has n cells positioned in a row, each of them contains some amount of diamonds. Let's make the problem more comfortable to work with and mark the cells with positive numbers from 1 to n from the left to the right.
Unfortunately, Joe didn't switch the last security system off. On the plus side, he knows the way it works.
Every minute the security system calculates the total amount of diamonds for each two adjacent cells (for the cells between whose numbers difference equals 1). As a result of this check we get an n - 1 sums. If at least one of the sums differs from the corresponding sum received during the previous check, then the security system is triggered.
Joe can move the diamonds from one cell to another between the security system's checks. He manages to move them no more than m times between two checks. One of the three following operations is regarded as moving a diamond: moving a diamond from any cell to any other one, moving a diamond from any cell to Joe's pocket, moving a diamond from Joe's pocket to any cell. Initially Joe's pocket is empty, and it can carry an unlimited amount of diamonds. It is considered that before all Joe's actions the system performs at least one check.
In the morning the bank employees will come, which is why Joe has to leave the bank before that moment. Joe has only k minutes left before morning, and on each of these k minutes he can perform no more than m operations. All that remains in Joe's pocket, is considered his loot.
Calculate the largest amount of diamonds Joe can carry with him. Don't forget that the security system shouldn't be triggered (even after Joe leaves the bank) and Joe should leave before morning.
Input
The first line contains integers n, m and k (1 ≤ n ≤ 104, 1 ≤ m, k ≤ 109). The next line contains n numbers. The i-th number is equal to the amount of diamonds in the i-th cell — it is an integer from 0 to 105.
Output
Print a single number — the maximum number of diamonds Joe can steal.
Examples
Input
2 3 1
2 3
Output
0
Input
3 2 2
4 1 3
Output
2
Note
In the second sample Joe can act like this:
The diamonds' initial positions are 4 1 3.
During the first period of time Joe moves a diamond from the 1-th cell to the 2-th one and a diamond from the 3-th cell to his pocket.
By the end of the first period the diamonds' positions are 3 2 2. The check finds no difference and the security system doesn't go off.
During the second period Joe moves a diamond from the 3-rd cell to the 2-nd one and puts a diamond from the 1-st cell to his pocket.
By the end of the second period the diamonds' positions are 2 3 1. The check finds no difference again and the security system doesn't go off.
Now Joe leaves with 2 diamonds in his pocket. | instruction | 0 | 33,071 | 8 | 66,142 |
Tags: greedy
Correct Solution:
```
import sys
from array import array # noqa: F401
def input():
return sys.stdin.buffer.readline().decode('utf-8')
n, m, k = map(int, input().split())
a = list(map(int, input().split()))
if n == 1:
print(min(a[0], m * k))
exit()
if n % 2 == 0:
print(0)
exit()
turn = (m // ((n + 1) // 2)) * k
print(min(turn, min(a[::2])))
``` | output | 1 | 33,071 | 8 | 66,143 |
Provide tags and a correct Python 3 solution for this coding contest problem.
It is nighttime and Joe the Elusive got into the country's main bank's safe. The safe has n cells positioned in a row, each of them contains some amount of diamonds. Let's make the problem more comfortable to work with and mark the cells with positive numbers from 1 to n from the left to the right.
Unfortunately, Joe didn't switch the last security system off. On the plus side, he knows the way it works.
Every minute the security system calculates the total amount of diamonds for each two adjacent cells (for the cells between whose numbers difference equals 1). As a result of this check we get an n - 1 sums. If at least one of the sums differs from the corresponding sum received during the previous check, then the security system is triggered.
Joe can move the diamonds from one cell to another between the security system's checks. He manages to move them no more than m times between two checks. One of the three following operations is regarded as moving a diamond: moving a diamond from any cell to any other one, moving a diamond from any cell to Joe's pocket, moving a diamond from Joe's pocket to any cell. Initially Joe's pocket is empty, and it can carry an unlimited amount of diamonds. It is considered that before all Joe's actions the system performs at least one check.
In the morning the bank employees will come, which is why Joe has to leave the bank before that moment. Joe has only k minutes left before morning, and on each of these k minutes he can perform no more than m operations. All that remains in Joe's pocket, is considered his loot.
Calculate the largest amount of diamonds Joe can carry with him. Don't forget that the security system shouldn't be triggered (even after Joe leaves the bank) and Joe should leave before morning.
Input
The first line contains integers n, m and k (1 ≤ n ≤ 104, 1 ≤ m, k ≤ 109). The next line contains n numbers. The i-th number is equal to the amount of diamonds in the i-th cell — it is an integer from 0 to 105.
Output
Print a single number — the maximum number of diamonds Joe can steal.
Examples
Input
2 3 1
2 3
Output
0
Input
3 2 2
4 1 3
Output
2
Note
In the second sample Joe can act like this:
The diamonds' initial positions are 4 1 3.
During the first period of time Joe moves a diamond from the 1-th cell to the 2-th one and a diamond from the 3-th cell to his pocket.
By the end of the first period the diamonds' positions are 3 2 2. The check finds no difference and the security system doesn't go off.
During the second period Joe moves a diamond from the 3-rd cell to the 2-nd one and puts a diamond from the 1-st cell to his pocket.
By the end of the second period the diamonds' positions are 2 3 1. The check finds no difference again and the security system doesn't go off.
Now Joe leaves with 2 diamonds in his pocket. | instruction | 0 | 33,072 | 8 | 66,144 |
Tags: greedy
Correct Solution:
```
#------------------------template--------------------------#
import os
import sys
from math import *
from collections import *
from fractions import *
from bisect import *
from heapq import*
from io import BytesIO, IOBase
def vsInput():
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
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")
ALPHA='abcdefghijklmnopqrstuvwxyz'
M=998244353
EPS=1e-6
def value():return tuple(map(int,input().split()))
def array():return [int(i) for i in input().split()]
def Int():return int(input())
def Str():return input()
def arrayS():return [i for i in input().split()]
#-------------------------code---------------------------#
# vsInput()
n,m,k=value()
a=array()
if(n%2==0):
print(0)
elif(m<ceil(n/2)):
print(0)
else:
maxLoot=inf
for i in range(n):
if(i%2==0):
maxLoot=min(maxLoot,a[i])
per_min=m//(ceil(n/2))
ans=per_min*k
print(min(ans,maxLoot))
``` | output | 1 | 33,072 | 8 | 66,145 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It is nighttime and Joe the Elusive got into the country's main bank's safe. The safe has n cells positioned in a row, each of them contains some amount of diamonds. Let's make the problem more comfortable to work with and mark the cells with positive numbers from 1 to n from the left to the right.
Unfortunately, Joe didn't switch the last security system off. On the plus side, he knows the way it works.
Every minute the security system calculates the total amount of diamonds for each two adjacent cells (for the cells between whose numbers difference equals 1). As a result of this check we get an n - 1 sums. If at least one of the sums differs from the corresponding sum received during the previous check, then the security system is triggered.
Joe can move the diamonds from one cell to another between the security system's checks. He manages to move them no more than m times between two checks. One of the three following operations is regarded as moving a diamond: moving a diamond from any cell to any other one, moving a diamond from any cell to Joe's pocket, moving a diamond from Joe's pocket to any cell. Initially Joe's pocket is empty, and it can carry an unlimited amount of diamonds. It is considered that before all Joe's actions the system performs at least one check.
In the morning the bank employees will come, which is why Joe has to leave the bank before that moment. Joe has only k minutes left before morning, and on each of these k minutes he can perform no more than m operations. All that remains in Joe's pocket, is considered his loot.
Calculate the largest amount of diamonds Joe can carry with him. Don't forget that the security system shouldn't be triggered (even after Joe leaves the bank) and Joe should leave before morning.
Input
The first line contains integers n, m and k (1 ≤ n ≤ 104, 1 ≤ m, k ≤ 109). The next line contains n numbers. The i-th number is equal to the amount of diamonds in the i-th cell — it is an integer from 0 to 105.
Output
Print a single number — the maximum number of diamonds Joe can steal.
Examples
Input
2 3 1
2 3
Output
0
Input
3 2 2
4 1 3
Output
2
Note
In the second sample Joe can act like this:
The diamonds' initial positions are 4 1 3.
During the first period of time Joe moves a diamond from the 1-th cell to the 2-th one and a diamond from the 3-th cell to his pocket.
By the end of the first period the diamonds' positions are 3 2 2. The check finds no difference and the security system doesn't go off.
During the second period Joe moves a diamond from the 3-rd cell to the 2-nd one and puts a diamond from the 1-st cell to his pocket.
By the end of the second period the diamonds' positions are 2 3 1. The check finds no difference again and the security system doesn't go off.
Now Joe leaves with 2 diamonds in his pocket.
Submitted Solution:
```
n, m, k = map(int, input().split())
cells = list(map(int, input().split()))
if n < 3:
print("0")
else:
print(min([cells[0] + cells[-1], m * (k // (n - 1))]))
``` | instruction | 0 | 33,073 | 8 | 66,146 |
No | output | 1 | 33,073 | 8 | 66,147 |
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