Codeforces Round #445 A. ACM ICPC【暴力】

A. ACM ICPC

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

In a small but very proud high school it was decided to win ACM ICPC. This goal requires to compose as many teams of three as possible, but since there were only 6 students who wished to participate, the decision was to build exactly two teams.

After practice competition, participant number i got a score of ai. Team score is defined as sum of scores of its participants. High school management is interested if it's possible to build two teams with equal scores. Your task is to answer that question.

Input

The single line contains six integers a1, ..., a6 (0 ≤ ai ≤ 1000) — scores of the participants

Output

Print "YES" (quotes for clarity), if it is possible to build teams with equal score, and "NO" otherwise.

You can print each character either upper- or lowercase ("YeS" and "yes" are valid when the answer is "YES").

Examples

input

1 3 2 1 2 1

output

YES

input

1 1 1 1 1 99

output

NO

Note

In the first sample, first team can be composed of 1st, 2nd and 6th participant, second — of 3rd, 4th and 5th: team scores are1 + 3 + 1 = 2 + 1 + 2 = 5.

In the second sample, score of participant number 6 is too high: his team score will be definitely greater.

【题意】：存在某三个和另外三个相等输出YES，否则输出NO

【分析】：暴力

【代码】：

#include <bits/stdc++.h>

using namespace std;
const int maxn = ;

int main()
{
    int a[],sum[];
    memset(sum,,sizeof(sum));
    for(int i=;i<=;i++)
    {
        cin>>a[i];
        if(a[]+a[]+a[]==a[]+a[]+a[])
        {
            printf("YES\n");
            return ;
        }

        if(a[]+a[]+a[]==a[]+a[]+a[])
        {
            printf("YES\n");
            return ;
        }

        if(a[]+a[]+a[]==a[]+a[]+a[])
        {
            printf("YES\n");
            return ;
        }

        if(a[]+a[]+a[]==a[]+a[]+a[])
        {
            printf("YES\n");
            return ;
        }

        if(a[]+a[]+a[]==a[]+a[]+a[])
        {
            printf("YES\n");
            return ;
        }

        if(a[]+a[]+a[]==a[]+a[]+a[])
        {
            printf("YES\n");
            return ;
        }

        if(a[]+a[]+a[]==a[]+a[]+a[])
        {
            printf("YES\n");
            return ;
        }

        if(a[]+a[]+a[]==a[]+a[]+a[])
        {
            printf("YES\n");
            return ;
        }

        if(a[]+a[]+a[]==a[]+a[]+a[])
        {
            printf("YES\n");
            return ;
        }

        if(a[]+a[]+a[]==a[]+a[]+a[])
        {
            printf("YES\n");
            return ;
        }

    }
    printf("NO\n");
    return ;
}

我的傻逼代码

#include <bits/stdc++.h>

using namespace std;
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<cstdio>
#include<iostream>
using namespace std;
int a[],sum=;

int main()
{
    for(int i=;i<;i++)
    {
        cin>>a[i];
        sum+=a[i];
    }
    for(int i=;i<;i++)
    {
        for(int j=i+;j<;j++)
        {
            for(int k=j+;k<;k++)
            {
                if(*(a[i]+a[j]+a[k])==sum)
                {
                    printf("YES\n");
                    return ;
                }
            }
        }
    }
    printf("NO\n");
}

枚举

#include <bits/stdc++.h>

using namespace std;

int main() {
    int a[];
    for(int i = ; i < ; i++)    cin >> a[i];
    sort(a, a + );
    do{
        if((a[] + a[] + a[]) == (a[] + a[] + a[])){
            cout << "YES";
            return ;
        }
    }while(next_permutation(a, a + ));
    cout << "NO";
    return ;
}

next_permutation的妙用