Description

Given a set of sticks of various lengths, is it possible to join them end-to-end to form a square?

Input

The first line of input contains N, the number of test cases. Each test case begins with an integer 4 ≤ M ≤ 20, the number of sticks. M integers follow; each gives the length of a stick - an integer between 1 and 10,000.

Output

For each case, output a line containing "yes" if is is possible to form a square; otherwise output "no".

• Sample
  Input
• Raw

3

4 1 1 1 1

5 10 20 30 40 50

8 1 7 2 6 4 4 3 5

• Sample
  Output
• Raw

yes

no

yes

#include<cstdio>

#include<algorithm>

#include<iostream>

#include<cstring>

using namespace std;

int data[250],state[250];

int ave,N,M,sum;

int dfs(int x,int pos,int len)

{

    int i;

    if(x==3)//如果有三条边都满足，结束

        return 1;

    for(i=pos;i>=0;i--){

        if(!state[i]){

            state[i]=1;

            if(len+data[i]<ave){

                if(dfs(x,i-1,len+data[i]))//i-1的作用是剪枝

                return 1;

            }

            if(len+data[i]==ave){

                if(dfs(x+1,M-1,0))

                return 1;

            }

            state[i]=0;

        }

    }

    return 0;

}

int main ()

{

    scanf("%d",&N);

    while(N--){

        sum=0;

        memset(state,0,sizeof(state));

        //memset(data,0,sizeof(data));

        scanf("%d",&M);

        for(int i=0;i<M;sum+=data[i],i++)

            scanf("%d",&data[i]);

        ave=sum/4;

        if(M<4||ave*4!=sum||ave<data[M-1]){

            printf("no\n");

            continue;

        }

        if(dfs(0,M-1,0))

            printf("yes\n");

        else

            printf("no\n");

    }

    return 0;

}