FOJ Problem 1099 Square

/*此题算是暴力求解，关键是如何剪枝的问题，此题是连续找出四条边即可，其实这么说有时给出的数据可以
凑出许多种的正方形，这些正方形中一定可调成一支特殊的正方形，就是每条边的组成线段都是递增的，然后

就一条边来说，就可以利用现在寻找的每条组成边的线段都是递增来剪枝，开始先排序一下，调用递归*/

Problem 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

3

4 1 1 1 1

5 10 20 30 40 50

8 1 7 2 6 4 4 3 5

Sample Output

yes

no

yes 

# include "stdio.h" 
# include "string.h"
# include <algorithm>
using namespace std;
int n, sum;
int a[30];
int used[30];
void init()
{
	for(int i=0; i<=29; i++)
	{
		used[i]=1;
	}
}
int dfs(int _n, int m, int num)
{
	if(m==sum)
	{
		m=0;
		_n=0;
		num++;
		if(num==4)
		return 1;
	}
	for(int i=_n+1; i<=n; i++)
	{
		if(m+a[i]<=sum&&used[i])
		{
			used[i]=0;
			if(dfs(i, m+a[i], num))
			return 1;
			used[i]=1;
		}
	}
	return 0;
}

int main()
{
	int t, i, j;
	scanf("%d", &t);
	for(i=1; i<=t; i++)
	{
		scanf("%d", &n);
		sum=0;
		for(j=1; j<=n; j++)
		{
			scanf("%d", &a[j]);
			sum=sum+a[j];
		}
		if(sum%4!=0)
		{
			printf("no\n");
		}
		else
		{
			sort(a+1,a+1+n);
			sum=sum/4;
			init();
			if(dfs(0, 0, 0))
			printf("yes\n");
			else
			{
				printf("no\n");
			}
		}
	}
	return 0;
}