本文主要是介绍FOJ Problem 1099 Square，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
/*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
Sample Output
# 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;
}
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