Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 11824    Accepted Submission(s): 3794

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

 有n支棍子，是否可以组成一个正方形

#include<cstdio>

#include<cstring>

#include<iostream>

#include<algorithm>

using namespace std;

int s[21],n,sum,target;

bool used[21];

int cmp(int a,int b)

{

	return a>b;

}

bool dfs(int curs,int curl,int pos)

{//curs：找到了几条正方形的边，curl：当前边已经拼凑的长度，只用pos之后的

	if(curs==3)

	return 1;//有了三条边，第四条肯定也是存在的

	for(int i=pos;i<n;i++)

	{

		if(used[i]==true)

		continue;

		if(curl+s[i]==target)

		{

			used[i]=true;

			if(dfs(curs+1,0,0)==true)//找到一条边之后就找下一条边

			return true;

			used[i]=false;

		}

		else if(curl+s[i]<target)

		{

			used[i]=true;

			if(dfs(curs,curl+s[i],i)==true)

			return true;

			used[i]=false;//回溯 ，一条边可用可不用

		}

	}

	return false;

}

int main()

{

	int t;

	scanf("%d",&t);

	while(t--)

	{

		memset(s,0,sizeof(s));

		scanf("%d",&n);

		sum=0;

		for(int i=0;i<n;i++)

		{

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

			sum+=s[i];

		}

		target=sum/4;

		if(sum%4!=0||n<4)

		cout<<"no"<<endl;

		else

		{

			memset(used,false,sizeof(used));

			sort(s,s+n,cmp);

			if(target<s[0])

			cout<<"no"<<endl;

			else if(dfs(0,0,0)==true)

			cout<<"yes"<<endl;

			else

			cout<<"no"<<endl;

		}

	}

	return 0;

}