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Square

Time Limit : 10000/5000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 75 Accepted Submission(s) : 15

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<iostream>
#include<algorithm>
using namespace std;
int N;//木块数量
int L[21];//每块木块的长度
int len;//形成正方形的边长
int s;//所有木块长度
bool used[21];
int cmp(const void * a,const void *b)
{
    return *(int *)b - *(int *)a;
}
bool find(int setnum,int depth,int sum=0)
{
    if(setnum ==1)
        return true;
    for(int i=depth;i<=N;++i)
    {
        if(used[i]==true)
            continue;
        if (i>1 && !L[i - 1] && L[i] == L[i-1])
            continue;
        int sum_now=L[i]+sum;
        if(sum_now < len)
        {
            used[i]=true;
            if(find(setnum,i+1,sum_now))
                return true;
            used[i]=false;
        }
        else
            if(sum_now==len)
            {
                used[i]=true;
                if(find(setnum-1,1,0))
                    return true;
                used[i]=false;
            }
            if(sum_now==len || sum ==0)
                break;
    }
    return false;
}
int main()
{
    int i,j,t;
    int MAX;
    cin>>t;
    for(i=1;i<=t;++i)
    {
        cin>>N;
        s=0;
        MAX=0;
        for(j=1;j<=N;++j)
        {
            cin>>L[j];
            s+=L[j];
            used[j]=false;//设置为未使用
            if(L[j]>MAX)
                MAX=L[j];//考虑单个L[j]>Len
        }
        if(s%4!=0)//总和不能被4整除
        {
            cout<<"no"<<endl;
            continue;
        }
        else
            len=s/4;
        if(MAX>len)//最大的数不有大于边长
        {
            cout<<"no"<<endl;
            continue;
        }
        qsort(L+1,N,sizeof(int),cmp);
        if(find(4,1,0))
            cout<<"yes"<<endl;
        else
            cout<<"no"<<endl;
    }
    return 0;
}