HDOJ 1518 Square

Square

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5443    Accepted Submission(s): 1732

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

 

Source

University of Waterloo Local Contest 2002.09.21

 

Recommend

LL

 

#include <iostream>

#include <cstring>

#include <cstdio>

#include <algorithm>

using namespace std;

int sum;

int n;

int a[25];

int vis[25];

bool dfs(int cur,int time,int k)

{

    if(time==3)

    {

       return true;

    }

    for(int i=k-1;i>=0;i--)

    {

        if(!vis )

        {

            vis =1;

            if(cur+a ==sum)

            {

                if(dfs(0,time+1,n))

                    return true;

            }

            else if(cur+a <sum)

            {

                if(dfs(cur+a ,time,i)) return true;

            }

            vis =0;

        }

    }

    return false;

}

int main()

{

    int T;

    scanf("%d",&T);

while(T--)

{

    memset(vis,0,sizeof(vis));

    sum=0;

    int maxn=-1;

    scanf("%d",&n);

    for(int i=0;i<n;i++)

    {

        scanf("%d",&a );

        sum+=a ;

        maxn=max(maxn,a );

    }

    if(sum%4!=0||maxn>sum/4)

    {

        puts("no");

        continue;

    }

    sum=sum/4;

    sort(a,a+n);

    if(dfs(0,0,n)) puts("yes");

    else puts("no");

}

    return 0;

}

转载于:https://www.cnblogs.com/CKboss/p/3351004.html