poj 2362

#include<iostream>
#include<iterator>
#include<algorithm>
using namespace std;
const int maxn=22;
int n;
int visited[maxn];
int dfs(int stick[],int curlen,int s,int deslen,int num)
{
   if(num==4)
    return 1;
   for(int i=s;i<n;i++)
   {
     if(!visited[i])
     {
        visited[i]=1;
        if(curlen+stick[i]==deslen)
         {
            if(dfs(stick,0,0,deslen,num+1))
              return 1;
           
         }
        else if(curlen+stick[i]<deslen)
        { 
             if(dfs(stick,curlen+stick[i],i,deslen,num))
               return 1;
        }
      visited[i]=0;   //這裏就是回朔要注意
     }
    
   }
return 0;
}
int cmp(int a,int b)
{
    return a>b;
}
int main()
{
    int nCase;
    int stick[22];
    cin>>nCase;
    while(nCase--)
    {
      cin>>n;
      int sum=0;
      for(int i=0;i<n;i++)
       {
       cin>>stick[i];
       sum+=stick[i];
       }
       memset(visited,0,sizeof(visited));
      sort(stick,stick+n,cmp);
      /*
       思路大致是這樣，首先判斷棍子是否能被4正除，若能，求出棍子的長度，然後在裏面找起組合，
       若能組成4條這樣的那麼就行了
      */
   //   copy(stick,stick+n,ostream_iterator<int>(cout,","));
      if(sum%4!=0)
       cout<<"no"<<endl;
      else
      {
          if(dfs(stick,0,0,sum/4,1))
          //這裏stick表示棍子，curlen表示當前長度，deslen表示目標長度，num表示當前以求出的數目，若==4，則行
           cout<<"yes"<<endl;
          else
           cout<<"no"<<endl;
      }
    }
    system("pause");
   
}