Square

Square
Time Limit : 10000/5000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 31 Accepted Submission(s) : 7
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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
//减掉最大值超过平均值，和sum无法整除的情况，然后每次当前值=aver时count+1就行了，当count==4时结束，否则dfs。

#include<iostream>#include<cstdio>#define max(a,b) (a)>(b)?(a):(b)using namespace std;int str[23],n,sum,aver,maxn;bool vist[23],flag;void dfs(int mySum,int Count,int t){    if(mySum==aver)    {       Count++;       if(Count==4)       {          flag=true;          return;       }       else mySum=0;       t=0;    }    if(flag) return;    for(int i=t;i<n;i++)    {       if(!vist[i]&&mySum+str[i]<=aver)       {          vist[i]=true;          dfs(mySum+str[i],Count,i);          vist[i]=false;       }    }}int main(){   int T,i;   scanf("%d",&T);   while(T--)   {      scanf("%d",&n);      sum=maxn=0;      for(i=0;i<n;i++)      {         scanf("%d",&str[i]);         sum+=str[i];         maxn=max(str[i],maxn);         vist[i]=false;      }      if(sum%4!=0||maxn>sum/4)//简单判断、不满足的直接给出结果          printf("no\n");      else      {         flag=false;         aver=sum/4;         dfs(0,0,0);         if(flag) printf("yes\n");         else printf("no\n");      }   }   return 0;}