天津大学TOJ题目，《square》

Given a set of sticks of various lengths, is it possible to join them end-to-end to form a square? 
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. 

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

 
Output for Sample Input
yes
no
yes

#include 
#include 
#include  //计时用的头文件
#include 
using namespace std;

#define size 20
int test_num;//TEST数目
int data[size]={};
bool findsquare = false;
int l1,l2,l3,l4;
int a[size]={};
int sum;
int TimeStart;
int TimeEnd;
int TimeUsed;

void handle(){
    //for(int i=0;i//    if(a[i]==1) {l1+=data[i+1];}
    //    if(a[i]==2) {l2+=data[i+1];}
    //    if(a[i]==3) {l3+=data[i+1];}
    //    if(a[i]==4) {l4+=data[i+1];}
    //}
    if(l1==l2&&l2==l3&&l3==l4) findsquare = true;
    
}

void dfs(int step){
    if(step ==data[0]+1){
      handle();
       return;
    }
     if(step ==0){
     for(int i=1;i0]+1;i++){
        sum +=data[i];
     }
     if(sum%4!=0){
        return;
     }
     }
     if(l1>sum/4||l2>sum/4||l3>sum/4||l4>sum/4){return;}
     if(l1==sum/4&&l2==sum/4&&l3==sum/4){findsquare=true;return;}

     a[step] = 1;
     l1 +=data[step+1];
     dfs(step+1);
     l1 -=data[step+1];

     a[step] = 2;
      l2 +=data[step+1];
     dfs(step+1);
      l2 -=data[step+1];

     a[step] = 3;
      l3 +=data[step+1];
     dfs(step+1);
     l3 -=data[step+1];

     a[step] = 4;
     l4 +=data[step+1];
     dfs(step+1);
     l4 -=data[step+1];
}


int main(){
   // time_t start,end,time; /*注意计时所用的变量名称*/
    //start=clock();
    int count=0;
    int data_num=0;
        //freopen("input.txt","r",stdin);
        scanf("%d",&test_num);
        for(int i=test_num;i>0;i--){
            scanf("%d ",&data[count]);
            count++;
            for(int j=data[0];j>0;j--){
            
                scanf("%d ",&data[count]);
                count++;
            }
            data_num = count;
            count =0;
            dfs(0);
            sum =0;
            if(findsquare == true){
            printf("yes\n");
            findsquare = false;
        }
        else{
        printf("no\n");
        }
        }//一个case的结束
        //end=clock();
        // time=end-start;//这里的时间是计算机内部时间
        // cout<<"time:"<// system("pause");

}

 

转载于:https://www.cnblogs.com/linux0537/p/6141588.html