hdu-1518 Square

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Square

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8922 Accepted Submission(s): 2903

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

34 1 1 1 15 10 20 30 40 508 1 7 2 6 4 4 3 5

Sample Output

yesnoyesi在不同地方定义有时也会产生不同的效果，甚至会让你wa,所以要注意每一点是很重要的代码如下：#include<stdio.h>#include<string.h>int sum,flag;int stick[30];int mark[30];int n;              //刚开始将i定义为全局变量，以为下边就不用定义了 void dfs(int side,int l,int k){ //这样应该很方便的的，但是这样却错了 int i;            //所以定义变量的时候位置也很重要 if(side==5){flag=1; return;}if(l==sum){dfs(side+1,0,0);if(flag)return ;}for(i=k;i<n;i++){ if(!mark[i]&&l+stick[i]<=sum){ mark[i]=1; dfs(side,stick[i]+l,i+1); if(flag)    return ; mark[i]=0; }}}int main(){int t;scanf("%d",&t);//printf("%d\n",t);while(t--){int i;  sum=0;memset(stick,0,sizeof(stick));memset(mark,0,sizeof(mark));scanf("%d",&n);for(i=0;i<n;i++){scanf("%d",&stick[i]);sum+=stick[i];}//printf("%d\n",sum);if(sum%4!=0){printf("no\n");continue;}sum/=4;for(i=0;i<n;i++){if(stick[i]>sum)   break;}if(i!=n){printf("no\n");continue;}flag=0;dfs(1,0,0);if(flag)   printf("yes\n");else   printf("no\n");}return 0;}

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