Square
Time Limit : 10000/5000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 88 Accepted Submission(s) : 37
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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
Source
University of Waterloo Local Contest 2002.09.21
#include <iostream> #include<cstdio> #include<algorithm> #include<cstring> using namespace std; int t,i,j,n,m,flag,sum,ans; int vis[25],a[25]; int cmp(int a,int b) { return a>b; } void dfs(int k,int l,int edge) { if(edge==4) { flag=1; return; } // if(flag) return; for(int i=k;i<=n;i++) if (!vis[i]) if (l+a[i]<=sum) { vis[i]=1; if (l+a[i]==sum) dfs(1,0,edge+1); else dfs(i+1,l+a[i],edge); if(flag) return; vis[i]=0; } return; } int main() { scanf("%d",&t); for(;t>0;t--) { scanf("%d",&n); sum=0; for(i=1;i<=n;i++) { scanf("%d",&a[i]); sum+=a[i]; } if (sum%4!=0) { printf("no\n"); continue; } sort(a+1,a+1+n,cmp); sum=sum/4; if (a[1]>sum) { printf("no\n"); continue; } memset(vis,0,sizeof(vis)); flag=0; dfs(1,0,1); if(flag) printf("yes\n"); else printf("no\n"); } return 0; }