标签：std length mes iostream freopen scan while cas output
InputThe 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.
OutputFor 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
1 #include <iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 using namespace std; 6 int t,m,num[22],sum,target; 7 bool vis[22]; 8 bool cmp(int a,int b) 9 { 10 return a<b; 11 } 12 bool dfs(int sum,int cnt,int k) 13 { 14 if(cnt==3) return true; 15 for(int i=k;i>=1;i--) 16 { 17 if(vis[i]==0) 18 { 19 vis[i]=1;//递归的精髓就是回溯 20 if(sum+num[i]==target 21 { 22 if(dfs(0,cnt+1,m)) return true; 23 } 24 else if(sum+num[i]<target) 25 { 26 f(dfs(sum+num[i],cnt,i-1)) return true; 27 } 28 vis[i]=0; 29 } 30 } 31 return false; 32 } 33 int main() 34 { 35 // freopen("cin.txt","r",stdin); 36 scanf("%d",&t); 37 while(t--) 38 { 39 scanf("%d",&m); 40 sum=0; 41 for(int i=1;i<=m;i++) { 42 scanf("%d",&num[i]); 43 sum+=num[i]; 44 } 45 memset(vis,0,sizeof(vis)); 46 target=sum/4; 47 if(m<=3||sum%4!=0) puts("no"); 48 else if(dfs(0,0,m)) puts("yes"); 49 else puts("no"); 50 } 51 return 0; 52 }
标签：std length mes iostream freopen scan while cas output
原文地址：https://www.cnblogs.com/linda-/p/13382931.html
