Trouble HDU - 4334

Hassan is in trouble. His mathematics teacher has given him a very difficult problem called 5-sum. Please help him. 
The 5-sum problem is defined as follows: Given 5 sets S_1,...,S_5 of n integer numbers each, is there a_1 in S_1,...,a_5 in S_5 such that a_1+...+a_5=0?

InputFirst line of input contains a single integer N (1≤N≤50). N test-cases follow. First line of each test-case contains a single integer n (1<=n<=200). 5 lines follow each containing n integer numbers in range [-10^15, 1 0^15]. I-th line denotes set S_i for 1<=i<=5.OutputFor each test-case output "Yes" (without quotes) if there are a_1 in S_1,...,a_5 in S_5 such that a_1+...+a_5=0, otherwise output "No".Sample Input

2
2
1 -1
1 -1
1 -1
1 -1
1 -1
3
1 2 3
-1 -2 -3
4 5 6
-1 3 2
-4 -10 -1

Sample Output

No
Yes

给你5个数组每个都有N个元素，然后在每一个数组里面取一个数看能否为0
这题5个for不用想肯定TEL 

分治的思想 将第一个数组和第二个数组合并 第三个和第四个合并 
这样就只有3个数组了。
下面的上代码 ，细节在代码里面体现

 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 using namespace std;
 long long  cnt[][];
 long long  a[],b[],c[];

 int main() {
     int t;
     scanf("%d",&t);
     while(t--){
         int n;
         scanf("%d",&n);
         for (int i= ;i< ;i++ ){
             for (int j= ;j<n ;j++){
                 scanf("%lld",&cnt[i][j]);
             }
         }
         int k=;
         for (int i= ;i<n ;i++ ){
             for (int j= ;j<n ;j++ ){
                 a[k]=cnt[][i]+cnt[][j];
                 k++;
             }
         }
         k=;
         for (int i= ;i<n ;i++ ){
             for (int j= ;j<n ;j++ ){
                 b[k]=cnt[][i]+cnt[][j];
                 k++;
             }
         }
         int len=;
         for (int i= ;i<n ;i++){
             c[len++]=cnt[][i];
         }
         sort(a,a+k);
         sort(b,b+k);
         sort(c,c+len);
         int flag=,temp1,temp2;
         for (int i= ;i<n ;i++){
             temp1=k-;
             temp2=;
             while(temp1>= &&temp2<k) {
                 if (a[temp1]+b[temp2]+c[i]==) {
                     flag=;
                     break;
                 }else if (a[temp1]+b[temp2]+c[i]>) {
                     temp1--;
                 }else temp2++;
             }
             if (flag==) break;
         }
         if (flag==) printf("Yes\n");
         else printf("No\n");
     }
     return ;
 }