Trouble HDU

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[6][205];
long long  a[40005],b[40005],c[40005];

int main() {
    int t;
    scanf("%d",&t);
    while(t--){
        int n;
        scanf("%d",&n);
        for (int i=0 ;i<5 ;i++ ){
            for (int j=0 ;j<n ;j++){
                scanf("%lld",&cnt[i][j]);
            }
        }
        int k=0;
        for (int i=0 ;i<n ;i++ ){
            for (int j=0 ;j<n ;j++ ){
                a[k]=cnt[1][i]+cnt[2][j];
                k++;
            }
        }
        k=0;
        for (int i=0 ;i<n ;i++ ){
            for (int j=0 ;j<n ;j++ ){
                b[k]=cnt[3][i]+cnt[4][j];
                k++;
            }
        }
        int len=0;
        for (int i=0 ;i<n ;i++){
            c[len++]=cnt[0][i];
        }
        sort(a,a+k);
        sort(b,b+k);
        sort(c,c+len);
        int flag=1,temp1,temp2;
        for (int i=0 ;i<n ;i++){
            temp1=k-1;
            temp2=0;
            while(temp1>=0 &&temp2<k) {
                if (a[temp1]+b[temp2]+c[i]==0) {
                    flag=0;
                    break;
                }else if (a[temp1]+b[temp2]+c[i]>0) {
                    temp1--;
                }else temp2++;
            }
            if (flag==0) break;
        }
        if (flag==0) printf("Yes
");
        else printf("No
");
    }
    return 0;
}