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
1 #include<cstdio>
2 #include<cstring>
3 #include<algorithm>
4 using namespace std;
5 long long cnt[6][205];
6 long long a[40005],b[40005],c[40005];
7
8 int main() {
9 int t;
10 scanf("%d",&t);
11 while(t--){
12 int n;
13 scanf("%d",&n);
14 for (int i=0 ;i<5 ;i++ ){
15 for (int j=0 ;j<n ;j++){
16 scanf("%lld",&cnt[i][j]);
17 }
18 }
19 int k=0;
20 for (int i=0 ;i<n ;i++ ){
21 for (int j=0 ;j<n ;j++ ){
22 a[k]=cnt[1][i]+cnt[2][j];
23 k++;
24 }
25 }
26 k=0;
27 for (int i=0 ;i<n ;i++ ){
28 for (int j=0 ;j<n ;j++ ){
29 b[k]=cnt[3][i]+cnt[4][j];
30 k++;
31 }
32 }
33 int len=0;
34 for (int i=0 ;i<n ;i++){
35 c[len++]=cnt[0][i];
36 }
37 sort(a,a+k);
38 sort(b,b+k);
39 sort(c,c+len);
40 int flag=1,temp1,temp2;
41 for (int i=0 ;i<n ;i++){
42 temp1=k-1;
43 temp2=0;
44 while(temp1>=0 &&temp2<k) {
45 if (a[temp1]+b[temp2]+c[i]==0) {
46 flag=0;
47 break;
48 }else if (a[temp1]+b[temp2]+c[i]>0) {
49 temp1--;
50 }else temp2++;
51 }
52 if (flag==0) break;
53 }
54 if (flag==0) printf("Yes\n");
55 else printf("No\n");
56 }
57 return 0;
58 }
原文地址：https://www.cnblogs.com/qldabiaoge/p/8511855.html