K - 4 Values whose Sum is 0(中途相遇法)

The SUM problem can be formulated as follows: given four lists A, B, C, D of integer values, compute how many quadruplet (a, b, c, d ) AxBxCxD are such that a + b + c + d = 0 . In the following, we assume that all lists have the same size n .

Input 

The input begins with a single positive integer on a line by itself indicating the number of the cases following, each of them as described below. This line is followed by a blank line, and there is also a blank line between two consecutive inputs.

The first line of the input file contains the size of the lists n (this value can be as large as 4000). We then have n lines containing four integer values (with absolute value as large as 2²⁸ ) that belong respectively to A, B, C and D .

Output 

For each test case, the output must follow the description below. The outputs of two consecutive cases will be separated by a blank line.

For each input file, your program has to write the number quadruplets whose sum is zero.

Sample Input 

1

6
-45 22 42 -16
-41 -27 56 30
-36 53 -37 77
-36 30 -75 -46
26 -38 -10 62
-32 -54 -6 45

Sample Output 

5

Sample Explanation: Indeed, the sum of the five following quadruplets is zero: (-45, -27, 42, 30), (26, 30, -10, -46), (-32, 22, 56, -46),(-32, 30, -75, 77), (-32, -54, 56, 30).

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cmath>
 5 #include <algorithm>
 6 #include <string>
 7 #include <vector>
 8 #include <stack>
 9 #include <queue>
10 #include <set>
11 #include <map>
12 #include <list>
13 #include <iomanip>
14 #include <cstdlib>
15 #include <sstream>
16 using namespace std;
17 typedef long long LL;
18 const int INF=0x5fffffff;
19 const double EXP=1e-8;
20 const int MS=4005;
21 int A[MS],B[MS],C[MS],D[MS],n,sum[MS*MS];
22 
23 int main()
24 {
25     int T;
26     scanf("%d",&T);
27     while(T--)
28     {
29         scanf("%d",&n);
30         for(int i=0;i<n;i++)
31             scanf("%d%d%d%d",&A[i],&B[i],&C[i],&D[i]);
32         int cnt=0;
33         for(int i=0;i<n;i++)
34             for(int j=0;j<n;j++)
35                 sum[cnt++]=A[i]+B[j];
36         sort(sum,sum+cnt);
37         LL ans=0;
38         for(int i=0;i<n;i++)
39             for(int j=0;j<n;j++)
40         {
41             ans+=upper_bound(sum,sum+cnt,-C[i]-D[j])-lower_bound(sum,sum+cnt,-C[i]-D[j]);
42         }
43         printf("%lld\n",ans);
44         if(T)
45             printf("\n");
46     }
47     return 0;
48 }