POJ 2785

4 Values whose Sum is 0

Time Limit: 15000MS		Memory Limit: 228000K
Total Submissions: 14475		Accepted: 4138
Case Time Limit: 5000MS

Description

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 ) ∈ A x B x C x D are such that a + b + c + d = 0 . In the following, we assume that all lists have the same size n .

Input

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

Sample Input

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

Hint

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).

Source

Southwestern Europe 2005

 

折半枚举，然后二分。

 

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <algorithm>

 using namespace std;

 const int MAX = ;

 typedef long long ll;

 int N;
 int a[][MAX];
 int cd[MAX * MAX];

 void solve() {
         ll res = ;

         for(int i = ; i <= N; ++i) {
                 for(int j = ; j <= N; ++j) {
                         cd[(i - ) * N + j] = a[][i] + a[][j];
                 }
         }
         sort(cd + ,cd + N * N + );

         for(int i = ; i <= N; ++i) {
                 for(int j = ; j <= N; ++j) {
                         int x = -(a[][i] + a[][j]);
                         res += upper_bound(cd + ,cd + N * N + ,x) -
                                 lower_bound(cd + ,cd + N * N + ,x);

                 }
         }

         printf("%I64d\n",res);

 }

 int main()
 {
    // freopen("sw.in","r",stdin);
     scanf("%d",&N);
     for(int i = ; i <= N; ++i) {
             for(int j = ; j <= ; ++j) {
                     scanf("%d",&a[j][i]);
             }
     }

     solve();

     return ;
 }