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 = 4005;

typedef long long ll;

int N;
int a[5][MAX];
int cd[MAX * MAX];

void solve() {
        ll res = 0;

        for(int i = 1; i <= N; ++i) {
                for(int j = 1; j <= N; ++j) {
                        cd[(i - 1) * N + j] = a[3][i] + a[4][j];
                }
        }
        sort(cd + 1,cd + N * N + 1);

        for(int i = 1; i <= N; ++i) {
                for(int j = 1; j <= N; ++j) {
                        int x = -(a[1][i] + a[2][j]);
                        res += upper_bound(cd + 1,cd + N * N + 1,x) -
                                lower_bound(cd + 1,cd + N * N + 1,x);

                }
        }

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


}

int main()
{
   // freopen("sw.in","r",stdin);
    scanf("%d",&N);
    for(int i = 1; i <= N; ++i) {
            for(int j = 1; j <= 4; ++j) {
                    scanf("%d",&a[j][i]);
            }
    }

    solve();

    return 0;
}