POJ 2785 4 Values whose Sum is 0（折半枚举）

4 Values whose Sum is 0

Time Limit: 15000MS		Memory Limit: 228000K
Total Submissions: 20123		Accepted: 6017
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).

折半枚举的思想+二分的查找的优点

#include <iostream>

#include <cstdio>

#include <cstring>

#include <algorithm>

#include <cmath>

using namespace std;

const int N = 4100;

int a
, b
, c
, d
, p[N*N];

int main()

{

    int n;

    while(scanf("%d",&n)!=EOF)

    {

        for(int i=0;i<n;i++)

        {

            scanf("%d %d %d %d", &a[i], &b[i], &c[i], &d[i]);

        }

        int cnt=0;

        for(int i=0;i<n;i++)

        {

            for(int j=0;j<n;j++)

            {

                p[cnt++]=a[i]+b[j];

            }

        }

        sort(p,p+cnt);

        int res=0;

        for(int i=0;i<n;i++)

        {

            for(int j=0;j<n;j++)

            {

                int tmp=c[i]+d[j];

                res+=(upper_bound(p,p+cnt,-tmp)-lower_bound(p,p+cnt,-tmp));

            }

        }

        printf("%d\n",res);

    }

    return 0;

}