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【折半枚举】POJ - 2785 4 Values whose Sum is 0

程序员文章站 2024-03-20 20:00:34
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4 Values whose Sum is 0 POJ - 2785 

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 2 28 ) 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).

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
const int maxn=4005;
int a[maxn],b[maxn],c[maxn],d[maxn],ab[maxn*maxn],cd[maxn*maxn];

int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        for(int i=0;i<n;i++)
            scanf("%d%d%d%d",&a[i],&b[i],&c[i],&d[i]);
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<n;j++)
            {
                cd[i*n+j]=c[i]+d[j];  //得到所有c,d的和的可能
            }
        }
        sort(cd,cd+n*n);
        int res=0;
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<n;j++)
            {
                int p=-(a[i]+b[j]);   //要求a,b的和跟cd中有相加可等0的
                res+=upper_bound(cd,cd+n*n,p)-lower_bound(cd,cd+n*n,p);
            }
        }
        printf("%d\n",res);
    }
    return 0;
}