4 Values whose Sum is 0(二分)

　　　　　　　　　　　　　　4 Values whose Sum is 0

Time Limit: 15000MS		Memory Limit: 228000K
Total Submissions: 21370		Accepted: 6428
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

 

// 题意: 比如例子 6 行，每行 4 个数，从第一，二，三，四列各选一个数，要使和为 0 ，有几种组合?

n 最大有4000，但还是可以暴力，加二分

a + b = -( c + d )

枚举 a + b ,对于 c + d ,枚举一次，用一个大数组保存和，sort ，就可以二分了。要注意的是，二分到了要对左右继续搜索一下，不同的位置代表有不同的组合 

 

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>

using namespace std;

int n;
int k;

int num[4002][4];
int s[16000005];

int erfen(int t)
{
    int l=0,r=k-1;
    while (l<=r)
    {
        int mid=(l+r)/2;
        if (s[mid]==t)
        {
            int e = mid-1;
            int  all=1;
            while (e>=0&&s[e]==t)
                e--,all++;
            e=mid+1;
            while (e<k&&s[e]==t)
                e++,all++;
            return all;
        }
        else if (s[mid]>t)
            r=mid-1;
        else
            l=mid+1;
    }
    return 0;
}

int main()
{
    while (scanf("%d",&n)!=EOF)
    {
        for (int i=0;i<n;i++)
            scanf("%d%d%d%d",&num[i][0],&num[i][1],&num[i][2],&num[i][3]);
        k=0;
        for (int i=0;i<n;i++)
            for (int j=0;j<n;j++)
            s[k++]=-(num[i][2]+num[j][3]);
        sort(s,s+k);
        int total=0;
        for (int i=0;i<n;i++)
        {
            for (int j=0;j<n;j++)
            {
                int left=num[i][0]+num[j][1];
                total+=erfen(left);
            }
        }
        printf("%d
",total);
    }
    return 0;
}