poj 2785 4 Values whose Sum is 0

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 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).

这道题我是使用了二分查找的方式；

开始抽了，忘记了多解情况，用了STL；

#include<iostream>
#include<cstdlib>
#include<string>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<stack>
#include<queue>
#include<iomanip>
using namespace std;
int sum1[16100000],sum2[16100000];
int binary(int left,int right,int k,int t)
{
    int i;
    while(left<=right)
    {
        int mid=(left+right)/2;
        int num=0;
        if(sum2[mid]==k)
        {
            num=1;
            for(i=mid-1;i>=0&&sum2[i]==k;--i) ++num;
            for(i=mid+1;i<t&&sum2[i]==k;++i) ++num;
            return num;
        }
        else if(sum2[mid]>k)
            right=mid-1;
        else left=mid+1;
    }
    return 0;
}
int main()
{
    int n,s[5000][4];
    scanf("%d",&n);
    for(int i=0;i<n;++i)
        for(int j=0;j<=3;++j)
    {
        scanf(" %d",&s[i][j]);
    }
    int t=0;
    for(int i=0;i<n;++i)
        for(int j=0;j<n;++j)
    {
        sum1[t]=s[i][0]+s[j][1];
        sum2[t++]=-(s[i][2]+s[j][3]);
    }
    sort(sum2,sum2+t);
    int ans=0;
    for(int i=0;i<t;++i)
    {
        ans+=binary(0,t,sum1[i],t);
    }
    printf("%d\n",ans);
    return 0;
}