4 Values whose Sum is 0 ｐｏｊ２７８５

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).

这道题刚拿到手的时候完全没思路，但是暴力绝逼超时，先是想着用ｓｅｔ后来发现不可行，二分尽然过了，，，，，

#include <iostream>
using namespace std;
#include<string.h>
#include<set>
#include<stdio.h>
#include<math.h>
#include<queue>
#include<map>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<cstring>
#include <cstdio>
#include <cstdlib>
#include<cstring>
int a[4010],b[4010],c[4010],d[4010];
int a1[16000000],b1[16000000];
set<int>TM;
int main()
{
    int t;
    cin>>t;
    for(int i=0;i<t;i++)
        cin>>a[i]>>b[i]>>c[i]>>d[i];
        int add=0;
        for(int i=0;i<t;i++)
            for(int j=0;j<t;j++)
        {
            b1[add]=(a[i]+b[j]);
            a1[add++]=c[i]+d[j];
        }
        sort(b1,b1+add);
        int sum=0;
        for(int i=0;i<add;i++)
        {
            int tou=0,wei=add,zhongjian;
            while(tou<=wei)
            {
                zhongjian=(tou+wei)/2;
                if(a1[i]+b1[zhongjian]==0)
                   {
                       sum++;
                for(int j=zhongjian-1;j>=0;j--)
                    {
                    if(b1[j]!=b1[j+1])
                    break;
                     else
                    sum++;
                    }
                for(int j=zhongjian+1;j<add;j++)
                   {
                    if(b1[j]!=b1[j-1])
                    break;
                     else
                    sum++;
                   }
                   break;
                   }
                   else if(a1[i]+b1[zhongjian]<0)
                       tou=zhongjian+1;
                   else
                    wei=zhongjian-1;
            }
        }
        cout<<sum<<endl;
    return 0;
}