4 Values whose Sum is 0_upper_bound&&ower_bound

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).

【题意】给出一个n*4的矩阵，每列上选一个数使得最后加起来为0，问有多少种取法

【思路】先用ab数组存a+b的所有组合，同理，存储cd数组，然后对cd数组进行排序，然后用upper_bound，lower_bound查找是否存在-ab[i],正好两者只差为1，即多了一种组合方式

#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<string.h>
using namespace std;
const int N=4000+10;
int n;
int a[N],b[N],c[N],d[N];
int ab[N*N],cd[N*N];
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        scanf("%d%d%d%d",&a[i],&b[i],&c[i],&d[i]);
    }
    int k=0;
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=n;j++)
        {
            ab[k]=a[i]+b[j];
            cd[k]=c[i]+d[j];
            k++;
        }
    }
    sort(cd,cd+k);
    long long ans=0;
    for(int i=0;i<k;i++)
    {
        int tmp=-ab[i];
        ans+=(long long )(upper_bound(cd,cd+k,tmp)-lower_bound(cd,cd+k,tmp));
    }
    printf("%I64d
",ans);
    return 0;
}