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).

思路：和之前更新的校赛的题目差不多，都是枚举+二分查找

代码：

#include<cstdio>

#include<cstring>

#include<algorithm>

#include<iostream>

using namespace std;

#define maxn 4040

int a[maxn],b[maxn],c[maxn],d[maxn],sum[maxn*maxn];

int main()

{

	int n,i,j;

	while(scanf("%d",&n)!=EOF)

	{

		for(i=0;i<n;++i)

		scanf("%d%d%d%d",&a[i],&b[i],&c[i],&d[i]);

		for(i=0;i<n;++i)

		{

			for(j=0;j<n;++j)

				sum[n*i+j]=c[i]+d[j];

		}

		sort(sum,sum+n*n);

		int ans=0;

		for(i=0;i<n;++i)

		{

			for(j=0;j<n;++j)

			{

				int k=-(a[i]+b[j]);

				ans+=upper_bound(sum,sum+n*n,k)-lower_bound(sum,sum+n*n,k);

			}

		}

		printf("%d\n",ans);

	}

	return 0;

}