4 Values whose Sum is 0（枚举+二分）

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
",ans);
	}
	return 0;
}