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 ) AxBxCxD are such that a + b + c + d = 0 . In the following, we assume that all lists have the same size n .

Input

The input begins with a single positive integer on a line by itself indicating the number of the cases following, each of them as described below. This line is followed by a blank line, and there is also a blank line between two consecutive inputs.

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 test case, the output must follow the description below. The outputs of two consecutive cases will be separated by a blank line.

For each input file, your program has to write the number quadruplets whose sum is zero.

Sample Input

1

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

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). 

#  include <algorithm>中：

upper_bound（a,  a+n, x）;在左闭右开区间a到a+n地址之间；找到大于x的最小值的位置：

lower_bound（a, a+n,x) ;在左闭右开区间a到a+n地址之间；找到大于或等于的值的位置;

//方法一：二分法求解;

# include <cstdio>
# include <string>
# include <cstring>
# include <iostream>
# include <algorithm>
using namespace std;

int a[4005],b[4005],c[4005],d[4005];
int sum[16000025],sum2[16000025];

int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        int n,i,j,len=0,ans=0;;
		cin>>n;
		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[len++]=a[i]+b[j];
		len=0;
		for(i=0;i<n;i++)
			for(j=0;j<n;j++)
			sum2[len++]=c[i]+d[j];
			sort(sum2,sum2+len);
		for(i=0;i<len;i++)
		{
			int l=0,r=len-1,mid;
			while(l<r)
			{
				mid=(r+l)>>1;
				if(sum[i]<-sum2[mid])    l=mid+1;
				else    r=mid;
			}
			while(sum2[l]==-sum[i]&&l<len)
			{
				ans++;
				l++;
			}
		}
		printf("%d\n",ans);
		if(t)    printf("\n");
    }
    return 0;
}

//方法二：用函数upper_boud（），lower_boud（）代替二分：
# include <iostream>
# include <cstdio>
# include <algorithm>
using namespace std;

int a[4005],b[4005],c[4005],d[4005];
int sum[16000005];

int main()
{
	int t;
    cin>>t;
    while(t--)
    {
		int n,i,j,cnt=0,ans=0;
		cin>>n;
		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[cnt++]=a[i]+b[j];
		sort(sum,sum+cnt);
		for(i=0;i<n;i++)
			for(j=0;j<n;j++)
			ans+=upper_bound(sum,sum+cnt,-c[i]-d[j])-lower_bound(sum,sum+cnt,-c[i]-d[j]);
		printf("%d\n",ans);
		if(t!=0)    printf("\n");
    }
    return 0;
}