POJ - 2785 ：4 Values whose Sum is 0 （二分）

Posted 2020-09-29 西北会法语

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).

 

输入四组数，问在四组数每组取一个数相加等于0的方法数。

将前两组和枚举排序，后两组和枚举。

二分查找排序后的数组即可，复杂度O(N^2)

 

源代码：

#include<iostream>
#include<algorithm>

using namespace std;

int a[4005],b[4005],c[4005],d[4005],e[16000005],f[16000005];

int main()
{
	int n;
	while(cin>>n)
	{
	
	int x=0,y=0,z,t;
	
	for(int i=0;i<n;i++)
		cin>>a[i]>>b[i]>>c[i]>>d[i];
	for(int i=0;i<n;i++)
		for(int j=0;j<n;j++)
		{
			e[x++]=a[i]+b[j];
			f[y++]=c[i]+d[j];
		}
	sort(e,e+x);
	long long int s=0;
	for(int i=0;i<y;i++)
	{
		s+=upper_bound(e,e+n*n,-f[i])-lower_bound(e,e+n*n,-f[i]);
	}
	cout<<s<<endl;
	}
	
	return 0;
}