POJ 2785：4 Values whose Sum is 0 二分

4 Values whose Sum is 0

Time Limit: 15000MS		Memory Limit: 228000K
Total Submissions: 18221		Accepted: 5363
Case Time Limit: 5000MS

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

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).

题意是给出4列数，从每一列中挑选一个数字，问有多少种方法使这四个数的和为0。

求出每两列的数的和，再二分查找。

代码：

#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <string>
#include <cstring>
#include <map>
#pragma warning(disable:4996)
using namespace std;

int a[4005],b[4005],c[4005],d[4005];
int sum1[4005*4005],sum2[4005*4005];
int n;

int main()
{	
	//freopen("i.txt","r",stdin);
	//freopen("o.txt","w",stdout);

	int i,j;
	
	scanf("%d",&n);

	for(i=1;i<=n;i++)
	{
		scanf("%d%d%d%d",a+i,b+i,c+i,d+i);
	}

	int num1=0;
	for(i=1;i<=n;i++)
	{
		for(j=1;j<=n;j++)
		{
			sum1[++num1] = -(a[i]+b[j]);
			sum2[num1] = (c[i]+d[j]);
		}
	}
	sort(sum1+1,sum1+1+num1);
	sort(sum2+1,sum2+1+num1);

	int ans=0;
	sum1[0]= -268435456*2 - 1;
	sum1[num1+1] = 268435456*2 + 1;
	for(i=1;i<=num1;i++)
	{
		int left = 0;
		int right = num1+1;
		int mid;
		while(left<right)
		{
			mid=(left+right)/2;
			if(sum2[i]<=sum1[mid])
			{
				right=mid;
			}
			else
			{
				left= mid+1;
			}
		}
		while(sum2[i]==sum1[right]&&right<=num1)
		{
			ans++;
			right++;
		}
	}

	printf("%d
",ans);
	//system("pause");
	return 0;
}