POJ 2785：4 Values whose Sum is 0 二分

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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 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

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\n",ans);

	//system("pause");

	return 0;

}