POJ - 2785 4 Values whose Sum is 0 二分

4 Values whose Sum is 0

Time Limit: 15000MS		Memory Limit: 228000K
Total Submissions: 25615		Accepted: 7697
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

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).

Source

Southwestern Europe 2005

题目大意：给4列数，每列有n个数字，从每列中取一个数，求四个数相加为零的情况有多少种。思路：用两个数组，分别记录左边两列和右边两列的数相加的和，再将其中一个数组从小到大排列，将另一数组遍历一边，用二分的方法判断个数。

#include<stdio.h>
#include<algorithm>
using namespace std;
int a[4005][4], sum1[16000001], sum2[16000001];
int main()
{
int n, mid;

while (~scanf("%d", &n))
{
int k = 0;
for (int i = 0; i < n; i++)
{
scanf("%d %d %d %d", &a[i][0], &a[i][1], &a[i][2], &a[i][3]);
}
for (int i = 0; i < n; i++)
for (int j = 0; j < n; j++)
{
sum1[k] = a[i][0] + a[j][1];
sum2[k++] = a[i][2] + a[j][3];
}
sort(sum1, sum1 + k);

int num = 0;
for (int i = 0; i < k; i++)
{
int left = 0, right = k - 1;
while (left <= right)
{
mid = (left + right) / 2;
if (sum2[i] + sum1[mid] == 0)
{
num++;
for (int j = mid + 1; j < k; j++)
{
if (sum2[i] + sum1[j] == 0)
num++;
else
break;
}
for (int j = mid - 1; j >= 0; j--)
{
if (sum2[i] + sum1[j] == 0)
num++;
else
break;
}
break;
}
if (sum2[i] + sum1[mid] > 0)
right = mid - 1;
else
left = mid + 1;
}
}
printf("%d\n", num);
}
return 0;
}