POJ-2785

 

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).

 

 

题意：就是给你4组数，然后让你在每组数里面找一个数，使其4个数和为零。问你有多少组。

题解：可以没两组数相加，和成两个数组，对其中一个数组排序。遍历两个数组（二分法，否则会TE），找相加为零的有多少组。

AC代码为：

 

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


int sum1[16000005], sum2[16000005];


int main()
{
int n, mid;

scanf("%d", &n);

int a[n][4];
for (int i = 0; i<n; i++)
{
scanf("%d%d%d%d", &a[i][0], &a[i][1], &a[i][2], &a[i][3]);
}

int k = 0;
int m = 0;
for (int i = 0; i<n; i++)
for (int j = 0; j<n; j++)
{
sum1[k++] = a[i][0] + a[j][1];
sum2[m++] = a[i][2] + a[j][3];
}

sort(sum2, sum2 + m);
int cnt = 0;

for (int i = 0; i<k; i++)
{
int left = 0;
int right = k - 1;

while (left <= right)
{
mid = (left + right) / 2;
if (sum1[i] + sum2[mid] == 0)
{
cnt++;
for (int j = mid + 1; j<k; j++)
{
if (sum1[i] + sum2[j] != 0)
break;
else
cnt++;
}
for (int j = mid - 1; j >= 0; j--)
{
if (sum1[i] + sum2[j] != 0)
break;
else
cnt++;
}
break;
}

if (sum1[i] + sum2[mid]<0)
left = mid + 1;
else
right = mid - 1;
}
}
cout<<cnt<<endl;

return 0;
}