D. Number of Parallelograms

D. Number of Parallelograms

原题链接

time limit per test

4 seconds

memory limit per test

256 megabytes

You are given n points on a plane. All the points are distinct and no three of them lie on the same line. Find the number of parallelograms with the vertices at the given points.

Input

The first line of the input contains integer n (1 ≤ n ≤ 2000) — the number of points.

Each of the next n lines contains two integers (xi, yi) (0 ≤ xi, yi ≤ 109) — the coordinates of the i-th point.

Output

Print the only integer c — the number of parallelograms with the vertices at the given points.

Example

input

4

0 1

1 0

1 1

2 0

output

1

题目大意：给出一堆点，求出这些点中能组成平行四边形的个数，且题目要求没有三个点在同一直线上。

思路分析：判断平行四边形的方法有两种，一种是两边平行且相等，另一种就是中点相等。如果对角线上的两点的中点坐标在同一点，即满足（x1+x2）/2==（x3+x4）/2&&（y1+y2)/2==(y3+y4)/2条件，则说明这四个点能够构成一个平行四边形。首先求出任意两点之间的中点，再使用map，即可快捷求出每个中点所对应的平行四边形个数。

AC代码：

#include <cstdio>
#include <map>
#include <iostream>
using namespace std;
const int maxn=2000+5;
pair<long long ,long long>b[maxn];
map<pair<int,int>,int>sum;
int main(void)
{
    int n;
    cin>>n;
    for(int i=0;i<n;i++)
    scanf("%d%d",&b[i].first,&b[i].second);
    int count=0;
    for(int i=0;i<n-1;i++)
        for(int j=i+1;j<n;j++)
        {
            int x=b[i].first+b[j].first;
            int y=b[i].second+b[j].second;
            count+=sum[make_pair(x,y)]++;
        }
    cout<<count<<endl;
}