Educational Codeforces Round 11 D. Number of Parallelograms 暴力

D. Number of Parallelograms

题目连接：

http://www.codeforces.com/contest/660/problem/D

Description

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.

Sample Input

4
0 1
1 0
1 1
2 0

Sample Output

1

Hint

题意

平面给你n个点，问你能够组成多少个平行四边形

保证三点不共线，点都是不相同的

题解：

能够成平行四边形的两条边的东西，一定是两个相同的向量

那么我们n^2把所有向量都计算出来就好了，注意得人为的去规定一下方向什么的，然后就完了。

代码

#include<bits/stdc++.h>
using namespace std;
const int maxn = 2005;
map<pair<int,int>,int> H;
pair<int,int>P[maxn];
int main()
{
    int n;scanf("%d",&n);
    for(int i=1;i<=n;i++)
        scanf("%d%d",&P[i].first,&P[i].second);
    long long ans = 0;
    for(int i=1;i<=n;i++)
    {
        for(int j=i+1;j<=n;j++)
        {
            pair<int,int> T;
            T.first=P[i].first-P[j].first;
            T.second=P[i].second-P[j].second;
            if(T.first<0)T.first=-T.first,T.second=-T.second;
            else if(T.second<0&&T.first==0)T.second=-T.second;
            ans+=H[T];
            H[T]++;
        }
    }
    cout<<ans/2<<endl;
}