CodeForces 660D Number of Parallelograms

D. Number of Parallelograms

time limit per test

4 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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

40 11 01 12 0

output

1

题意：给你n个点 求能组成平行四边形的个数       

就是模拟   求四个点的横坐标之差和纵坐标之差

两两相等就能构成平行四边形 

也可以用向量来求，不过要用到STL  map  pair 什么的

#include<stdio.h>#include<algorithm>struct aa{    int x, y;    int a,b;}s[2000010],cc[2001];int cmp(aa xx,aa yy){    return xx.x<yy.x||(xx.x==yy.x&&xx.y<yy.y);//排序便于求相等}int cnp(aa x,aa y){    return x.a<y.a||(x.a==y.a&&x.b<y.b);//点的坐标排序}using namespace std;int main(){    int n;    int a[2001],b[2001];        scanf("%d",&n);    for(int i=0;i<n;i++)    {        scanf("%d%d",&cc[i].a,&cc[i].b);    }    sort(cc,cc+n,cnp);    int k=0;    for(int i=0;i<n-1;i++)    {        for(int j=i+1;j<n;j++)        {            s[k].x=cc[j].a-cc[i].a;            s[k++].y=cc[j].b-cc[i].b;        }    }    sort(s,s+k,cmp);    int pos=0,m=0,sum=0;    for(int i=1;i<k;i++)    {        if(s[i].x!=s[i-1].x||s[i].y!=s[i-1].y)        {            m=i-pos;            pos=i;            sum+=(m*(m-1))/2;//两两组合的个数        }    }    printf("%d\n",sum/2);//要四条边两两互相平行相等}