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There are n distinct points in the plane, given by their integer coordinates. Find the number of parallelograms whose vertices lie on these points. In other words, find the number of 4-element subsets of these points that can be written as {A, B, C, D} such that AB || CD, and BC || AD. No four points are in a straight line.

Input
Input starts with an integer T (≤ 15), denoting the number of test cases.

The first line of each test case contains an integer n (1 ≤ n ≤ 1000). Each of the next n lines, contains 2 space-separated integers x and y (the coordinates of a point) with magnitude (absolute value) of no more than 1000000000.

Output
For each case, print the case number and the number of parallelograms that can be formed.

Sample Input
2

6

0 0

2 0

4 0

1 1

3 1

5 1

7

-2 -1

8 9

5 7

1 1

4 8

2 0

9 8

Sample Output
Case 1: 5

Case 2: 6

题意：给你n个点，任选四个点为一组，问能找到多少组能构成平行四边形的点，点可以重复使用。数据保证任意四点不共线，n的范围是1000.点的坐标范围是1e9；

思路：用题目给的判断两直线平行肯定是要超时的。
还有一个判定平行四边形的方法就是
如果两条线段的中点重合，那么这两条线段的端点可以组成一个平行四边形。 (初中知识……)
代码：

#include<cstdio>#include<cstring>#include<algorithm>#include<string>#include<cmath>#include<iostream>using namespace std;#define ll long longconst ll Max=1e6+9;struct nod{    double x;    double y;}a[1009],noe[Max];//a数组存题目给的点的坐标。//noe数组存任意两点构成的线段的中点的坐标bool cmp(nod a,nod b){    if(a.x<b.x)        return 1;    else if(a.x==b.x&&a.y<b.y)        return 1;    return 0;}int main(){    int t;    scanf("%d",&t);    int k=0;    while(t--)    {        k++;        int n;        scanf("%d",&n);        for(int i=1;i<=n;i++)        {            scanf("%lf%lf",&a[i].x,&a[i].y);        }        int pos=1;        for(int i=1;i<=n;i++)        {            for(int j=i+1;j<=n;j++)            {                noe[pos].x=(a[i].x+a[j].x)/2;                noe[pos++].y=(a[i].y+a[j].y)/2;            }        }        sort(noe+1,noe+pos,cmp);//排序后一遍遍历就可以找到所有的相同的中点坐标        //for(int i=1;i<pos;i++)        //    cout<<noe[i].x<<" "<<noe[i].y<<endl;        ll num=1;//记录当前相同的中点坐标的个数        ll ans=0;//记录最终结果        for(int i=2;i<pos;i++)        {            if(noe[i].x==noe[i-1].x&&noe[i].y==noe[i-1].y)            {                num++;            }            else            {                ans+=(num)*(num-1)/2;//当前有num个相同的中点坐标，那么答案就加上C(num,2)个，组合数。                num=1;//num要记得重置            }        }        if(num>1)//遍历完还要判断一下num的值            ans+=(num)*(num-1)/2;        printf("Case %d: %lld\n",k,ans);    }    return 0;}