1058 - Parallelogram Counting    PDF (English)StatisticsForum Time Limit: 2 second(s)Memory Limit: 32 MB

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 than1000000000.

Output

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

Sample Input

Output for 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

Case 1: 5

Case 2: 6

很明显看了别人的想法才知道这样子，其实平行四边形你会发现，2条对角线的中点相等，我们就可 以应用这个特点，中点坐标出现一次就代表着一条对角线，只有出现2次中点坐标才会有一个平行四边形 当然出现3次中点坐标，任意的2个就可以组成一个平行四边形，我们可以运用组合

#include<cstdio> #include<algorithm> using namespace std; #define maxn 1000*1000/2 struct node { double x; double y; }arr[maxn];//存中点坐标 double a[1000+11]; double b[1000+11]; bool cmp(node c,node d)//对中点坐标进行排序 { if(c.x==d.x) return c.y<d.y; return c.x<d.x; } int main() { int t,test=1; scanf("%d",&t); while(t--) { int n,i,j; scanf("%d",&n); for(i=0;i<n;++i) scanf("%lf%lf",&a[i],&b[i]); int k=0; for(i=0;i<n;++i) { for(j=i+1;j<n;++j) { arr[k].x=(a[i]+a[j])/2.0; arr[k++].y=(b[i]+b[j])/2.0; } } sort(arr,arr+k,cmp); int sum=1,ans=0; for(i=0;i<k;) { for(j=i+1;j<k;++j) { if(arr[i].x==arr[j].x&&arr[i].y==arr[j].y) ++sum; else break; } i=j; if(sum>=2) ans+=sum*(sum-1)/2; sum=1; } printf("Case %d: %d\n",test++,ans); } return 0; }