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

 

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SPECIAL THANKS: JANE ALAM JAN (DESCRIPTION, SOLUTION, DATASET) 平行四边形的性质，对角线互相平分，，求任意两个点的中点。找中点一样的线段，任意取两条就是平行四边形了。 #include<stdio.h> #include<string.h> #include<algorithm> using namespace std; struct node { int x,y; }t[1100],t1[550000]; bool cmp(node a,node b) { if(a.x!=b.x) return a.x<b.x; return a.y<b.y; } int main() { int T,n,i,j; scanf("%d",&T); int cnt=1; while(T--) { scanf("%d",&n); for(i=0;i<n;i++) scanf("%d%d",&t[i].x,&t[i].y); int k=0; for(i=0;i<n-1;i++) for(j=i+1;j<n;j++) { t1[k].x=t[i].x+t[j].x; t1[k].y=t[i].y+t[j].y; k++; } sort(t1,t1+k,cmp); int l=1; int sum=0; for(i=1;i<k;i++) { if(t1[i].x==t1[i-1].x&&t1[i].y==t1[i-1].y) { l++; } else { sum+=l*(l-1)/2; l=1; } } printf("Case %d: %d\n",cnt++,sum); } return 0; }