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
#include<iostream>
#include<cstdio>
#include<algorithm>
#define maxn 1200
#define maxm 1000005
#define ll long long
using namespace std;
int t;
int n;
int cnt;
struct node
{
ll x,y;
}node1[maxn],node2[maxm];
bool cmp(node a,node b)
{
if(a.x==b.x)
return a.y<b.y;
return a.x<b.x;
}
int main()
{
scanf("%d",&t);
int w=0;
while(t--)
{w++;
cnt=0;
scanf("%d",&n);
for(int i=1;i<=n;i++)
{scanf("%lld%lld",&node1[i].x,&node1[i].y);
}
for(int i=1;i<=n;i++)
for(int j=i+1;j<=n;j++)
{
node2[++cnt].x=node1[i].x+node1[j].x;
node2[cnt].y=node1[i].y+node1[j].y;
}
sort(node2+1,node2+cnt+1,cmp);
ll ans=1,sum=0;
for(int i=1;i<cnt;i++)
{if(node2[i].x==node2[i+1].x&&node2[i].y==node2[i+1].y)
++ans;
else
{sum+=(ans-1)*ans/2;
ans=1;
}
}
printf("Case %d: %lld\n",w,sum);
}
return 0;
}