Squares

/*Squares
A square is a 4-sided polygon whose sides have equal length and adjacent sides form 90-degree angles. It is also a polygon such that
rotating about its centre by 90 degrees gives the same polygon. It is not the only polygon with the latter property, however, as a
regular octagon also has this property.
So we all know what a square looks like, but can we find all possible squares that can be formed from a set of stars in a night sky?
To make the problem easier, we will assume that the night sky is a 2-dimensional plane, and each star is specified by its x and y
coordinates.
Input
The input consists of a number of test cases. Each test case starts with the integer n (1 <= n <= 1000) indicating the number of points
to follow. Each of the next n lines specify the x and y coordinates (two integers) of each point. You may assume that the points are
distinct and the magnitudes of the coordinates are less than 20000. The input is terminated when n = 0.
Output
For each test case, print on a line the number of squares one can form from the given stars.
Sample Input
4
1 0
0 1
1 1
0 0
9
0 0
1 0
2 0
0 2
1 2
2 2
0 1
1 1
2 1
4
-2 5
3 7
0 0
5 2
0
Sample Output
1
6
1*/
//大致题意：有一堆平面散点集，任取四个点，求能组成正方形的不同组合方式有多少。
//相同的四个点，不同顺序构成的正方形视为同一正方形。

#include<iostream>
#include<stdio.h>
#include<queue>
#include<stack>
#include<string.h>
#include<algorithm>
#include<math.h>
#define INF 0x3f3f3f3f;
using namespace std;
const int N=997;
struct point
{
    int x,y;
}points[N+10];
struct node
{
    int x,y;
    int next;
}T[N+10];
int head[N+10];
int top;
void build(point p)
{
    int sum=(p.x*p.x+p.y*p.y)%N;
    T[top].x=p.x;
    T[top].y=p.y;
    T[top].next=head[sum];
    head[sum]=top++;
}
bool Query(point p)
{
    int sum=(p.x*p.x+p.y*p.y)%N;
    int q=head[sum];
    while(q!=-1)
    {
        if(T[q].x==p.x&&T[q].y==p.y)
            return 1;
        q=T[q].next;
    }
    return 0;
}
int main()
{
    int n,x,y;
    while(scanf("%d",&n),n)
    {
        top=0;
        memset(head,-1,sizeof(head));
        for(int i=0;i<n;i++)
        {
            scanf("%d%d",&x,&y);
            points[i].x=x<<1;
            points[i].y=y<<1;
            build(points[i]);
        }
        int g,h,u,v;
        point a,b;
        int ans=0;
        for(int i=0;i<n;i++)
        {
            for(int j=i+1;j<n;j++)
            {
                g=(points[i].x+points[j].x);
                h=(points[i].y+points[j].y);
                u=(points[j].x-points[i].x);
                v=(points[i].y-points[j].y);
                a.x=(g-v)/2;
                a.y=(h-u)/2;
                b.x=(v+g)/2;
                b.y=(h+u)/2;
                if(Query(a)&&Query(b))
                    ans++;
            }
        }
        printf("%d\n",ans/2);
    }
    return 0;
}