Description
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
给出点的坐标求出能连成平行四边形的数量。
思路：
用结构体记录每两个点的x坐标和与y坐标和（ 相当于记录这两个点为对角线的情况 ）然后循环判断多少个对角线的被交点平分（结构体中x，y的值相等）求出对角线的数量，然后C（sum，2）。
1 #include<cstdio>
2 #include<algorithm>
3 using namespace std;
4 int x[1010],y[1010];
5 struct stu
6 {
7 int x,y;
8 }st[1010*1010/2];
9 bool cmp(stu a,stu b)
10 {
11 if(a.x != b.x)
12 return a.x < b.x;
13 else
14 return a.y < b.y;
15 }
16 int main()
17 {
18 int t,ss=0;
19 scanf("%d",&t);
20 while(t--)
21 {
22 int n,i,j;
23 scanf("%d",&n);
24 for(i = 0 ; i < n ; i++)
25 {
26 scanf("%d %d",&x[i],&y[i]);
27 }
28 int k=0;
29 for(i = 0 ; i < n ; i++)
30 {
31 for(j = i+1 ; j < n ; j++)
32 {
33 st[k].x=x[i]+x[j];
34 st[k++].y=y[i]+y[j];
35 }
36 }
37 sort(st,st+k,cmp);
38 int num=0,sum=1,ans=0;
39 for(i = 1 ; i < k ; i++)
40 {
41 if(st[num].x == st[i].x && st[num].y == st[i].y)
42 sum++; //sum代表线的数量
43 else
44 {
45 ans+=(sum*(sum-1)/2); //这里是C(sum,2)
46 num=i;
47 sum=1;
48
49 }
50 }
51 printf("Case %d: %d\n",++ss,ans);
52 }
53 }