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
题意：给你n个点的坐标，求能构成的平行四边形的个数。平行四边形的对角线的中点肯定是重合的。所以把对角线中点求出来。排序，便利查找相等的个数，再排列组合。求出结果。
#include<stdio.h> #include<string.h> #include<algorithm> using namespace std; struct node { int x; int y; }stu[1100],mid[1100000]; int cmp(node a,node b) { if(a.x!=b.x) return a.x<b.x; else return a.y<b.y; } int main() { int t,n,i,j,cas=0; int cnt,ans,k; scanf("%d",&t); while(t--) { scanf("%d",&n); for(i=1;i<=n;i++) scanf("%d%d",&stu[i].x,&stu[i].y); k=0; for(i=1;i<n;i++) for(j=i+1;j<=n;j++) { mid[k].x=stu[i].x+stu[j].x; mid[k].y=stu[i].y+stu[j].y; k++; } sort(mid,mid+k,cmp); cnt=1; ans=0; for (i=0;i<k-1;i++) { if(mid[i].x==mid[i+1].x&&mid[i].y==mid[i+1].y) cnt++; else { ans=ans+(cnt*(cnt-1)/2); cnt=1; } } printf("Case %d: %d\n",++cas,ans); } return 0; }