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 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.
For each case, print the case number and the number of parallelograms that can be formed.
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
SPECIAL THANKS: JANE ALAM JAN (DESCRIPTION, SOLUTION, DATASET)
题目大意：给你若干个点的坐标，问这些点随机组合之后，能得到多少个平行四边形。
解题思路：按照斜率、向量的思路的话会重复或遗漏，这里想想，平行四边形两条对角线交点固定，即两两点中点值一样，那么就好办了，如果n对点中点值一样的话，那么从中任选两对，c（n，2），就能求出当前组合的平行四边形个数了。
代码如下：
#include <cstdio> #include <algorithm> using namespace std; struct node { int x,y; }dian[1010]; node mid[500010]; bool cmp(struct node a,struct node b)//排序 { if(a.x==b.x) { return a.y<b.y; } else { return a.x<b.x; } } int main() { int t; scanf("%d",&t); int k=1; while(t--) { int n; scanf("%d",&n); for(int i=0;i<n;i++) { scanf("%d%d",&dian[i].x,&dian[i].y); } int num=0; for(int i=0;i<n;i++)//两两点组合成对 { for(int j=i+1;j<n;j++) { mid[num].x=dian[i].x+dian[j].x; mid[num].y=dian[i].y+dian[j].y; num++; } } sort(mid,mid+num,cmp);//排下序，以为要计算n对中点相同的点的对数 int geshu=1; int ans=0; int flag=0;//作为区间的起点 for(int i=1;i<num;i++) { if(mid[i].x==mid[flag].x&&mid[i].y==mid[flag].y)//统计n为多少 { geshu++; } else { ans=ans+geshu*(geshu-1)/2; geshu=1;//再次初始化 flag=i;//更新flag } } printf("Case %d: %d\n",k++,ans); } return 0; }