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 2space-separated integers x and y (the coordinates of a point) with magnitude (absolute value) of no more than1000000000.
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
根据平行四边形性质，两对角线中点一定重合，所以先取任意两点间的中点，然后在对这些中点排序sort一下，然后用组合数公式C（n，2）求出任意相同的点中任意两点组合的个数求和即可。 #include<cstdio> #include<iostream> #include<algorithm> #include<cstring> using namespace std; const int N = 1005; struct node { int x; int y; } s[N],mid[N*N]; bool cmp(node A,node B) { if(A.x==B.x) return A.y<B.y; return A.x<B.x; } int main() { int T,p=0; scanf("%d",&T); while(T--) { int n; scanf("%d",&n); for(int i=0; i<n; i++) { scanf("%d%d",&s[i].x,&s[i].y); } int cnt=0; for(int i=0; i<n; i++) { for(int j=i+1; j<n; j++) { mid[cnt].x=s[i].x+s[j].x;//不能除二，可能为奇数导致误差 mid[cnt].y=s[i].y+s[j].y; cnt++; } } sort(mid,mid+cnt,cmp); int count=1,sum=0; for(int i=0; i<cnt;i++) { if(mid[i].x==mid[i+1].x&&mid[i].y==mid[i+1].y) { count++; } else { sum=sum+count*(count-1)/2; count=1; } } printf("Case %d: %d\n",++p,sum); } return 0; }
