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 than1000000000.
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
给n点看有多少个平行四边形 定理：中点相同构成一个平行四边形 代码： #include<cstdio> #include<algorithm> using namespace std; struct edge { int x,y; }a[1010],mid[1000010]; bool cmp(edge a,edge b) { if(a.x==b.x) return a.y<b.y; return a.x<b.x; } int main() { int t,n,num,sum,k=1; scanf("%d",&t); while(t--) { scanf("%d",&n); for(int i=0;i<n;i++) { scanf("%d%d",&a[i].x,&a[i].y); } int cnt=0; for(int i=0;i<n;i++) { for(int j=i+1;j<n;j++) { mid[cnt].x=a[i].x+a[j].x; mid[cnt].y=a[i].y+a[j].y; cnt++; } } sort(mid,mid+cnt,cmp); num=1; sum=0; for(int i=1;i<cnt;i++) { if(mid[i].x==mid[i-1].x&&mid[i].y==mid[i-1].y) num++; else { sum+=num*(num-1)/2; num=1; } } printf("Case %d: %d\n",k++,sum); } return 0; }
