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
The first line of the input contains a single integer t (1 <= t <= 10), the number of test cases. It is followed by the input data for each test case. 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
Output should contain t lines. Line i contains an integer showing the number of the parallelograms as described above for test case i.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 8Sample Output
5 6用一个新的点记录两点的x与y的坐标和，即相当于对角线的坐标，平行四边形ABCD两条对角线的两顶点和相等即AC = BD的坐标和，暴力统计所有坐标和，看有多少一样的，n条线两个为一组，一共多少组，标准的组合数C(n， 2)
#include<cstdio> #include<cstring> #include<algorithm> #define ll long long #define maxn 1010 #define inf 0x3f3f3f3f #define mes(a, b) memset(a, b, sizeof(a) using namespace std; int WW[maxn], LL[maxn]; struct node{ int L, W; }num[maxn*maxn]; bool cmp(node a, node b){ if(a.L == b.L){ return a.W < b.W; } return a.L < b.L; } int main(){ int t, n; scanf("%d", &t); while(t--){ scanf("%d", &n); for(int i = 1; i <= n; i++){ scanf("%d%d", &LL[i], &WW[i]); } int c, k, ans; c = ans = 0; for(int i = 1; i <= n-1; i++){ for(int j = i+1; j <= n; j++){ num[c].L = LL[i] + LL[j]; num[c++].W = WW[i] + WW[j]; } } sort(num, num+c, cmp); k = 1; for(int i = 1; i < c; i++){ if(num[i].L == num[i-1].L && num[i].W == num[i-1].W){ k++; } else{ ans += k*(k-1)/2; k = 1; } } printf("%d\n", ans); } return 0; }
