VJ组队赛->Dancing Stars on Me（5533）

Problem Description

The sky was brushed clean by the wind and the stars were cold in a black sky. What a wonderful night. You observed that, sometimes the stars can form a regular polygon in the sky if we connect them properly. You want to record these moments by your smart camera. Of course, you cannot stay awake all night for capturing. So you decide to write a program running on the smart camera to check whether the stars can form a regular polygon and capture these moments automatically.
Formally, a regular polygon is a convex polygon whose angles are all equal and all its sides have the same length. The area of a regular polygon must be nonzero. We say the stars can form a regular polygon if they are exactly the vertices of some regular polygon. To simplify the problem, we project the sky to a two-dimensional plane here, and you just need to check whether the stars can form a regular polygon in this plane.

Input

The first line contains a integer T indicating the total number of test cases. Each test case begins with an integern, denoting the number of stars in the sky. Following n lines, each contains 2 integers xi,yi, describe the coordinates of n stars.

1≤T≤300
3≤n≤100
−10000≤xi,yi≤10000
All coordinates are distinct.

Output

For each test case, please output "`YES`" if the stars can form a regular polygon. Otherwise, output "`NO`" (both without quotes).

Sample Input

330 01 11 040 00 11 01 150 00 10 22 22 0

Sample Output

NOYESNO

题目大意：给你n个点，看看这几个点能不能构成一个正n边形
题意很明确，但当时没想到怎么做啊QAQ，其实思路很简单的QAQ，找出最小的一条边len就是边长，然后判断长度等于len的条数是不是n个，即为正n边形

#include <iostream>#include <algorithm>#include <cstdio>#include <cmath>using namespace std;const int N = 101;double x[N],y[N];double distance(int i,int j){return sqrt((x[i]-x[j])*(x[i]-x[j]) + (y[i]-y[j])*(y[i]-y[j]));}int main(){int t,n;cin >> t;while(t--){cin >> n;int i,j;for(i = 0; i < n; i++)scanf("%lf%lf",&x[i],&y[i]);double len = 1e9;int num = 0;for(i = 1; i < n; i++)        len = min(len,distance(0,i));for(i = 0; i < n-1; i++){for(j = i+1; j < n; j++)if(distance(i,j) == len)num ++;}if(num == n)cout << "YES\n";elsecout << "NO\n";}return 0;}