Dancing Stars on Me HDU - 5533

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本文介绍了一种编程方法，用于通过智能手机相机自动检测夜空中星星是否能组成正多边形，以记录这些天文现象。程序简化了星空投影到二维的问题，仅需判断星星在平面上能否构成等边角和等长边的多边形。

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 TT indicating the total number of test cases. Each test case begins with an integer nn, denoting the number of stars in the sky. Following nn lines, each contains 22 integers xi,yixi,yi, describe the coordinates of nn stars.

1≤T≤3001≤T≤300
3≤n≤1003≤n≤100
−10000≤xi,yi≤10000−10000≤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

Sample Output

NO
YES
NO

#include<iostream>
#include<cmath>
#include<cstring>
using namespace std;
struct Point{
    int x,y;
}point[10010];
double d(Point a,Point b){
    return pow((a.x-b.x),2)+pow((a.y-b.y),2);
}
int main(){
    int t;
    cin>>t;
    while(t--){
        int n;
        cin>>n;
        for(int i=0;i<n;i++) cin>>point[i].x>>point[i].y;
        if(n!=4) cout<<"NO"<<endl;
        else{
            if(d(point[0],point[1])==d(point[2],point[3])&&
               d(point[0],point[2])==d(point[1],point[3])&&
               d(point[0],point[3])==d(point[1],point[2]))
                cout<<"YES"<<endl;
            else cout<<"NO"<<endl;
        }
    }
}