HDU 5533 Dancing Stars on Me（数学+水题）

Dancing Stars on Me

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)

Total Submission(s): 1302    Accepted Submission(s): 730

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

 

题意：给定n个点坐标(x,y)，问这几个点是否能恰好组成正n边形。

思路：感觉董小姐的思路真是碉堡了。直接算均值（x,y）则为正n边形的圆心，然后求其它点到该圆心距离是否相等，主要注意精度问题。

#include <iostream>#include <algorithm>#include <string.h>#include <stdio.h>#include <math.h>using namespace std;int main(){        int t;    scanf("%d",&t);    while(t--){        int n;        scanf("%d",&n);        int x[105];        int y[105];        double rx=0;        double ry=0;        for(int i=0;i<n;i++){            scanf("%d %d",&x[i],&y[i]);            rx+=x[i];            ry+=y[i];        }        rx=rx/(1.0*n);        ry=ry/(1.0*n);        int flag=0;        double dis=(x[0]*1.0-rx)*(x[0]*1.0-rx)+(y[0]*1.0-ry)*(y[0]*1.0-ry);        //cout<<"dis"<<dis<<endl;        for(int i=1;i<n;i++){            double dis1=(x[i]*1.0-rx)*(x[i]*1.0-rx)+(y[i]*1.0-ry)*(y[i]*1.0-ry);        //    cout<<"dis1"<<dis1<<endl;            if(fabs(dis-dis1)>1e-5){                flag=1;                break;            }        }        if(flag)            cout<<"NO"<<endl;        else            cout<<"YES"<<endl;     }     return 0;}