hdu 5533 Dancing Stars on Me（数学，水）

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 integer n, 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

3
3
0 0
1 1
1 0
4
0 0
0 1
1 0
1 1
5
0 0
0 1
0 2
2 2
2 0

 

 

 

Sample Output

NO 
YES 
NO

 

 

 

Source

2015ACM/ICPC亚洲区长春站-重现赛（感谢东北师大）

 

题意：给你几个点，判断这几个点能否组成正n边形

先算出两两之间的距离存于mp[][]数组，然后找出最小的距离minnn，如果mp数组刚好有n个等于minnn，则是正多边形

 1 #pragma comment(linker, "/STACK:1024000000,1024000000")
 2 #include<iostream>
 3 #include<cstdio>
 4 #include<cstring>
 5 #include<cmath>
 6 #include<math.h>
 7 #include<algorithm>
 8 #include<queue>
 9 #include<set>
10 #include<bitset>
11 #include<map>
12 #include<vector>
13 #include<stdlib.h>
14 #include <stack>
15 using namespace std;
16 #define PI acos(-1.0)
17 #define max(a,b) (a) > (b) ? (a) : (b)
18 #define min(a,b) (a) < (b) ? (a) : (b)
19 #define ll long long
20 #define eps 1e-10
21 #define MOD 1000000007
22 #define N 106
23 #define inf 1<<26
24 int n;
25 double x[N],y[N];
26 double mp[N][N];
27 double cal(int i,int j){
28    return ((x[i]-x[j])*(x[i]-x[j])+(y[i]-y[j])*(y[i]-y[j]));
29 }
30 int main()
31 {
32    int t;
33    scanf("%d",&t);
34    while(t--){
35       scanf("%d",&n);
36       for(int i=0;i<n;i++){
37          scanf("%lf%lf",&x[i],&y[i]);
38       }
39       double minnn=inf;
40       for(int i=0;i<n;i++){
41          for(int j=i+1;j<n;j++){
42             mp[i][j]=cal(i,j);
43             if(minnn>mp[i][j]){
44                minnn=mp[i][j];
45             }
46          }
47       }
48       int ans=0;
49       for(int i=0;i<n;i++){
50          for(int j=i+1;j<n;j++){
51             if(mp[i][j]==minnn){
52                ans++;
53             }
54          }
55       }
56       if(ans==n){
57          printf("YES\n");
58       }else{
59          printf("NO\n");
60       }
61    }
62     return 0;
63 }

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