Dancing Stars on Me

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

Total Submission(s): 1184 Accepted Submission(s): 651

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

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

Recommend

hujie | We have carefully selected several similar problems for you:

给一些点问组成的图形是不是正的

瞎暴力一下就可以了（其实遇到这种题基本也只会暴力

把所有点互相的距离都求出来

然后排下序看最小的点和第n小的点是否相等就可以

#include 
      
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             #include 
             
              #include 
              using namespace std;#define FIN freopen("input.txt","r",stdin);#define FOUT freopen("output.txt","w",stdout);#define INF 0x3f3f3f3f#define INFLL 0x3f3f3f3f3f3f3f#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1typedef long long LL;typedef pair
               
                 PII;const int MAXN = 1e2 + 5;int T, n;int a[MAXN], b[MAXN];double dis[MAXN * MAXN];int main(){ //FIN scanf("%d", &T); while(T--) { scanf("%d", &n); for(int i = 1; i <= n; i++) { scanf("%d%d", &a[i], &b[i]); } int cas = 0; for(int i = 1; i <= n - 1; i++) { for(int j = i + 1; j <= n; j++) { dis[++cas] = (double) sqrt((a[j] - a[i])*(a[j] - a[i]) + (b[j] - b[i])*(b[j] - b[i])); } } sort(dis + 1, dis + cas + 1);// for(int i = 1; i <= cas; i++) {// cout << dis[i] << endl;// }// cout << "cas=" << cas << endl; if(dis[1] == dis[n]) printf("YES\n"); else printf("NO\n"); } return 0;}

转载于:https://www.cnblogs.com/Hyouka/p/5875299.html

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