【codeforce】几何

I - I

Crawling in process... Crawling failed Time Limit:2000MS    Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

Submit Status

Description

Gerald is setting the New Year table. The table has the form of a circle; its radius equalsR. Gerald invited many guests and is concerned whether the table has enough space for plates for all those guests. Consider all plates to be round and have the same radii that equalr. Each plate must be completely inside the table and must touch the edge of the table. Of course, the plates must not intersect, but they can touch each other. Help Gerald determine whether the table is large enough forn plates.

Input

The first line contains three integers n,R and r (1 ≤ n ≤ 100,1 ≤ r, R ≤ 1000) — the number of plates, the radius of the table and the plates' radius.

Output

Print "YES" (without the quotes) if it is possible to placen plates on the table by the rules given above. If it is impossible, print "NO".

Remember, that each plate must touch the edge of the table.

Sample Input

Input

4 10 4

Output

YES

Input

5 10 4

Output

NO

Input

1 10 10

Output

YES

Sample Output

      

        NO
YES 
      

 

Hint

The possible arrangement of the plates for the first sample is: 小圆挨着大圆，是否可以放下n个

#include<cstdio>
#include<algorithm>
using namespace std;
#define pi acos(-1.0)
#include<cmath>
int main()
{
	int n,R,r;
	while(~scanf("%d%d%d",&n,&R,&r))
	{
		if(r>R){
			if(n==0) printf("YES\n");
			else printf("NO\n");
			continue;
		}
		 else if(r==R){
			if(n<=1) printf("YES\n");
			else printf("NO\n");
			continue;
		}
		else if(4*r>2*R){
			if(n<=1) printf("YES\n");
			else printf("NO\n");
			continue;
		}
		double m=(2.0*(R-r)*(R-r)-(2.0*r)*(2.0*r))/(2.0*(R-r)*(R-r));
		//余弦定理 
		double a=acos(m); 
		if((n-2.0*pi/a)<1e-10)
		printf("YES\n");
		else
		printf("NO\n"); 
	 } 
	return 0;
 }