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codeforces-140【A几何】【精度】

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A. New Year Table time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output

Gerald is setting the New Year table. The table has the form of a circle; its radius equals R. 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 equal r. 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 for n 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 place n 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.

Examples input

4 10 4
      

output

YES
      

input

5 10 4
      

output

NO
      

input

1 10 10
      

output

YES
      

Note

The possible arrangement of the plates for the first sample is:

codeforces-140【A几何】【精度】

大意:判断在一个半径为 R 的桌子上放 n 个半径为 r 的盘子,使得各个盘子不互相重叠,是否合法

思路:

codeforces-140【A几何】【精度】

如图所示,将此问题转化为数学问题。我们求出  ∠BCD 即可,知道 BC = R - r , BD = r 利用 arcsin() 即可。

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
int n,R,r;
int main()
{
	while(~scanf("%d%d%d",&n,&R,&r))
	{
		if(R<r)
		{
			puts("NO");
			continue;
		}
		if(2*r>R) // 一个盘子就过圆心了,所以只能放一个 
		{
			if(n==1)
				puts("YES");
			else
				puts("NO");
			continue;
		}
		double angle=asin(r*1.0/(R-r));
		if(2*acos(-1.0)-angle*2*n>-1e-10) // 这里卡精度 
			puts("YES");
		else
			puts("NO");
	}
	return 0;
 }