Codeforces--140A--New Year Table（思维）



New Year Table

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 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.

Sample Input

Input

4 10 4

Output

YES

Input

5 10 4

Output

NO

Input

1 10 10

Output

YES

Hint

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



Source
Codeforces Round #100
将n个半径为r盘子放在半径为R的桌子上，盘子要紧挨着桌沿问是否可以实现

#include <cstdio>
#include <cstring>
#include <cmath>
#include <map>
#include <iostream>
using namespace std;
#define PI acos(-1)
int main()
{
int n;
double R,r;
while(scanf("%d%lf%lf",&n,&R,&r)!=EOF)
{
if(R<r)
printf("NO\n");
else if(R<2*r)
{
if(n==1)
printf("YES\n");
else
printf("NO\n");
}
else if(R==2*r)
{
if(n<=2)
printf("YES\n");
else
printf("NO\n");
}
else if(R==3*r)
{
if(n<=6)
printf("YES\n");
else
printf("NO\n");
}
else
{
double d=R-r;
double s=sqrt(d*d-r*r);
double co=(d*d+s*s-r*r)/(2*d*s);
double q=acos(co);
int k=PI/q;
//			printf("%d %d\n",k,n);
if(k>=n)
printf("YES\n");
else
printf("NO\n");
}
}
}