Time Limit:2000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u
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
4 10 4
YES
5 10 4
NO
1 10 10
YES
Hint
The possible arrangement of the plates for the first sample is
题解：小圆贴着大圆的边，问是否能放n个；注意精度；
代码：
#include<iostream>
#include<cstdio>
#include<cmath>
#in\
clude<algorithm>
using namespace std;
const double PI = acos(-1.0);
int main(){
int n, R, r;
while(~scanf("%d%d%d", &n, &R, &r)){
if(r > R){
if(n == )
puts("YES");
else
puts("NO");
continue;
}
else if(r == R){
if(n <= )
puts("YES");
else
puts("NO");
continue;
} else if( * r > R){
if(n <= )
puts("YES");
else
puts("NO");
continue;
}
double cosa = (2.0 * (R - r) * (R - r) - (2.0*r) * (2.0*r))/ (2.0*(R-r)*(R-r));
double a = acos(cosa);
//printf("%lf\n", 2.0 * PI / a);
if((n - 2.0 * PI / a) <= 1e-){
puts("YES");
}
else{
puts("NO");
}
}
}