CodeForces - 140A New Year Table （数学几何&精度）大圆内能放几个小圆

CodeForces - 140A

New Year Table

Time Limit: 2000MS

Memory Limit: 262144KB

64bit IO Format: %I64d & %I64u

SubmitStatus

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

Hint

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

Source

Codeforces Round #100

//题意：输入n,R,r

表示给你一个半径为R的大圆，问沿着大圆的边界放不相交（可以相切）的半径为r的小圆，问是否可以放n个。

//思路：

通过求圆心角大小来求最多可以放几个小圆（画个图模拟一下就可以理解了），加上几个特判就过了，然后就是精度问题（1e-9的精度）。

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#include<iostream>
#define PI acos(-1.0)
using namespace std;
int main()
{
	int n;
	double r,R;	
	while(scanf("%d%lf%lf",&n,&R,&r)!=EOF)
	{
		int k=0;
		double l1,l2,l3;
		double y,x1;
		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
		{
			l1=R-r;l2=r;l3=sqrt(l1*l1-l2*l2);
			y=(l1*l1+l3*l3-l2*l2)/(2*l1*l3);
			x1=acos(y);
			k+=(PI+0.000000001)/x1;
			if(k<n)
				printf("NO\n");
			else
				printf("YES\n");
		}
	}
	return 0;
}

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#include<iostream>
#define PI acos(-1.0)
using namespace std;
int main()
{
	int n;
	double r,R;	
	while(scanf("%d%lf%lf",&n,&R,&r)!=EOF)
	{
		int k=0;
		double l1,l2,l3;
		double y,x1;
		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)//这块是刚开始时不知是精度问题，加的一个特判，没想到过了（就那一组卡到了1e-9的精度），然后根据这组数据推出的这是个精度问题)
		{
			if(n<=6)
				printf("YES\n");
			else
				printf("NO\n");
		}
		else
		{
			l1=R-r;l2=r;l3=sqrt(l1*l1-l2*l2);
			y=(l1*l1+l3*l3-l2*l2)/(2*l1*l3);
			x1=acos(y);
			k+=PI/x1;
			if(k<n)
				printf("NO\n");
			else
				printf("YES\n");
		}
	}
	return 0;
}