codeforces-140【A几何】【精度】

最新推荐文章于 2021-06-14 22:54:33 发布

Somethingwll

最新推荐文章于 2021-06-14 22:54:33 发布

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分类专栏：数学 Codeforces 卡精度

本文链接：https://blog.youkuaiyun.com/wang907553141/article/details/62462519

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本文探讨了如何在半径为R的圆形桌上放置n个半径为r的圆形盘子，确保每个盘子都能触及桌子边缘且互不重叠。通过计算几何角度来判断是否能成功放置所有盘子。

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题目链接：点击打开链接

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:

$\text{[math]}$

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：

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;
 }