D - New Year Table CodeForces - 140A——double精度+弧度角度转化

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本文详细解析了CodeForces 140A 题目“New Year Table”，介绍了如何判断能否将n个半径为r的小圆盘完全放置在一个半径为R的大圆盘上，并使每个小圆盘都与大圆盘边缘相切且不相互重叠的方法。

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Think:
1题意：输入n, R, r代表小圆个数，大圆半径，小圆半径，询问可否将所有小圆完全放入大圆中，且小圆不重合且小圆与大圆内切
2反思：
1>double精度知识几乎空白
2>没有考虑当r > R/2情况下不能通过asin(r/(R-r))公式进行判断小圆在大圆中的所求弧度与r, R的关系
3知识+：PI = acos(-1), asin(), double精度……

D - New Year Table CodeForces - 140A

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.

Example
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:  ![这里写图片描述](https://img-blog.youkuaiyun.com/20170607130404336?watermark/2/text/aHR0cDovL2Jsb2cuY3Nkbi5uZXQvQmxlc3NpbmdYUlk=/font/5a6L5L2T/fontsize/400/fill/I0JBQkFCMA==/dissolve/70/gravity/SouthEast)

以下为Accepted代码

#include <bits/stdc++.h>
#define PI acos(-1.0)

using namespace std;

int main()
{
    int n;
    double R, r;
    while(scanf("%d %lf %lf", &n, &R, &r) != EOF)
    {
        if(r <= R)
        {
            if(r*2 <= R)
            {
                double x = asin(r/(R-r));
                if(n*x - PI > 1e-8)
                    printf("NO\n");
                else
                    printf("YES\n");
            }
            else {
                if(n <= 1)
                    printf("YES\n");
                else
                    printf("NO\n");
            }
        }
        else
            printf("NO\n");
    }
    return 0;
}