2018 ACM 国际大学生程序设计竞赛上海大都会赛重现赛 C Thinking Bear magic

链接：https://www.nowcoder.com/acm/contest/163/D
来源：牛客网
 

题目描述

In order to become a magical girl, Thinking-Bear are learning magic circle.
He first drew a regular polygon of N sides, and the length of each side is a.
He want to get a regular polygon of N sides, and the polygon area is no more than L.
He doesn't want to draw a new regular polygon as it takes too much effort.
So he think a good idea, connect the midpoint of each edge and get a new regular polygon of N sides.
How many operations does it need to get the polygon he want?

输入描述:

The first line of the input is T(1≤ T ≤ 100), which stands for the number of test cases you need to solve.
The first line of each case contains three space-separated integers N, a and L (3 ≤ N ≤ 10, 1 ≤ a ≤ 100, 1 ≤ L ≤ 1000).

输出描述:

For each test case, output a single integer.

示例1

输入

复制

1
4 2 3

输出

复制

1

正多边形面积公式为  S=1/2*sin （2*pi/n）*n*R*R;

R = a/(   2*sin(pi/n)  )

然后，就一直更新面积啦，看需要更新几次，就是答案啦。

用公式写，复杂的也没多大啦QWQ

#include<iostream>
#include<cmath>
#include<algorithm>
using namespace std;
const double eps = 1e-6;
#define pi acos(-1.0)

int main()
{
	int T,sum=0;;
	double N,s,a,b,L;
	cin>>T;
	while (T--)
	{
		sum=0;
		cin>>N>>a>>L;
		s=0.5*sin(2*pi/N)*N*(  (a/(2*sin(pi/N))  )  *   (a/(2*sin(pi/N))  )   );
		while (s>L)
		{
			a = a * sin( (N-2) * (pi/ (N*2) )    );
			s=0.5*sin(2*pi/N)*N*(  (a/(2*sin(pi/N))  )  *   (a/(2*sin(pi/N))  )   );
			//cout<<"s= "<<s<<endl;
			sum++;
		}
		cout<<sum<<endl;
	}
	
}