Thinking-Bear magic（多边形面积）

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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.

输入

1
4 2 3

输出

1

S = sin(2 * pi / n) * n * R * R( R:  外接圆半径);

R = a / ( sin(pi / n) * 2 );

多边形内角和：（n - 2）* 180;

相邻中点连线构成的多边形面积比：(R1 * R1) / (R2 * R2);

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const double pi = acos(-1.0);
const double eps = 1e-5;
int main()
{
    int T;
    scanf("%d", &T);
    for(int Case = 1; Case <= T; Case++){
    	double l, a;
		int n;
    	scanf("%d%lf%lf", &n, &a, &l);
    	double t = sin((n - 2) * pi / n / 2.0);
    	t *= t;
    	double r = a / sin(pi / n) / 2.0;
		double s = 0.5 * sin(2.0 * pi / n) * n * r * r;
		int i = 0;
		while(1){
			if(s <= l){
				printf("%d\n", i);
				break;
			}
			else{
				s *= t;
			}
			i++;
		}
    }
    return 0;
}