2018 ACM上海大都会赛 D-Thinking-Bear magic(中点正多边形面积变换)

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

1
4 2 3

输出

1

题目大意：给出一个正n边形和它的边长，给出一个面积L，每次操作：连接每边中点构成一个小正n边形，求几次操作后正n边形面积小于或等于L。

思路：读懂题就感觉有问题，一定有公式在里面。事实证明确实如此，因为是正多边形，所以可以直接使用面积公式：

S=1/4*（r²）*n*tan(1/2θ)，然后每次操作后的面积也有公式：操作后的面积与原面积比为：【cos(180°/n)】²(公式证明，看看？)

然后，，，这题就没了.......

代码如下：

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<cmath>
#include<string>
#include<string.h>
using namespace std;
#define pi acos(-1.0)
int main()
{
    int t,k=0,i;
	cin>>t;
    while(t--)
	{
        k++;
        double n,a,l,b,s;
		cin>>n>>a>>l;
        b=cos(pi/n)*cos(pi/n);
        s=(double)n*a*a/tan(pi/n)/4;
        for(i=0;;i++)
		{
            if(s<=l)
			{
			   cout<<i<<endl;break;
			}
            s*=b;
        }
    }
    return 0;
}