Thinking-Bear magic（正n边形面积）

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

1
4 2 3

输出

1

题解：题意很简单就是不断计算变小的正n边形的面积，看是否小于等于一开始给出的面积，小于时输出此时进行的轮数；

           tips：内层边长为：len=sin((n-2)*M_PI/2/n)*a;   （a为原正n边形面积）

                    sin（）函数中放入的是弧度需转化，M_PI是预定义好的π；

              面积计算可以用公式去推：设正n边形的半径为R，边长为an，中心角为αn，边心距为rn，则αn=360°÷n，an=2Rsin(180°÷n)，rn=Rcos(180°÷n)，R^2=r n^2+(an÷2)^2，周长pn=n×an，面积Sn=pn×rn÷2。

                   在储存数据的时候最好用double，其他队用的float可能因为精度问题WA了；

#include<bits/stdc++.h>
using namespace std;
int main()
{
    double area,jiao,rn,pn,len,de,a;
    int n,t,k=0;
    cin>>t;
    while(t--)
    {
        cin>>n>>a>>de;
        area=n*a*a/(4*tan(M_PI/n));//一开始的正n边形的面积
        len=a;
        if(area<=de)
        {
            cout<<"0"<<endl;
        }
        else
        {
            for(int i=1;; i++)
            {
                len=sin((n-2)*M_PI/2/n)*len;//随着轮数变化的边长
                area=n*len*len/(4*tan(M_PI/n));//面积
                if(area<=de)
                  {
                    k=i;
                    break;

                  }
            }
            cout<<k<<endl;
        }

    }
}