Thinking-Bear magic（正n邊形面積）

鏈接：https://www.nowcoder.com/acm/contest/163/D
來源：牛客網
 

題目描述

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?

輸入描述:

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).

輸出描述:

For each test case, output a single integer.

示例1

輸入

1
4 2 3

輸出

1

題解：題意很簡單就是不斷計算變小的正n邊形的面積，看是否小於等於一開始給出的面積，小於時輸出此時進行的輪數；

           tips：內層邊長爲：len=sin((n-2)*M_PI/2/n)*a;   （a爲原正n邊形面積）

                    sin（）函數中放入的是弧度需轉化，M_PI是預定義好的π；

              面積計算可以用公式去推：設正n邊形的半徑爲R，邊長爲an，中心角爲αn，邊心距爲rn，則αn=360°÷n，an=2Rsin(180°÷n)，rn=Rcos(180°÷n)，R^2=r n^2+(an÷2)^2，周長pn=n×an，面積Sn=pn×rn÷2。

                   在儲存數據的時候最好用double，其他隊用的float可能因爲精度問題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));//一開始的正n邊形的面積
        len=a;
        if(area<=de)
        {
            cout<<"0"<<endl;
        }
        else
        {
            for(int i=1;; i++)
            {
                len=sin((n-2)*M_PI/2/n)*len;//隨着輪數變化的邊長
                area=n*len*len/(4*tan(M_PI/n));//面積
                if(area<=de)
                  {
                    k=i;
                    break;

                  }
            }
            cout<<k<<endl;
        }

    }
}