题面

Time limit per test: 2 seconds

Memory limit per test: 256 megabytes

Description

You are given a regular polygon with 2⋅n vertices (it's convex and has equal sides and equal angles) and all its sides have length 1. Let's name it as 2n-gon.

Your task is to find the square of the minimum size such that you can embed 2n-gon in the square. Embedding 2n-gon in the square means that you need to place 2n-gon in the square in such way that each point which lies inside or on a border of 2n-gon should also lie inside or on a border of the square.

You can rotate 2n-gon and/or the square.

Input

The first line contains a single integer T (1≤T≤200) — the number of test cases.

Next T lines contain descriptions of test cases — one per line. Each line contains single even integer n (2≤n≤200). Don't forget you need to embed 2n-gon, not an nn-gon.

Output

Print T real numbers — one per test case. For each test case, print the minimum length of a side of the square 2n-gon can be embedded in. Your answer will be considered correct if its absolute or relative error doesn't exceed 10⁻⁶.

Example

input

output

1.000000000
2.414213562
127.321336469

题意

给定一个边长为 1 的正 2n 边形

求外接正方形的最小面积

解题思路

因为本题给定的 n 是偶数

故正方形与正 2n 边形恰好可以有4条边重合

正八边形如下：

正十二边形如下：

Z12

以正十二边形为例，根据对称性，可以先只看右上的部分：

根据正十二边形的每条边再画出水平和竖直的两条边，组成一个直角三角形

可以得到，最上边截得半条边，长度为 1/2

在所有斜边中，因为正多边形角度平分，所以可以得到每条斜边与水平边的夹角是平分90°角的

所以这张图中的水平长度为

1/2 + 1*cos30° + 1*cos60°

实际正方形边长为 2*(1/2 + 1*cos30° + 1*cos60°)

推广出去，对于正十六边形

在四分之一图中共有三条斜边

所以将 90° 角分为 4 份，每份 22.5°

实际边长为 2*(1/2 + 1*cos22.5° + 1*cos45° + 1*cos67.5°)

可以发现，从四边形开始，每增加四条边，斜边数量+1

故斜边数量可以通过 n/2-1 得到，每次增加 90/(n/2) °

完整程序

#include<bits/stdc++.h>
using namespace std;
const double PI=acos(-1.0);
const double epsilon=PI/180.0; //角度转弧度

void solve()
{
    int n;
    cin>>n;
    double tri=90.0/(n/2),trid=0.0,ans=0;
    for(int i=1;i<n/2;i++)
    {
        trid+=tri;
        ans+=cos(trid*epsilon);
    }
    cout<<fixed<<setprecision(9)<<(ans+0.5)*2<<'\n';
}
int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);cout.tie(0);
    int T;cin>>T;
    while(T--)
        solve();
    return 0;
}

Codeforces 1354C1 - Simple Polygon Embedding （几何）

题面