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 odd integer n (3≤n≤199). Don't forget you need to embed 2n-gon, not an n-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

3
3
5
199

output

1.931851653
3.196226611
126.687663595

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 是奇数

又因为外接正方形要求面积最小

所以不会有边与正方形重合，只存在四个点会位于外接正方形的四条边上

以正十边形为例

外接正方形的作图方法为

选择一对 连线能够通过正十边形中心点 的点对，连接两点，并向两端延长

然后从正十边形中心点开始作垂直于前面那条边的边，并向两端延长

然后以这两条线为准作一个斜45°正方形

为将正方形缩小到最小，所以要保证正十边形上有点会落在正方形上

故作出来的图如上图所示

只看其四分之一部分

首先可以通过 360°/(2*n) 来求出每份角的角度 ang

因为边长为 1 ，所以可以通过一个顶角为 ang ，底边为 1 的等腰三角形来求出腰长 x = 0.5/sin(ang/2)

因为总共有 10 个角，所以在四分之一图中平均有 10/4 = 2.5个角

因为这些正多边形都是凸多边形，所以与正方形的交点一定是最接近于中间位置的那个点

所以实际的 θ = round(2.5/2)*ang

公式里表示为 ang2 = round(n/4)*ang

对于ans1，根据正弦定理

可以得到 ans1 / sinθ = x / sin45°

得到 ans1 = x / sin45° * sinθ

对于ans2，根据正弦定理

可以得到 ans2 / sin(90°-θ) = x / sin45°

得到 ans2 = x / sin45° * sin(90°-θ)

相加即为正方形边长

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 ang=180.0/n;
    double ang2=round(n/4.0)*ang;
    double x=1.0/(2.0*sin(ang/2.0*epsilon));
    double ans1=x/sin(45.0*epsilon)*sin(ang2*epsilon);
    double ans2=x/sin(45.0*epsilon)*sin((90.0-ang2)*epsilon);
    cout<<fixed<<setprecision(9)<<ans1+ans2<<'\n';
}
int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);cout.tie(0);
    int T;cin>>T;
    while(T--)
        solve();
    return 0;
}