Educational Codeforces Round 87 (Rated for Div. 2) C2.Not So Simple Polygon Embedding

Educational Codeforces Round 87 (Rated for Div. 2) C2.Not So Simple Polygon Embedding

題目鏈接
The statement of this problem is the same as the statement of problem C1. The only difference is that, in problem C1, n is always even, and in C2, n is always odd.

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

此處擺一張題解的圖片，當我們正常放置正 2-n 形時，如圖最左邊，此時若將 2-n 邊形繞對稱中心旋轉 $\frac{Π}{4*n}$，如中間那幅圖所示，就能得到最小邊長，首先求得對稱中心到任一點的距離爲 $\frac{0.5}{sin(Π/2n)}$，然後用一個餘弦定理可求得外接矩形邊長的一半 $\frac{cos(Π/4n)}{2*sin(Π/2n)}$，最後的答案即爲 $\frac{cos(Π/4n)}{sin(Π/2n)}$
至於爲什麼旋轉 $\frac{Π}{4*n}$ 後矩形邊長最短，題解說是二分，我覺得我們可以猜測一下，不轉就是最左邊的圖，轉 $\frac{Π}{2*n}$ 就是最右邊的圖，兩個圖的矩形邊長一樣，那麼可以推測矩形邊長的變化是一個凹函數，極值點一定在中間的位置，那麼用兩個角度求平均即可得到 $\frac{Π}{4*n}$，AC代碼如下：

#include <bits/stdc++.h>
typedef long long ll;
using namespace std;
#define pi acos(-1)
main(){
    int t,n;
    cin>>t;
    while(t--){
        cin>>n;
        printf("%.9f\n",cos(pi/double(4*n))/sin(pi/double(2*n)));
    }
}