Educational Codeforces Round 87 (Rated for Div. 2) C1.Simple Polygon Embedding

Educational Codeforces Round 87 (Rated for Div. 2) C1.Simple Polygon Embedding

题目链接
The statement of this problem is the same as the statement of problem C2. 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 even integer n (2≤n≤200). 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
2
4
200

output

1.000000000
2.414213562
127.321336469

看了半天以为是规律题，后来发现还是要死算，我发现这个其实不太好讲，我摆个图吧

图画得不好，还请见谅
不难发现，我们设答案 $ans$ 初始值为1，变化的角为 $ANG$，内角为 $ang$，求答案的过程就是一个递推的公式：
$$\{\begin{array}{l}ans=ans+2\ast cos(ANG)\\ ANG=ang+ANG-pi\end{array}$$
AC代码如下:

#include
using namespace std;
typedef long long ll;
#define pi acos(-1)
main(){
    int t;
    cin>>t;
    while(t--){
        double n;
        cin>>n;
        double ANG=0,ang=pi*(n-1)/n;
        double ans=1.0;
        for(int i=0;i<ll(n)/2-1;i++){
            ANG=ang+ANG-pi;
            ans+=2*cos(ANG);
        }
        printf("%.9f\n",ans);
    }
}