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
思路：比第二道題目簡單的不是一點半點。主要用到了余弦定理。
代碼如下：

#include<bits/stdc++.h> #define ll long long #define PI acos(-1.0) using namespace std;int n;int main() {int t;scanf("%d",&t);while(t--){scanf("%d",&n);int m=n;n*=2;double z=1.0/(2.0-2.0*cos(PI/(m*1.0)));double y=z+z-2.0*z*cos(((m-2)*1.0/n)*PI);double x=sqrt(y/2.0);x=2.0*x+1.0;printf("%.9lf\n",x);}return 0; }

努力加油a啊，(^o)/~

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