Simple Polygon Embedding
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 $$$2 \cdot n$$$ 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.
The first line contains a single integer $$$T$$$ ($$$1 \le T \le 200$$$) — the number of test cases.Next $$$T$$$ lines contain descriptions of test cases — one per line. Each line contains single even integer $$$n$$$ ($$$2 \le n \le 200$$$). Don't forget you need to embed $$$2n$$$-gon, not an $$$n$$$-gon.
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 $$$10^{-6}$$$.
import math

def simplePolygonEmbedding(n):
    ans = math.tan((n - 1) * math.pi / (2 * n))
    return ans
for x in range(int(input())):
    print(simplePolygonEmbedding(int(input())))
import math
k = int(input())
for _ in range(k):
    n = int(input())
    print(1 / math.tan(math.pi / (2 * n)))
import math

def minimum_square_side(n):
    return 1 / math.tan(math.pi / n)
T = int(input())
for _ in range(T):
    n = int(input())
    side_length = minimum_square_side(2 * n)
    print('{:.9f}'.format(side_length))