OpenKattis
Kattis Set 08

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2020-03-09 05:15 AKDT

## Kattis Set 08

#### End

2020-03-16 01:30 AKDT
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# Problem JStirling's Approximation

The value of $n! = 1\cdot 2 \cdot 3 \cdot \ldots \cdot (n-1) \cdot n$ can be quite large, and it can be tedious to calculate. Fortunately for us, the French mathematician Abraham de Moivre (1667-1754) and the Scottish mathematician James Stirling (1692-1770) came up with a nice approximation:

$n!\approx S(n)=\sqrt {2\pi n}\frac{n^ n}{e^ n}.$

This is known as Stirling’s approximation.

For this problem, determine how good Stirling’s approximation is for various $n$.

## Input

Input consists of up to $500$ test cases, one per line. Each test case has a single integer $0 < n \leq 10^5$. Input ends when $n=0$.

## Output

For each test case, print out $n! / S(n)$ with an absolute error at most $10^{-8}$.

Sample Input 1 Sample Output 1
1
10
150
0

1.084437551419227546611577313
1.008365359132400245905553271
1.000555709081632955425046604