Problem:
Score : 100 points

Problem Statement

Given is an integer N.

Takahashi chooses an integer a from the positive integers not greater than N with equal probability.

Find the probability that a is odd.

Constraints

1 \leq N \leq 100

Input

Input is given from Standard Input in the following format:

N

Output

Print the probability that a is odd.
Your output will be considered correct when its absolute or relative error from the judge's output is at most 10^{-6}.

Sample Input 1

4

Sample Output 1

0.5000000000

There are four positive integers not greater than 4: 1, 2, 3, and 4. Among them, we have two odd numbers: 1 and 3. Thus, the answer is \frac{2}{4} = 0.5.

Sample Input 2

5

Sample Output 2

0.6000000000

Sample Input 3

1

Sample Output 3

1.0000000000

Solution in Rust (1.15.1) (judge verdict: Accepted)


#[allow(dead_code)]
fn rd() -> String {
    let mut input = String::new();
    std::io::stdin().read_line(&mut input).unwrap();
    input.trim().to_owned()
}

/// 1行の文字を空白で分割してVec<T>にする。
#[allow(dead_code)]
fn rd_vec<T : std::str::FromStr>() -> Vec<T> {
    rd().split_whitespace().map(|x| x.parse().ok().unwrap()).collect()
}

/// 1つの要素を入力で受け取る
#[allow(dead_code)]
fn rd_one<T : std::str::FromStr>() -> T {
    rd().parse().ok().unwrap()
}

/// 複数回read_splitを行う
#[allow(dead_code)]
fn rd_vec_mul<T : std::str::FromStr>(count : i32) -> Vec<Vec<T>> {
    (0..count).map(|_| rd_vec()).collect()
}

/// Vecを1列で出力する
#[allow(dead_code)]
fn op_vec<T : std::fmt::Display>(item : Vec<T>, splitter : char) {
    let cnt = item.len();
    let mut c = 0;
    for i in item.iter() {
        c+=1;
        if cnt > c {print!("{}{}", i, splitter);} else {println!("{}", i);}
    }
}

fn main() {
    let input : f32 = rd_one();
    let b = input as i32;
    println!("{}", (((b + 1) / 2) as f32) / input);
}