Problem:
Score : 200 points

Problem Statement

Given is an integer N. Find the number of positive integers less than or equal to N that have an odd number of digits (in base ten without leading zeros).

Constraints

1 \leq N \leq 10^5

Input

Input is given from Standard Input in the following format:

N

Output

Print the number of positive integers less than or equal to N that have an odd number of digits.

Sample Input 1

11

Sample Output 1

9

Among the positive integers less than or equal to 11, nine integers have an odd number of digits: 1, 2, \ldots, 9.

Sample Input 2

136

Sample Output 2

46

In addition to 1, 2, \ldots, 9, another 37 integers also have an odd number of digits: 100, 101, \ldots, 136.

Sample Input 3

100000

Sample Output 3

90909

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

fn read_line() -> String {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).unwrap();
    s.trim_right().to_owned()
}

trait Integer {
    fn count_digit(&self) -> u32;
    fn is_odd(&self) -> bool;
}

impl Integer for u32 {
    fn count_digit(&self) -> u32 {
        (*self as f32).log10().floor() as u32 + 1_u32
    }
    
    fn is_odd(&self) -> bool {
        (*self) & 1 == 1
    }
}

fn main() {
    let n: u32 = read_line().parse().unwrap();
    
    let count = (1..(n + 1))
        .filter(|&x| x.count_digit().is_odd())
        .count();

    println!("{}", count);
}