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 get_num_of_digit(v: isize) -> isize {
    let fv = v as f64;
    (fv.log10() + 1.0).floor() as isize
}

fn main() {
    let mut line = String::new();
    std::io::stdin().read_line(&mut line).unwrap();
    let n: isize = line.trim().parse().unwrap();

    println!("{}", (1..n+1).filter(|&i| get_num_of_digit(i) % 2 == 1).count());
}