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)

use std::io;

fn main() {
    let stdin = io::stdin();

    let mut buf = String::new();
    stdin.read_line(&mut buf).ok();
    let num: usize = buf.trim().parse().unwrap();

    let mut keta = 0;

    for i in 1..(num + 1) {
        if (i.to_string().len() == 1 || i.to_string().len() == 3 || i.to_string().len() == 5) {
            keta += 1;
        }
    }

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