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 main() {
    let n = read::<u32>();

    if n < 10 {
        println!("{}", n);
    } else if 10 <= n && n < 10u32.pow(2){
        println!("{}", 9);
    } else if 10u32.pow(2) <= n && n < 10u32.pow(3) {
        println!("{}", 9 + (n - 100 + 1));
    } else if 10u32.pow(3) <= n && n < 10u32.pow(4) {
        println!("{}", 9 + (999 - 100 + 1));
    } else if 10u32.pow(4)<= n && n < 10u32.pow(5) {
        println!("{}", 9 + (999 - 100 + 1) + (n - 10000 + 1));
    } else if n == 10u32.pow(5) {
        println!("{}", 9 + (999 - 100 + 1) + (99999 - 10000 + 1));
    } else {
        //nop
    }
}

fn read<T: std::str::FromStr>() -> T {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim().parse().ok().unwrap()
}