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)

#![allow(non_snake_case)]

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

fn main() {
    let N: i32 = read_line().parse().unwrap();
    let len = N.to_string().len() as u32;
    let mut ans = 0;
    let mut i = 1;
    while i < len {
        ans += 9 * 10_i32.pow(i-1);
        i += 2;
    }
    if len % 2 == 1 {
        ans += N - 10_i32.pow(len-1) + 1
    }
    println!("{}", ans);
}