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 (1.42.0) (judge verdict: Wrong Answer)

use std::io::Read;
 
fn main() {
    let mut buf = String::new();
    let stdin = std::io::stdin();
    stdin.lock().read_to_string(&mut buf).unwrap();
    let mut iter = buf.split_whitespace();
    let n = iter.next().unwrap().parse::<usize>().unwrap();

    if n < 100 {
        println!("{}", std::cmp::min(9, n));
    } else if n < 1000 {
        println!("{}", std::cmp::min(999-99+9, n-99+9));
    } else {
        println!("{}", std::cmp::min(99999-9999+999-99+9, n-9999+999-99+9));
    }
}