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 count_digit(n: &i32) -> i32 {
    let mut x = *n;
    let mut count = 0;
    while 0 < x {
        x /= 10;
        count += 1;
    }
    count
}

fn main() {
    let mut input = String::new();
    stdin().read_line(&mut input);

    let n: i32 = input.trim().parse().unwrap();
    let ans = (1..n+1).filter(|x| count_digit(x) % 2 == 1).count();
    println!("{}", ans);
}