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: Accepted)

use proconio::input;

#[allow(unused_mut)]
fn main() {
    input! {
        n: usize
    }
    let mut res = 0;
    for i in 0..n + 1 {
        if get_digit(i) % 2 == 1 {
            res += 1;
        }
    }
    println!("{}", res);
}

fn get_digit(n: usize) -> usize {
    let mut n = n;
    let mut count = 0;
    while n != 0 {
        n /= 10;
        count += 1;
    }
    count
}