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)

#[allow(unused_imports)]
use proconio::{fastout, input};

#[fastout]
fn main() {
    input!(n: usize);
    let mut count = 0;
    let mut div = 1;
    loop {
        if n / div >= 1 {
            if n < div * 10 {
                count += (n % ((div) * 10)) - div + 1;
            } else {
                count += div * 9;
            }
        } else {
            break;
        }
        div *= 100;
    }
    println!("{}", count);
}