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;
use std::io::BufRead;

fn main() {
    let stdin = io::stdin();
    let mut stdin = stdin.lock();
    
    let n = {
        let mut n = String::new();
        stdin.read_line(&mut n).unwrap();
        n.trim().parse::<i32>().unwrap()
    };

    if n == 1 {
        println!("1");
        return;
    }

    let k = (n as f64).log10().floor() as u32;
    
    let mut answer = {
        if k%2==0 {
            n - 10i32.pow(k) + 1
        } else {
            0
        }
    };

    for i in 0..k {
        if i%2 == 0 {
            answer += 9 * 10i32.pow(i);
        }
    }

    println!("{}", answer );
}