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::str::FromStr;

fn read<T: FromStr> () -> T {
    let stdin = std::io::stdin();
    let stdin = stdin.lock();
    let token: String = stdin
        .bytes()
        .map(|c| c.expect("failed to read char") as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect();
    token.parse().ok().expect("failed to parse token")
}

fn digits_odd(n: usize) -> bool {
    let div = 10;
    let mut m = n;
    let mut count = 0;
    while m > 0 {
        m /= 10;
        count += 1;
    }
    if count % 2 == 0 {
        false
    }
    else{
        true
    }
}

fn main(){
    let mut N = read::<usize>();
    let mut ans: usize = 0;

    for i in 0..N+1 {
        if digits_odd(i) {
            ans += 1;
        }
        else{
            ;
        }
    }
   
   println!("{}", ans);
}