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::*;

fn read<T: str::FromStr>() -> T {
    let mut buf = String::new();
    io::stdin().read_line(&mut buf).ok();
    buf.trim().parse().ok().unwrap()
}

fn read_vec<T: str::FromStr>() -> Vec<T> {
    read::<String>()
        .split_whitespace()
        .map(|a| a.parse().ok().unwrap())
        .collect()
}

fn read_vec2d<T: str::FromStr>(n: u32) -> Vec<Vec<T>> {
    (0..n).map(|_| read_vec()).collect()
}

fn digit(n: u32) -> u32 {
    if n < 10 {
        1
    } else {
        digit(n / 10) + 1
    }
}

fn main() {
    let n: u32 = read();
    let ans: u32 = (1..n + 1).map(|n| digit(n) % 2).sum();
    println!("{}", ans);
}