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(non_snake_case)]
#![allow(unused_imports)]
#![allow(dead_code)]

use proconio::{input, fastout};
use proconio::marker::*;
use whiteread::parse_line;
use std::collections::*;
use num::*;
use num_traits::*;
use superslice::*;
use std::ops::*;

use itertools::Itertools;
use itertools_num::ItertoolsNum;


#[fastout]
fn solve() {
    const MOD: usize = 1_000_000_007;
    const INF: usize = std::usize::MAX;
    
    input!{
        n: usize,
    }

    let mut ans = 0;

    for i in 1..=n {
        if i.to_string().len() % 2 != 0 {
            ans += 1;
        }
    }

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

fn main() {
    solve()
}

#[cfg(test)]
mod test {
    use super::solve;
}