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(dead_code, unused_variables, unused_imports)]
use proconio::marker::Chars;
use proconio::{fastout, input};
use std::cmp::{max, min};
use std::collections::{HashMap, HashSet};

#[fastout]
fn main() {
    input!{
        n:usize,
    }
    
    let mut ans = 0;
    for i in 1..=n{
        if cnt_digit(i)%2 == 1{
            ans += 1;
        }
    }
    println!("{}",ans);
}   

fn cnt_digit(mut n:usize) -> usize{
    let mut cnt = 0;
    while n > 0{
        cnt += 1;
        n /= 10;
    }
    cnt
}