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(unused, non_snake_case, dead_code, non_upper_case_globals)]
use {
    proconio::{marker::*, *},
    std::{cmp::*, collections::*, mem::*},
};

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

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