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)

// -*- coding:utf-8-unix -*-

use proconio::input;

fn main() {
    input! {
        n: i64,
    }
    let mut ans = 0;
    let d = n.to_string().len();
    for m in (1..d + 1).step_by(2) {
        if m < d {
            ans += 10i64.pow((m - 1) as u32) * 9;
        } else {
            ans += n - 10i64.pow((d - 1) as u32) + 1;
        }
    }

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