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)

use proconio::input;

fn is_ok(mut n: usize) -> bool {
  let mut m = 0;
  while n != 0 {
    n /= 10;
    m += 1;
  }
  if m % 2 == 0 { false } else { true }
}

fn main() {
    input! {
        n: usize,
    }

    let mut count = 0;
    for i in 1..=n {
      if is_ok(i) {
        count += 1;
      }
    }

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