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_imports)]
use proconio::{fastout, input};
 
#[fastout]
fn main() {
  input!(n: isize);
  let mut optional = Some(n);
  let mut count = 0;
  let mut iter_cnt = 0;
  let mut multiplyer = 10;
  while let Some(i) = optional {
    if i >= 10 {
      if iter_cnt % 2 == 0 {
        count += multiplyer - multiplyer / 10;
      }
      multiplyer *= 10;
      optional = Some(i / 10);
      iter_cnt += 1;
    } else if i > 0 {
      if iter_cnt % 2 == 0 {
        count += n - multiplyer / 10 + 1;
      }
      multiplyer *= 10;
      optional = Some(i / 10);
      iter_cnt += 1;
    } else {
      break;
    }
  }
  println!("{}", count);
}