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 Kotlin (1.0.0) (judge verdict: Accepted)

import java.util.*
 
fun main(args: Array<String>) {
  var num = readLine()!!

  when(num.length) {
    1 -> {
      println(num.toInt())
    }
    2 -> {
      println(9)
    }
    3 -> {
      println(9 + num.toInt() - 99)
    }
    4 -> {
      println(9 + 900)
    }
    5 -> {
      println(9 + 900 + num.toInt() - 9999)
    }
    6 -> {
      println(9 + 900 + 90000)
    }
  }
}