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 Scala (2.11.7) (judge verdict: Accepted)

object Main {
  def main(args:Array[String]) {
    val N = readLine().toInt

    var ans = 0
    for(i <- 1 to N) {
      var count = 0
      var num = i
      while(num > 0) {
        num = num / 10
        count += 1
      }
      if(count % 2 != 0) ans += 1
    }
    println(ans)
    }
}