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 C# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
namespace Answer {
  class Program {
    public static void Main(string[] args) {
      var N = int.Parse(Console.ReadLine());
      var digitNum = N.ToString().Length;
      var ans = 0;
      int init, last;
      var digit = 1;
      var lessThanN = true;
      while (lessThanN) {
        init = (int)Math.Pow(10, digit - 1);
        last = (int)Math.Pow(10, digit) - 1;
        if (init > N) break;
        if (last >= N) {
          lessThanN = false;
          last = N;
        }
        ans += (int)last - (int)init + 1;
        digit += 2;
      }
      Console.WriteLine(ans);
    }
  }
}