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: Runtime Error)

using System;
using System.Collections.Generic;
using System.Linq;

namespace UnevenNumbers
{
    class Program
    {
        static void Main(string[] args)
        {
            string n_str = Console.ReadLine();
            List<char> n_char = n_str.ToCharArray().ToList();

            int digit = n_str.Length;


            double result = 0;
            for (int index = 1; index < digit; index++)
            {
                if (index % 2 != 0)
                {
                    result += Math.Pow(10, index);
                }
            }

            //add max digit
            if (digit % 2 != 0)
            {
                n_char.RemoveAt(0);
                string aaa = "";
                foreach (char v in n_char)
                {
                    aaa += v.ToString();
                }

                result += (int.Parse(aaa) + 1);
            }



            //except 0
            Console.WriteLine(result - 1);

        }
    }
}