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: Wrong Answer)

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

namespace ConsoleApplication1
{
    public class Program
    {
        static void Main(string[] args)
        {
            try
            {
                var str = Console.ReadLine();
                long n = str.Length;
                long src = long.Parse(str);
                long number = 0;

                if (n % 2 == 0)
                {
                    for (long i = 1; i < n; i += 2)
                    {
                        number += (long)Math.Pow(10, i) - (long)Math.Pow(10, i - 1);
                    }
                }
                else
                {
                    for (long i = 1; i < n; i += 2)
                    {
                        number += (long)Math.Pow(10, i) - (long)Math.Pow(10, i - 1);
                    }

                    number += (src % (long)Math.Pow(10,n-1)) + 1;
                }

                Console.WriteLine(number);

            }
            catch (Exception ex)
            {
                Console.WriteLine(ex.Message);
            }

            Console.WriteLine("終了");
            Console.Read();
        }


    }
}