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;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
//using ProCon.Util;

namespace ProCon
{
    class Program
    {
        static void Main(string[] args)
        {
            AC136B();
        }

        static void AC136B()
        {
            //https://atcoder.jp/contests/abc136/tasks

            var n = int.Parse(Console.ReadLine());
            int count;

            if(n < 10)
            {
                count = n;
            }
            else if(n < 100)
            {
                count = 9;
            }
            else if(n < 1000)
            {
                count = n - 100 + 1 + 9;
            }
            else if (n < 10000)
            {
                count = 900 + 9;
            }
            else if( n < 100000)
            {
                count = n - 10000 + 1 + 900 + 9;
            }
            else
            {
                count = 90000 + 900 + 9;
            }

            Console.WriteLine(count);
            Console.ReadKey();

        }

    }
}