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;

class Program
{
    static void Main(string[] args)
    {
        string s = Console.ReadLine();
        int n = int.Parse(s);

        int len = s.Length;//桁数

        int ret = 0;
        if (len == 1)
        {
            ret = n;
        }
        else if (len == 2)
        {
            ret = 9;
        }
        else if (len == 3)
        {
            ret = n - 99 + 9;
        }
        else if (len == 4)
        {
            ret = 909;
        }
        else if (len == 5)
        {
            ret = n - 9000 - 90;
        }
        else 
        {
            ret = 90909;
        }



        Console.WriteLine(ret.ToString());

    }
}