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 Task3
{

    public static void Main(string[] args)
    {
        int N = int.Parse(Console.ReadLine());
        int len = N.ToString().Length;
        int ans = 0;

        for (int i = 0; i < len; i++)
        {
            if ((i + 1) % 2 != 0)
            {
                ans += 9 * (Convert.ToInt32(Math.Pow(10, i)));
            }
            else
            {
                ans += 0;
            }
        }
        if (len % 2 != 0)
        {
            ans += (N + 1) - (Convert.ToInt32(Math.Pow(10, len)));
        }
        Console.WriteLine(ans);
        
    }
}