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;

class Program
{
    static void Main(string[] args)
    {
        var N = Int32.Parse(Console.ReadLine());
        var l = ((int)Math.Log10(N));
        var ans = 0;
        if(10000 <= N && N < 100000)
        {
            ans += ((N - 10000) + 1);
        }else if (100 <= N && N < 1000)
        {
            ans+= ((N - 100) + 1);
        }else if (1 <=N && N <10 )
        {
            ans += N;
        }
        if (l % 2 == 0) l--;
        while (-1 < l)
        {
            if (l % 2 == 0) ans += (9 * ((int)Math.Pow(10,(l))));
            l--;
        }
        Console.WriteLine(ans);
    }

}