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;

namespace abc136_b
{
    static class Program
    {
        static void Main(string[] args)
        {
            var n = Console.ReadLine().ToInteger();
            // 何桁まであるか？
            var maxDigit = (int)Math.Log(n, 10) + 1;

            var count = 0;
            for (int i = 1; i <= maxDigit; i++)
            {
                // 奇数桁
                if (i % 2 != 0)
                {
                    if (maxDigit != i)
                    {
                        // 個数
                        count += (int)Math.Pow(10, i) - (int)Math.Pow(10, i - 1);
                    }
                    else
                    {
                        // 最終桁
                        count += n - (int)Math.Pow(10, i - 1) + 1;
                    }
                }
            }

            Console.WriteLine($"{count}");
        }

        static int ToInteger(this string s)
        {
            return int.Parse(s);
        }
    }
}