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: Wrong Answer)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

namespace _136B
{
    class Program
    {
        static void Main(string[] args)
        {
            string strN = Console.ReadLine();
            int n = int.Parse(strN);

            double ans = 0;
            int keta = strN.Length;
            int a = 1;

            for (; a <= keta; a+=2)
            {
                if (Math.Pow(10, a) < n)
                {
                    ans += (Math.Pow(10, a) - 1) - (Math.Pow(10, a - 1)) + 1;
                }
            }

            if (keta % 2 == 1)
            {
                ans += n - (Math.Pow(10, a - 3)) + 1;
            }

            Console.WriteLine(ans);
        }
    }
}