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;

namespace abc136B
{
    class MainClass
    {
        public static void Main(string[] args)
        {
            string A = Console.ReadLine();
            int C = int.Parse(A);
            int length = A.Length;
            
            double Ans = 0;
            for(int i = 1; i < length; i+=2)
            {
                Ans = Ans + (Math.Pow(10,i)-Math.Pow(10,i-1));
                
            }
            
            double B = 0;
            if (C % 10 != 0)
                B = Ans + (C - (Math.Pow(10, (length - 1))) + 1);
            else
                B = Ans;
            Console.WriteLine(B);
        }
    }
}