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 AC_5
{
    class Program
    {
        static void Main(string[] args)
        {
            string[] input = Console.ReadLine().Split(' ');
            int A = int.Parse(input[0]);
            if (A < 10)
            {
                Console.WriteLine(A);
            }
            else if (10 <= A && A < 100)
            {
                Console.WriteLine(9);
            }
            else if (100 <= A && A < 1000)
            {
                Console.WriteLine((A - 100) + 10);
            }
            else if (1000 <= A && A < 10000)
            {
                Console.WriteLine(909);
            }
            else if (10000 <= A && A < 11000)
            {
                Console.WriteLine((A - 10000) + 911);
            }
            else if (11000 <= A && A < 100000)
            {
                Console.WriteLine(11910);
            }
            else if (A == 100000)
            {
                Console.WriteLine(90909);
            }
         
        }
    }
}