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;
class ProgramB
    {
        static void Main(string[] args)
        {
            string input = Console.ReadLine();
            int n = int.Parse(input);
            
            if(n <= 9)
                Console.WriteLine(n);
            else if(n <= 99)
                Console.WriteLine("9");
            else if(n <= 999)
                Console.WriteLine(n -90);
            else if(n <= 9999)
                Console.WriteLine("909");
            else if(n <= 99999)
                Console.WriteLine(n -9090);
            else
                Console.WriteLine("90909");
        }
    }