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 AtCoderPractice
{
    class Program
    {
        static void Main(string[] args)
        {

            string[] line = Console.ReadLine().Split(' ');
            int n = int.Parse(line[0]);

            // n以下の
            // 5桁の数 … 10000~99999 90000個
            // 3桁の数 … 100~999 900個
            // 1桁の数 … 1~9 9個
            int result = 0;

            if (n < 100)
            {
                result = n >= 9 ? 9 : n;
            }
            else if (n < 10000)
            {
                result = 9 + (n >= 999 ? 900 : n - 100 + 1);
            }
            else
            {
                result = 9 + 900 + (n >= 99999 ? 90000 : n - 10000 + 1);
            }

            Console.WriteLine(result);
        }
    }
}