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;
using System.Linq;
using System.Collections.Generic;

namespace ForAtcoder
{
    class Program
    {
        static void Main(string[] args)
        {
            var n = long.Parse(Console.ReadLine());
            long num;
            var result = 0;
            for (var i = 1; i <= n; i++)
            {
                var count = 0;
                num = i;
                while (num > 0)
                {
                    num = num / 10;
                    count++;
                }
                if (count % 2 == 1) result++;
            }
            Console.WriteLine(result);
        }
    }
}