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.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

namespace ConsoleApp2
{
    class Program
    {
        static void Main(string[] args)
        {
            var N = Console.ReadLine().Split().Select(s => int.Parse(s)).ToArray()[0];
            var count = 0;
            for(var i = N;0<i;i--)
            {
                if (0 < i && i < 10)
                {
                    count++;
                    continue;
                }
                if (99 < i && i < 1000)
                {
                    count++;
                    continue;
                }
                if (9999 < i && i < 100000)
                {
                    count++;
                    continue;
                }
            }
            Console.WriteLine(count);
        }
    }
}