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# (.NET Core 3.1.201) (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)
        {
            int N = int.Parse(Console.ReadLine());
            int cnt = 0;

            for (int i = 1; i <= N; i++)
            {
                if (0 < i && i < 10)
                    cnt++;
                else if (99 < i && i < 1000)
                    cnt++;
                else if (9999 < i && i < 100000)
                    cnt++;
            }
            Console.WriteLine(cnt);
        }
    }
}