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;

namespace B
{
    class Program
    {
        static void Main(string[] args)
        {
            int n = int.Parse(Console.ReadLine());
            int ans = 0;
            for(int i = 1; i <= n; i++)
            {
                if (Keta(i) % 2 == 1) ans++;
            }
            Console.WriteLine(ans);
        }
        static public int Keta(int i)
        {
            int cnt =0;
            while (i != 0)
            {
                i /= 10;
                cnt++;
            }
            return cnt;
        } 
    }
}