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-csc 3.5.0) (judge verdict: Wrong Answer)

using System;

namespace study11
{
    class Program
    {
        static void Main(string[] args)
        {
            int N = int.Parse(Console.ReadLine());
            int ans = 0;
            for (int K = 1; K <= N; K++)
            {
                var NA = (int)(Math.Log10(K) + 1);
                if (NA % 2 == 1)
                {
                    ans++;
                    Console.WriteLine("ansが加算されました");
                }
            }
            Console.WriteLine(ans);
        }
    }
}