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;

class Program
{
    static void Main()
    {
        var n = int.Parse(Console.ReadLine());
        var ans = 0;
        for (int i = 1; i <= n; i++)
        {
            if (NumberOfDigits(i) % 2 != 0) 
            {
                ans++;
            }
        }
        Console.WriteLine(ans);
    }

    static int NumberOfDigits(int n)
    {
        int cnt = 1;
        while ((n /= 10) != 0)
        {
            cnt++;
        }
        return cnt;
    }
}