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 ABC136
{
    class Program
    {
        static void Main(string[] args)
        {
            int n = int.Parse(Console.ReadLine());
            int ans = 0;
            if (n <= 9) { ans = n; }
            else if (n <= 99) { ans = 9; }
            else if (n <= 999) { ans = 9 + (n - 99); }
            else if (n <= 9999) { ans = 9 + 900; }
            else if (n <= 99999) { ans = 9 + 900 + (n - 9999); }
            else { ans = 9 + 900 + 90000; }
            Console.WriteLine(ans);
        }
    }
}