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;
using System.Collections.Generic;
using System.IO;
using System.Linq;

namespace B
{
    public class Program
    {
        static void Main(string[] args)
        {
            var sw = new StreamWriter(Console.OpenStandardOutput()) { AutoFlush = false };
            Console.SetOut(sw);
            Solve();
            Console.Out.Flush();
        }

        public static void Solve()
        {
            var n = int.Parse(Console.ReadLine());
            var count = 0;
            var num = 1;
            while (num <= n)
            {
                if (num.ToString().Length % 2 == 0)
                {
                    num *= 10;
                    continue;
                }
                count++;
                num++;
            }
            Console.WriteLine(count);
        }
    }
}