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: Wrong Answer)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

namespace ABC_Training
{

    class Program
    {
        static void Main(string[] args)
        {
            var ns = Console.ReadLine();
            var nf = ns.Length;
            var n = int.Parse(ns);
            if (nf % 2 == 0)
            {
                if (nf == 2)
                {
                    Console.WriteLine("9");
                }else if (nf == 4)
                {
                    Console.WriteLine("909");
                }else
                {
                    Console.WriteLine("90909");
                }
            }
            else
            {
                if (nf == 1)
                {
                    Console.WriteLine(n);
                }else if (nf == 3)
                {
                    Console.WriteLine(n % 100 + 1 + 9);
                }else 
                {
                    Console.WriteLine(n % 10000 + 1 + 909);
                }
            }
        }
    }
}