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.IO;
using static System.Console;
using static System.Math;


namespace AtCodeeeer
{
    class AtCOOOOOOOOOOOder
    {
        static void Main(string[] args)
        {
            int n = int.Parse(ReadLine());
            int ans = 0;
            for (int i = 0; i < n; i++)
            {
                if (cal(i)%2 == 1)
                {
                    ans++;
                }
            }
            WriteLine(ans);
        }

        static int cal(int num)
        {
            int tmp = 0;
            while (num != 0)
            {
                num /= 10;
                tmp++;
            }
            return tmp;
        }
    }
}