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;

namespace AtCoder0804
{
    class Program
    {
        static void Main(string[] args)
        {
            string str=Console.ReadLine();
            int l = str.Length;
            int n = Int32.Parse(str);
            int ans = 0;
            for(int i = 0; i <l; i++)
            {
                if (i==l-1　&& i%2==0)
                {
                    ans += n - (int)Math.Pow(10, i) + 1;
                }
                else if (i%2==1 && i!=l-1)
                {
                    ans += (int)Math.Pow(10, i)- (int)Math.Pow(10, i-1);
                }
            }
           Console.WriteLine(ans);
        }
    }
}