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 n = Int32.Parse(str);
            int ans = 0;
            bool isOdd = true;
            for(int i = 1; i <= n; i *= 10)
            {
                if (n > i && n < i * 10 && isOdd)
                {
                  int a=n/i==1?n%i+1:n-n%i;
                  ans += a;
                }
                else if (n >= i && !isOdd)
                {
                    ans += i-i/10;
                }
                isOdd = isOdd == true ? false : true;
            }
                          Console.WriteLine(ans);
        }
    }
}