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 question2
{
    class Program
    {
        static void Main(string[] args)
        {
            int N = int.Parse(Console.ReadLine());
            int digit = 0;
            int amari = 0;
            int[] numbers = new int[7];
            int[] all = new int[7];
            numbers[0] = 1;
            numbers[1] = 10;
            numbers[2] = 100;
            numbers[3] = 1000;
            numbers[4] = 10000;
            numbers[5] = 100000;
            numbers[6] = 1000000;
            all[0] = 9;
            all[1] = 0;
            all[2] = 900;
            all[3] = 0;
            all[4] = 90000;
            all[5] = 0;
            all[6] = 9000000;
            int res = 0;
            for (int i =1; i < 7; i++)
            {
                if (N < numbers[i])
                {
                    digit = i-1;
                    amari = N % numbers[i-1];
                    if (all[i-1]!=0) res += 1 + amari;
                    break;
                }
                res += all[i-1];
            }
            Console.WriteLine(res.ToString());
            //Console.ReadLine();
        }
    }
}