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: Accepted)

using System;

namespace ABC136
{
    class A
    {
        static void Main(string[] args)
        {
            int N = int.Parse(Console.ReadLine());

            int sum = 0;
            if (1 <= N && N <= 9)
            {
                Console.WriteLine(N);
            }

            sum += 9;

            if (10 <= N && N <= 99)
            {
                Console.WriteLine(sum);
            }

            if (100 <= N && N <= 999)
            {
                Console.WriteLine(sum + N - 100 + 1);
            }

            sum += (999 - 100 + 1);

            if (1000 <= N && N <= 9999)
            {
                Console.WriteLine(sum);
            }

            if (10000 <= N && N <= 99999)
            {
                Console.WriteLine(sum + N - 10000 + 1);
            }

            sum += (99999 - 10000 + 1);

            if (N == 100000)
            {
                Console.WriteLine(sum);
            }
        }
    }
}