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)

//ABC136 B - Uneven Numbers
//20190905

using System;
using System.Linq;

namespace B
{
    class Program
    {
        static void Main(string[] args)
        {
            // Console.WriteLine("Hello World!");
            int N = int.Parse(Console.ReadLine());
            double OUTPUT = 0;

            if (N < 10)
            {
                // OUTPUT = (int)(N / 2.0 + 0.5);
                OUTPUT = N;
            }
            else if (N < 100)
            {
                // OUTPUT = 5;
                OUTPUT = 9;
            }
            else if (N < 1000)
            {
                // OUTPUT = 5 + (int)((N - 100) / 2.0 + 0.5);
                OUTPUT = 9 + N - 100 + 1;
            }
            else if (N < 10000)
            {
                // OUTPUT = 5 + 4950;
                OUTPUT = 9 + 900;
            }
            else if (N < 100000)
            {
                // OUTPUT = 5 + 4950 + (int)((100000 - 10000) / 2.0 + 0.5);
                OUTPUT = 9 + 900 + N - 1000 + 1;
            }
            else
            {
                OUTPUT = 9 + 900 + 90000 ;
            }

            Console.WriteLine(OUTPUT);

        }
    }
}