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;
using System.Collections.Generic;
using System.Linq;

class Program
{
    static void Main(string[] args)
    {
        var input = Console.ReadLine();
        int N = int.Parse(input);
        int ans = 0;
        if (N == 100000) { ans = 90909; }
        else if (N >= 10000) { ans = N - 9999+900+9; }
        else if (N >= 1000) { ans = 909; }
        else if (N >= 100) { ans = N - 99 + 9; }
        else if (N >= 10) { ans = 9; }
        else { ans = N; }
        Console.WriteLine(ans);
    }
}