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;
class Program
{
  static void Main (string[] args)
  {
    int n = int.Parse(Console.ReadLine());
    int count = 0;
    
    if (n < 10) { count = n; }
    if (10 <= n && n < 100) { count = (n % 10); }
    if (100 <= n && n < 1000) { count = ((n % 100) + 9); }
    if (1000 <= n && n < 10000) { count = ((n % 100) + 9); }
    if (10000 <= n && n < 100000) { count = ((n % 10000) + 9 + 100); }
    if (100000 <= n) { count = ((n % 10000) + 9 + 100); }
    Console.WriteLine(count);
  }
}