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 Ruby (2.3.3) (judge verdict: Accepted)

n = gets.chomp

digit_count = n.size

res = 0
(1..digit_count).each do |i|
  i.even? && next

  if i == 1
    if digit_count > 1
      res += 9
    else
      res += n.to_i
    end
    next
  end

  if i == digit_count
    x = '1'
    (1..i-1).each { x += '0' }
    res += n.to_i - x.to_i + 1
    next
  end

  res += 10**(i - 1) * 9
end

puts res