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)

=begin
－－－－－－－－－－－－－－
  N = 入力値
  cnt = 桁数
  ans = 答え
－－－－－－－－－－－－－－
  digitsメソッドを作成(num)
    1: cnt += 1
    2: num /10
    3. num < 0 になったら cnt を返す
  メソッド－－－－－－－－－－

  1: 1 ~ N まで繰り返す(i)
  2: digits(i) % 2 が 0 以外なら ans += 1
  3: ans を出力
=end

def digits(num)
  cnt = 0;
  while num > 0
    cnt += 1;
    num /= 10;
  end
  return cnt;
end

N = gets.to_i;
ans = 0;
(1..N).each do |i|
  ans += 1 unless digits(i) % 2 == 0
end

puts ans