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.to_i
ans = 0

def keta(m)
  ret = 0
  for i in 1..(m-1) do
    if i.even?
      ret += 10 ** i - 10 ** (i-1)
    end
  end
  ret
end

def keta2(m)
  ret = 0
  for i in 1..(m-1) do
    if i.odd?
      ret += 10 ** i - 10 ** (i-1)
    end
  end
  ret
end

for m in 1..6 do
  if n / 10 ** m == 0
    if m.odd?
      if m == 1
        ans = n
      else
        ans = n % 10 ** m - keta(m)
      end
    else
      ans = keta2(m)
    end
    break
  end
end

puts ans