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)

def func(num,total=0)
    if num < 1
        0
    elsif num == 1
        total + 9
    else
        func(num-2,total+10**(num-1)*9)
    end
end

num = gets.to_i
digits = num.to_s.length
ans = 0
if digits %2 == 0
    digits -= 1
    ans += func digits
else
    ans += num - 10**(digits-1) + 1
    ans += func digits-2
end
puts ans