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)

include Math
n = gets.chomp.to_i
digit_cout = log10(n).to_i
ans = 0
case digit_cout
when 5
  ans = 90909
when 4
  ans = (10000..n).count + (100..999).count + (1..9).count
when 3
  ans = (100..999).count + (1..9).count
when 2
  ans = (100..n).count + (1..9).count
when 1
  ans = (1..9).count
when 0
  ans = (1..n).count
end
p ans