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 bb
	n = gets.chomp.to_i
	if    n <=      9 then puts n
	elsif n <=     99 then puts 9
	elsif n <=    999 then puts 9 +   n-100+1
	elsif n <=   9999 then puts 9 + 999-100+1
	elsif n <=  99999 then puts 9 + 999-100+1 +     n-10000+1
	elsif n <= 999999 then puts 9 + 999-100+1 + 99999-10000+1
#	elsif n <= 100000 then puts 90909
	end
end

bb