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.chomp.to_i

if N == 100000
  p 90000 + 900 + 9
elsif N <= 99999 && N >= 10000
  p (N - 9999) + 900 + 9
elsif N <= 9999 && N >= 1000
  p 900 + 9
elsif N <= 999 && N >= 100
  p (N - 99) + 9
elsif N <= 99 && N >= 10
  p 9
else
  p N
end