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, A, B = gets.split(" ").map(&:to_i)
# S = gets.chomp
N = gets.to_i

ans = 0
case
when N >= 100000
  ans = (100000 - 10000) + (1000 - 100) + (10 - 1)
when N >= 10000
  ans = (N + 1 - 10000) + (1000 - 100) + (10 - 1)
when N >= 1000
  ans = (1000 - 100) + (10 - 1)
when N >= 100
  ans = (N + 1 - 100) + (10 - 1)
when N >= 10
  ans = (10 - 1)
when N >= 1
  ans = N - 1 + 1
end
puts ans