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)

#1行目
n = gets.chomp.to_i

unchi = 0

if n < 10 #1桁
  unchi = n
elsif n >= 10 and n < 100
  unchi = 9
elsif n >= 100 and n < 1000
  unchi = n - 99 + 9
elsif n >= 1000 and n < 10000
  unchi = 909
elsif n >= 10000 and n < 100000
  unchi = n -9999 + 909
else
  unchi = 90909
end

print unchi