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 Fortran (gfortran v4.8.4) (judge verdict: Accepted)

program main
  implicit none

  integer                :: n, odd
  
  read *, n

!  select case ( n )
!     
!  case ( 1:9 )
!     odd = n
!  case ( 10:99 )
!     odd = 9
!  case ( 100:999 )
!     odd = 9 + n - 99
!  case ( 1000:9999 )
!     odd = 9 + 900
!  case ( 10000:99999 )
!     odd = 9 + 900 + n - 9999
!  case ( 100000:999999 )
!     odd = 9 + 900 + 90000
!     
!  end select
!

  if ( n <= 9 ) then
     odd = n
  else if ( 10 <= n .and. n <= 99 ) then
     odd = 9
  else if ( 100 <= n .and. n <= 999 ) then
     odd = 9 + n - 99
  else if ( 1000 <= n .and. n <= 9999 ) then
     odd = 9 + 900
  else if ( 10000 <= n .and. n <= 99999 ) then
     odd = 9 + 900 + n - 9999
  else if ( 100000 <= n .and. n <= 999999 ) then
     odd = 9 + 900 + 90000
     
  end if
  
  print *, odd
  
end program main