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 nn
  implicit none
  integer :: N,digi,i
  integer :: alle

  read(*,*) N

  digi=int(log(N*1.0)/log(10.0))+1
  alle=0
!  write(*,*) digi                                                              
  if(mod(digi,2).eq.0) then
     do i=1,digi/2
        alle=alle+9*100**(i-1)
!        write(*,*) alle                                                        
     enddo
  else
    do i=1,digi/2
        alle=alle+9*100**(i-1)
     enddo
     alle=alle+N-(10**(digi-1))+1
  endif

  write(*,'(i0)') int(alle)
end program nn