program main

!*****************************************************************************80
!
!! task_division_test() tests task_division().
!
!  Discussion:
!
!    This program simply demonstrates how one might automate the
!    assignment of T tasks to P processors, assuming that the assignment
!    is to be beforehand.
!
!    In that case, we just want to make sure that we assign each task
!    to a processor, that we assign about the same number of tasks
!    to each processor, and that we assign each processor a contiguous
!    range of tasks, say tasks I_LO to I_HI.
!
!    The routine that is called simulates this process.
!
!  Licensing:
!
!    This code is distributed under the MIT license.
!
!  Modified:
!
!    23 October 2011
!
!  Author:
!
!    John Burkardt
!
  implicit none

  integer proc_first
  integer proc_last
  integer task_number

  call timestamp ( )
  write ( *, '(a)' ) ' '
  write ( *, '(a)' ) 'task_division_test():'
  write ( *, '(a)' ) '  FORTRAN90 version'
  write ( *, '(a)' ) '  Demonstrate how to automate the division of'
  write ( *, '(a)' ) '  T tasks among a range of P processors'
  write ( *, '(a)' ) '  indexed from PROC_FIRST to PROC_LAST.'

  task_number = 23
  proc_first = 0
  proc_last = 3
  call task_division ( task_number, proc_first, proc_last )

  task_number = 17
  proc_first = 1
  proc_last = 6
  call task_division ( task_number, proc_first, proc_last )

  task_number = 17
  proc_first = 4
  proc_last = 6
  call task_division ( task_number, proc_first, proc_last )

  task_number = 5
  proc_first = -2
  proc_last = 6
  call task_division ( task_number, proc_first, proc_last )

  task_number = 5
  proc_first = 0
  proc_last = 4
  call task_division ( task_number, proc_first, proc_last )

  task_number = 5
  proc_first = 0
  proc_last = 0
  call task_division ( task_number, proc_first, proc_last )

  task_number = 1000
  proc_first = 1
  proc_last = 17
  call task_division ( task_number, proc_first, proc_last )
!
!  Terminate.
!
  write ( *, '(a)' ) ' '
  write ( *, '(a)' ) 'TASK_DIVISION_TEST():'
  write ( *, '(a)' ) '  Normal end of execution.'
  write ( *, '(a)' ) ' '
  call timestamp ( )

  stop 0
end
subroutine timestamp ( )

!*****************************************************************************80
!
!! timestamp() prints the current YMDHMS date as a time stamp.
!
!  Example:
!
!    31 May 2001   9:45:54.872 AM
!
!  Licensing:
!
!    This code is distributed under the MIT license.
!
!  Modified:
!
!    06 August 2005
!
!  Author:
!
!    John Burkardt
!
!  Parameters:
!
!    None
!
  implicit none

  character ( len = 8 ) ampm
  integer d
  integer h
  integer m
  integer mm
  character ( len = 9 ), parameter, dimension(12) :: month = (/ &
    'January  ', 'February ', 'March    ', 'April    ', &
    'May      ', 'June     ', 'July     ', 'August   ', &
    'September', 'October  ', 'November ', 'December ' /)
  integer n
  integer s
  integer values(8)
  integer y

  call date_and_time ( values = values )

  y = values(1)
  m = values(2)
  d = values(3)
  h = values(5)
  n = values(6)
  s = values(7)
  mm = values(8)

  if ( h < 12 ) then
    ampm = 'AM'
  else if ( h == 12 ) then
    if ( n == 0 .and. s == 0 ) then
      ampm = 'Noon'
    else
      ampm = 'PM'
    end if
  else
    h = h - 12
    if ( h < 12 ) then
      ampm = 'PM'
    else if ( h == 12 ) then
      if ( n == 0 .and. s == 0 ) then
        ampm = 'Midnight'
      else
        ampm = 'AM'
      end if
    end if
  end if

  write ( *, '(i2,1x,a,1x,i4,2x,i2,a1,i2.2,a1,i2.2,a1,i3.3,1x,a)' ) &
    d, trim ( month(m) ), y, h, ':', n, ':', s, '.', mm, trim ( ampm )

  return
end