C     This goal of program is to use a subroutine which can be used to
C     display a 1D array of real numbers.  The subroutine takes
C     arguments which give the stating address of the array, the number
C     of elements in the array, and the number of elements which should
C     be displayed per line of output.  The subroutine also shows on the
C     right side of the line the range of indices of the elements being
C     shown on that particular line.
C
C     John Schneider
C     CptS 203, HW 8, Problem 2.
C--------------------------------------------------------------------------
      program disp
      implicit none
      real a(200)
      integer i, nelem, perlin
      character filnam*50
      
C.....
C     Read the elements of the array from the specified file.  We
C     are allowed to assume no more than 200 elements will be entered.
C.....
      print *, 'Enter the file name:'
      read '(a)',filnam
      open(12,file=filnam,status='old')

      i=1
 10   read(12,*,end=20) a(i)
         i=i+1
      goto 10
 20   continue
      nelem=i-1

C.....
C     Report the number of elements read and prompt user for number of 
C     elements to be shown per line.
C.....
      print *, 'Number of elements ',nelem,'.'

      print *, 'Enter number of elements per line:'
      read *, perlin

      call showit(a,nelem,perlin)

      stop
      end

C.....
C     Subroutine to display one-dimensional array "a" with "perlin"
C     values per line of output.  The array has nelem elements.
C.....
      subroutine showit(a,nelem,perlin)
      implicit none
      real a(*)
      integer nelem, perlin, i, j, low, high

C.....
C     Write all the "fully populate" lines of output.  If perlin is one,
C     handle that as an exception to make output a bit neater.
C.....
      do 10 i=1,nelem/perlin
         if (perlin .ne. 1) then
            low  = (i-1)*perlin + 1
            high = low + perlin - 1
            print *,low,' --',high,': ',(a((i-1)*perlin+j),j=1,perlin)
         else
            print *,i,': ',a(i)
         endif
 10   continue

C.....
C     Check if the number of elements is not a multiple of perlin (could
C     also check this be comparing mod(nelem,perlin) to zero).  If not,
C     there are any left-over elements that we still have to display.
C.....
      if (nelem .ne. perlin*(nelem/perlin)) then
C........Find the indices for the elements still left to be displayed.
         low  = perlin*(nelem/perlin) + 1
         high = nelem
C........
C        Handle things differently depending on if there are one element or
C        more than elements to be displayed.
C........
         if (high .ne. low) then
            print *,low,' --',high,': ',(a(j),j=low,high)
         else
            print *,low,': ',(a(j),j=low,high)
         endif
      endif

      return
      end