How to resolve the algorithm Discordian date step by step in the Fortran programming language

Published on 12 May 2024 09:40 PM

How to resolve the algorithm Discordian date step by step in the Fortran programming language

Table of Contents

Problem Statement

Convert a given date from the   Gregorian calendar   to the   Discordian calendar.

Let's start with the solution:

Step by Step solution about How to resolve the algorithm Discordian date step by step in the Fortran programming language

Source code in the fortran programming language

program discordianDate
    implicit none
    ! Declare variables
    character(32) :: arg
    character(15) :: season,day,holyday
    character(80) :: Output,fmt1,fmt2,fmt3
    character(2) :: dayfix,f1,f2,f3,f4
    integer :: i,j,k,daysofyear,dayofweek,seasonnum,yold,dayofseason,t1,t2,t3
    integer,dimension(8) :: values
    integer, dimension(12) :: daysinmonth
    logical  :: isleapyear, isholyday, Pleapyear
     
    ! Get the current date
    call date_and_time(VALUES=values)
    ! Set some values up to defaults
    daysinmonth = (/ 31,28,31,30,31,30,31,31,30,31,30,31 /)
    isleapyear = .false.
    isholyday = .false.
    ! process any command line arguments
    ! using arguments dd mm yyyy 
    j = iargc()
    do i = 1, iargc()
      call getarg(i, arg)  ! fetches argument as a character string
      if (j==3) then
        if (i==1) then
          read(arg,'(i2)') values(3)  ! convert to integer
        endif
        if (i==2) then
          read(arg,'(i2)') values(2)  ! convert to integer
        endif
        if (i==3) then
          read(arg,'(i4)') values(1)  ! convert to integer
 
        endif
      endif
      if (j==2) then  ! arguments dd mm
        if (i==1) then
          read(arg,'(i2)') values(3)  ! convert to integer
        endif
        if (i==2) then
          read(arg,'(i2)') values(2)  ! convert to integer
        endif
      endif
      if (j==1) then ! argument dd
        read(arg,'(i2)') values(3)  ! convert to integer
      endif
    end do
 
    !Start the number crunching here 

    yold = values(1) + 1166
    daysofyear = 0
    if (values(2)>1) then 
    do i=1 , values(2)-1 , 1
        daysofyear = daysofyear + daysinmonth(i)
     end do
   end if
    daysofyear = daysofyear + values(3)
    isholyday = .false.
    isleapyear = Pleapyear(yold)
    dayofweek = mod (daysofyear, 5)
    seasonnum = ((daysofyear - 1) / 73) + 1 
    dayofseason = daysofyear - ((seasonnum - 1)  * 73)
    k = mod(dayofseason,10)  ! just to get the day number postfix
    select case (k)
      case (1)
         dayfix='st'
      case (2)
         dayfix='nd'
      case (3)
         dayfix='rd'
      case default
         dayfix='th'
    end select
    ! except between 10th and 20th  where we always have 'th'
    if (((dayofseason > 10) .and. (dayofseason < 20)) .eqv. .true.) then
      dayfix = 'th'
    end if
    select case (Seasonnum)
      case (1) 
        season ='Choas'
        f4 = '5'
      case (2) 
        season ='Discord'
        f4 = '7'
      case (3) 
        season ='Confusion'
        f4 = '9'
      case (4) 
        season ='Bureaucracy'
        f4 = '10'
      case (5) 
        season ='The Aftermath'
        f4 = '13'
    end select
    select case (dayofweek)
       case (0) 
         day='Setting Orange'
         f2 = '14'
       case (1) 
         day ='Sweetmorn'
         f2 = '9'
       case (2) 
         day = 'Boomtime'
         f2 = '8'
       case (3) 
         day = 'Pungenday'
         f2 = '9'
       case (4) 
         day = 'Prickle-Prickle'
         f2 = '15'
    end select
    ! check for holydays
    select case (dayofseason)
      case (5)
        isholyday = .true.
        select case (seasonnum)
          case (1)
             holyday ='Mungday'
             f1 = '7'
          case (2)
             holyday = 'Mojoday'
             f1 = '7'
          case (3)
            holyday = 'Syaday'
            f1 = '6'
          case (4)
             holyday = 'Zaraday'
             f1 = '7'
          case (5)
            holyday = 'Maladay'
            f1 = '7'
         end select
      Case (50)
        isholyday = .true.
        select case (seasonnum)
          case (1)
            holyday = 'Chaoflux'
            f1 = '8'
          case (2)
            holyday = 'Discoflux'
            f1 = '9'
          case (3)
            holyday = 'Confuflux'
            f1 = '9'
          case (4)
            holyday = 'Bureflux'
            f1 = '8'
          case (5)
            holyday = 'Afflux'
            f1 = '6'
         end select
    end select
    

    ! Check if it is St. Tibbs day
    if (isleapyear .eqv. .true.) then 
      if ((values(2) == 2) .and. (values(3) == 29)) then
         isholyday = .true.
      end if
    end if
    ! Construct our format strings
    f3 = "2"
    if (dayofseason < 10) then
      f3 = "1"
    end if
    fmt1 = "(a,i4)"
    fmt2 = "(A,a" // f1 // ",A,A" // f2 // ",A,I" // f3 // ",A2,A,A" // f4 // ",A,I4)"
    fmt3 = "(A,A" // f2 // ",A,I" // f3 //",A2,A,A" // f4 // ",A,I4)"
    ! print an appropriate line
    if (isholyday .eqv. .true.) then
      if (values(3) == 29) then
         print fmt1,'Celebrate for today is St. Tibbs Day in the YOLD ',yold
       else
         print fmt2, 'Today is ',holyday, ' on ',day,' the ',dayofseason,dayfix,' day of ',season,' in the YOLD ',yold 
       end if
     else   ! not a holyday
         print fmt3, 'Today is ',day,' the ',dayofseason,dayfix, ' day of ',season,' in the YOLD ',yold
     end if
    end program discordianDate

    ! Function to check to see if this is a leap year returns true or false!

    function Pleapyear(dloy) result(leaper)
    implicit none
    integer, intent(in) :: dloy
    logical :: leaper
    leaper = .false.
    if (mod((dloy-1166),4) == 0)  then
      leaper = .true.
    end if
    if (mod((dloy-1166),100) == 0) then
        leaper = .false.
        if (mod((dloy-1166),400)==0) then
           leaper = .true.
        end if
    end if 
    end function Pleapyear


  

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