The Longest Chain (SPOJ)

Well, certainly I don’t have time to work on this right now, but this exercise has caught my eye, and it will be the first one I will look into when I have some spare time. Getting a low score on this one is very easy, but getting a maximum score of 1000 ….
bch

Anti-blot system (SPOJ)

I just decided a couple of days ago to start a daily “programming gym” session, to keep my programming skills sharp, since otherwise there are long periods of time in which I don’t do any programming, and the “trade” suffers.

Languages to practice will vary, but at present the focus will be in Fortran. And the problems to solve will come from Sphere Online Judge.

The first program to try has been ABSYS, and here is the code for it:

PROGRAM ABSYS
IMPLICIT NONE

CHARACTER :: mas, igual
CHARACTER (LEN=50) :: dato1,dato2,resultado
INTEGER :: sets,count,idato1,idato2,iresultado

READ*, sets

DO count=1,sets
READ*, dato1,mas,dato2,igual,resultado
CALL convert(dato1,idato1)
CALL convert(dato2,idato2)
CALL convert(resultado,iresultado)

IF (idato1 .EQ. -1) WRITE(*,’(I0,A,I0,A,I0)’) iresultado-idato2, ” + “, idato2, ” = “, iresultado
IF (idato2 .EQ. -1) WRITE(*,’(I0,A,I0,A,I0)’) idato1, ” + “, iresultado-idato1, ” = “, iresultado
IF (iresultado .EQ. -1) WRITE(*,’(I0,A,I0,A,I0)’) idato1, ” + “, idato2, ” = “, idato1+idato2
END DO

CONTAINS

SUBROUTINE convert(dato,idato)
CHARACTER (LEN=*) :: dato
INTEGER :: idato, pos

pos = INDEX(dato,”machula”)
IF (pos .EQ. 0) THEN
READ (dato,’(I50)’), idato
ELSE
idato = -1
END IF

END SUBROUTINE convert

END PROGRAM ABSYS

The program is trivial, but there are a couple of things to note:

In Fortran, to print a number with just the minimum necessary width, the edit descriptor is I0.

There are no intrinsic functions to convert a string to a number or viceversa, but instead we use what it is called an internal file, and we can write something like “READ (dato,’(I50)’), idato”