Converting numbers into roman numerals..

[dave]

I was wondering if anyone had an easy way I could go at this...

I need to go up to 2003, and I could do it in 2027 lines of code [bunch of if statements, lol] but I know you could do it with a DO loop.

If anyone has any ideas for me, please reply.

[Agamemnus]

http://www.geocities.com/pisforpi/flyingsoft

"Roman Numerals". Downloads section. Has source code.

[oracle]

I need to go up to 2003, and I could do it in 2027 lines of code [bunch of if statements, lol] but I know you could do it with a DO loop.

Sounds suspiciously like:

Code:

IF num$ = 1 THEN PRINT "I"
IF num$ = 2 THEN PRINT "II"
...
IF num$ = 2003 THEN PRINT "MMIII"

:rotfl:

[dave]

you got it

[Sterling Christensen]

Woah, you didn't do all of it by hand, did you?

[Agamemnus]

My program is 100% bug free....

[relsoft]

I need to go up to 2003, and I could do it in 2027 lines of code [bunch of if statements, lol] but I know you could do it with a DO loop.

Sounds suspiciously like:

Code:

IF num$ = 1 THEN PRINT "I"
IF num$ = 2 THEN PRINT "II"
...
IF num$ = 2003 THEN PRINT "MMIII"

:rotfl:

Dang!!! I just remembered that I posted something like that for an odd even number problem.

Welcome to the club!!!

[oracle]

I remember that!

Code:

IF num = 1 THEN PRINT "Odd."
IF num = 2 THEN PRINT "Even."
IF num = 3 THEN PRINT "Odd."
...
IF num = 600000 THEN PRINT "Even."

:rotfl:

I seem to recall that the person asking for it had trouble compiling it ;*)

[dave]

First off I was just gonna do it in an array with all the set roman numerals like M, D, L, V etc...and whenever it wasnt one of those then do the math to get the number..

I had like

DIM ones$(9)
LET ones$(1) = "I"
LET ones$(2) = "II"
LET ones$(3) = "III"
....

DIM hundreds$(900)
....
LET hundreds$(500) = "M"
....


you get the point.

[oracle]

Nightmare... :lol:

Is agamemnus's program good for you?

ps: LET is ancient. You can completely omit the word LET in front of a variable, and QB will understand.

[dave]

Agamemnus's program is great..

Thanks for the tip, too.

[Agamemnus]

Great, I'm glad my program has made a difference in the world.

[Antoni Gual]

Code:

DECLARE SUB roman (a%, I%)
'
DIM SHARED b$
b$ = "IVXLCDM"
DO
DO
INPUT "Enter an integer >0 and <4000 (0 to exit)"; a%
IF a% < 1 THEN END
LOOP UNTIL a% < 4000
PRINT
roman a%, 1
PRINT
LOOP
END

SUB roman (a%, I%)
IF a% = 0 THEN EXIT SUB
n% = a% MOD 10
roman a% \ 10, I% + 2
SELECT CASE n%
CASE 9:
 PRINT MID$(b$, I%, 1) + MID$(b$, I% + 2, 1);
CASE 8, 7, 6:
 PRINT MID$(b$, I% + 1, 1) + STRING$(n% - 5, ASC(MID$(b$, I%, 1)));
CASE 5:
 PRINT MID$(b$, I% + 1, 1);
CASE 4:
 PRINT MID$(b$, I%, 1) + MID$(b$, I% + 1, 1);
CASE 3, 2, 1:
 PRINT STRING$(n%, ASC(MID$(b$, I%, 1)));
END SELECT
END SUB

[Neo]

Looks great! Only beware that QB doesn't have such a good stack-handler so don't recurse too much...

[Antoni Gual]

Three levels is not too much... It would be different if i was calculating a factorial