I'm trying to output every number that evenly divides the input (so there is no remainder).
For example: if 60 is input, then the output should be:
1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60
I have programmed code that outputs 10 in this example: 50 / 5. But I cannot figure out how to modify this, so that I get every factor. I would appreciate some help!
IN
STO DIVID
IN
STO DIVIS
LOOP1 LDA count
ADD one
STO count
LDA DIVID
SUB DIVIS
STO DIVID
OUT DIVID
BRP LOOP1
HLT
DIVID DAT 000
DIVIS DAT 000
one DAT 001
count DAT 000
You'll need an extra, outer loop to manage the value of DIVIS, increasing it from 1 to DIVID.
Here is how that could look in a runnable snippet:
#input:60
LDA ONE
STO divisor
IN
BRZ invalid
STO dividend
next LDA dividend
loop BRZ output
STO remainder
LDA remainder
SUB divisor
BRP loop
BRA inc
output LDA divisor
OUT
inc LDA divisor
SUB dividend
BRP done
LDA divisor
ADD ONE
STO divisor
BRA next
invalid OUT ; zero
done HLT
ONE DAT 1 ; constant
divisor DAT
dividend DAT
remainder DAT
<script src="https://cdn.jsdelivr.net/gh/trincot/lmc@v0.74/lmc.js"></script>In case the input is 0 this program will consider it as "invalid" and output 0 and halt.
This is a quite basic algorithm, and not optimised for efficiency. Several optimisations are possible, which would make the code longer. For instance, every number has divisor 1, so you could output it immediately, and start the main loop with divisor=2 instead of 1. Similarly, once the divisor becomes greater than half of the dividend, you could exit the loop and output the dividend itself, as there are no other divisors to be found between dividend/2 and dividend.
More efficiency is achieved by implementing the long division algorithm.