Code-Golf: Modulus Divide - language-agnostic

Question

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Challenge:
Without using the modulus divide operator provided already by your language, write a program that will take two integer inputs from a user and then displays the result of the first number modulus divided number by the second number. Assume all input is positive.
Example:
Input of first number:2
Input of second number:2
Result:0
Who wins:
In case you don't know how Code Golf works, the winner is the person who writes this program in the least amount of characters.

Answer 1

CSS: 107 chars :)
CSS (ungolfed):
li {
counter-increment: a;
}
li:after {
content: counter(a);
}
li:nth-child(3n) { /* replace 3 with 2nd input ("b" in "a % b") */
counter-reset: a;
counter-increment: none;
}
Accompanying HTML:
<ol> <li></li> <li></li> <li></li> <!-- etc. --> </ol>
This doesn't work in IE (surprise surprise!).

Answer 2

J, 10 characters
([-]*<.#%)
Usage:
10 ([-]*<.#%) 3
1
J, 17 characters (with input as a list)
({.-{:*[:<.{.%{:)
Usage:
({.-{:*[:<.{.%{:) 10 3
1
({.-{:*[:<.{.%{:) 225 13
4
Explanation:
I took a totem pole and turned it into a smiley, and it worked.

Answer 3

Golfscript, 6 7 13 chars:
2*~/*-
Usage (only way to input into golfscript):
echo 14 3 | ruby golfscript.rb modulo.gs
2
Explanation:
2*~ #double the input string and eval (so now 14 3 14 3 are on the stack)
/ #int divide 14 / 3, gives quotient
*- #multiply that result by 3, subtract from 14, gives remainder

Answer 4

RePeNt, 5 chars
2?/*-
Run using:
RePeNt mod.rpn 17 3
RePeNt "2?/*-" 17 3
RePeNt is a stack-based toy language I made myself where every operator/command/loop is entered in Reverse Polish Notation (RPN). I will release the interpreter when I have tidied it up a bit.
Command Explanation Stack
------- ----------- -----
n/a The program takes 2 parameters ( 17 3 ) and pushes them 17 3
onto the stack
2 Pushes a 2 onto the stack 17 3 2
? Pops a number (x) off the stack + copies the last x 17 3 17 3
stack items onto the stack
/ Divides on stack 17 3 5
* Multiplies on stack 17 15
- Subtracts on stack 2

Answer 5

Ruby (32):
p(a=gets.to_i)-a/(b=gets.to_i)*b

Answer 6

Sure I won't win, but here goes nothing:
<?php
$a=readline("#1:");
$b=readline("#2:");
while($b<=$a)$a-=$b;
echo "Result: $a";

Answer 7

I know there's already two Ruby answers, but why not; getting the input this way is a different enough approach to knock off a few characters.
Ruby 1.8.7+, 29 chars
a,n=*$*.map(&:to_i);p a-a*n/n
$ ruby a.rb 10 3
1

Answer 8

C: 52
main(a,b){scanf("%d%d",&a,&b);printf("%d",a-a/b*b);}

Answer 9

Python: 25 chars
Behaves with negative numbers, identically to modulus operator. Takes two comma-separated numbers.
x,y=input()
print x-x/y*y

Answer 10

Clojure: 30 characters
#(if(>%2%1)%1(recur(-%1%2)%2)))

Answer 11

Unefunge-98: 14 13 22 chars
&:7p&:' \/*-.#
Unefunge is the 1-dimensional instance of Funge-98: http://quadium.net/funge/spec98.html
Explanation (Command <- Explaination [Stack]):
& <- Get integer input of value A and store on stack.
[A]
: <- Duplicate top of stack.
[A A]
7 <- Push 7 on stack. Used for the `p` command.
[A A 7]
p <- Pop top two values (7 then A). Place the character whose ASCII value
is A at position 7 in the code (where the space is).
[A]
& <- Get integer input of value B and store on stack.
[A B]
: <- Duplicate top of stack.
[A B B]
' <- Jump over next character and grap the ASCII value of the jumped character.
[A B B A]
<- Because of the `p` command, this is actually the character whose ASCII
value is A at this point in the code. This was jumped over by the
previous instruction.
\ <- Swap top two values of stack.
[A B A B]
/ <- Pop top two values (B then A). Push (A/B) (integer division) onto stack.
[A B (A/B)]
* <- Pop top two values ((A/B) then B). Push (B*(A/B)) onto stack.
[A (B*(A/B))]
- <- Pop top two values ((B*(A/B)) then A). Push (A-(B*(A/B))) onto stack.
[(A-(B*(A/B)))]
. <- Pop top value and print it as an integer.
[]
# <- Exit program.
Code tested is this incomplete (but complete enough) Unefunge-98 interpreter I wrote to test the code:
module Unefunge where
import Prelude hiding (subtract)
import qualified Data.Map as Map
import Control.Exception (handle)
import Control.Monad
import Data.Char (chr, ord)
import Data.Map (Map)
import System.Environment (getArgs)
import System.Exit (exitSuccess, exitFailure, ExitCode (..))
import System.IO (hSetBuffering, BufferMode (..), stdin, stdout)
-----------------------------------------------------------
iterateM :: (Monad m) => (a -> m a) -> m a -> m b
iterateM f m = m >>= iterateM f . f
-----------------------------------------------------------
data Cell = Integer Integer | Char Char
-----------------------------------------------------------
newtype Stack = Stack [Integer]
mkStack = Stack []
push :: Integer -> Stack -> Stack
push x (Stack xs) = Stack (x : xs)
pop :: Stack -> Stack
pop (Stack xs) = case xs of
[] -> Stack []
_:ys -> Stack ys
top :: Stack -> Integer
top (Stack xs) = case xs of
[] -> 0
y:_ -> y
-----------------------------------------------------------
data Env = Env {
cells :: Map Integer Cell
, position :: Integer
, stack :: Stack
}
withStack :: (Stack -> Stack) -> Env -> Env
withStack f env = env { stack = f $ stack env }
pushStack :: Integer -> Env -> Env
pushStack x = withStack $ push x
popStack :: Env -> Env
popStack = withStack pop
topStack :: Env -> Integer
topStack = top . stack
-----------------------------------------------------------
type Instruction = Env -> IO Env
cellAt :: Integer -> Env -> Cell
cellAt n = Map.findWithDefault (Char ' ') n . cells
currentCell :: Env -> Cell
currentCell env = cellAt (position env) env
lookupInstruction :: Cell -> Instruction
lookupInstruction cell = case cell of
Integer n -> pushInteger n
Char c -> case c of
'\''-> fetch
'\\'-> swap
'0' -> pushInteger 0
'1' -> pushInteger 1
'2' -> pushInteger 2
'3' -> pushInteger 3
'4' -> pushInteger 4
'5' -> pushInteger 5
'6' -> pushInteger 6
'7' -> pushInteger 7
'8' -> pushInteger 8
'9' -> pushInteger 9
' ' -> nop
'+' -> add
'-' -> subtract
'*' -> multiply
'/' -> divide
'#' -> trampoline
'&' -> inputDecimal
'.' -> outputDecimal
':' -> duplicate
'p' -> put
'#' -> stop
instructionAt :: Integer -> Env -> Instruction
instructionAt n = lookupInstruction . cellAt n
currentInstruction :: Env -> Instruction
currentInstruction = lookupInstruction . currentCell
runCurrentInstruction :: Instruction
runCurrentInstruction env = currentInstruction env env
nop :: Instruction
nop = return
swap :: Instruction
swap env = return $ pushStack a $ pushStack b $ popStack $ popStack env
where
b = topStack env
a = topStack $ popStack env
inputDecimal :: Instruction
inputDecimal env = readLn >>= return . flip pushStack env
outputDecimal :: Instruction
outputDecimal env = putStr (show n ++ " ") >> return (popStack env)
where
n = topStack env
duplicate :: Instruction
duplicate env = return $ pushStack (topStack env) env
pushInteger :: Integer -> Instruction
pushInteger n = return . pushStack n
put :: Instruction
put env = return env' { cells = Map.insert loc c $ cells env'}
where
loc = topStack env
n = topStack $ popStack env
env' = popStack $ popStack env
c = Char . chr . fromIntegral $ n
trampoline :: Instruction
trampoline env = return env { position = position env + 1 }
fetch :: Instruction
fetch = trampoline >=> \env -> let
cell = currentCell env
val = case cell of
Char c -> fromIntegral $ ord c
Integer n -> n
in pushInteger val env
binOp :: (Integer -> Integer -> Integer) -> Instruction
binOp op env = return $ pushStack (a `op` b) $ popStack $ popStack env
where
b = topStack env
a = topStack $ popStack env
add :: Instruction
add = binOp (+)
subtract :: Instruction
subtract = binOp (-)
multiply :: Instruction
multiply = binOp (*)
divide :: Instruction
divide = binOp div
stop :: Instruction
stop = const exitSuccess
tick :: Instruction
tick = trampoline
-----------------------------------------------------------
buildCells :: String -> Map Integer Cell
buildCells = Map.fromList . zip [0..] . map Char . concat . eols
eols :: String -> [String]
eols "" = []
eols str = left : case right of
"" -> []
'\r':'\n':rest -> eols rest
_:rest -> eols rest
where
(left, right) = break (`elem` "\r\n") str
data Args = Args { sourceFileName :: String }
processArgs :: IO Args
processArgs = do
args <- getArgs
case args of
[] -> do
putStrLn "No source file! Exiting."
exitFailure
fileName:_ -> return $ Args { sourceFileName = fileName }
runUnefunge :: Env -> IO ExitCode
runUnefunge = iterateM round . return
where
round = runCurrentInstruction >=> tick
main :: IO ()
main = do
args <- processArgs
contents <- readFile $ sourceFileName args
let env = Env {
cells = buildCells contents
, position = 0
, stack = mkStack
}
mapM_ (`hSetBuffering` NoBuffering) [stdin, stdout]
handle return $ runUnefunge env
return ()

Answer 12

Ruby: 36 chars
a,b=gets.split.map(&:to_i);p a-a/b*b

Answer 13

Scheme: 38
(define(m a b)(- a(*(quotient a b)b)))

Answer 14

JavaScript, 11 chars
a-b*(0|a/b)
Assumes input integers are contained the variables a and b:
a = 2;
b = 2;
alert(a-b*(0|a/b)); // => 0

Answer 15

PHP, 49 chars
Assuming query string input in the form of script.php?a=27&b=7 and short tags turned on:
<?echo($a=$_GET['a'])-(int)($a/$b=$_GET['b'])*$b;
(That could be shortened by four by taking out the single-quotes, but that would throw notices.)
With the vile register_globals turned on you can get it down to 25 chars:
<?echo $a-(int)($a/$b)*b;

Answer 16

Perl, 33 chars
Reading the inputs could probably be shortened further.
($a,$b)=#ARGV;print$a-$b*int$a/$b
Usage
$ perl -e "($a,$b)=#ARGV;print$a-$b*int$a/$b" 2457 766
159

Answer 17

Java. Just for fun
Assuming that s[0] and s[1] are ints. Not sure this is worth anything but it was a bit of fun.
Note that this won't suffer from the loop effect (large numbers) but will only work on whole numbers. Also this solution is equally fast no matter how large the numbers are. A large percentage of the answers provided will generate a huge recursive stack or take infinitely long if givin say a large number and a small divisor.
public class M
{
public static void main(String [] s)
{
int a = Integer.parseInt(s[0]);
int b = Integer.parseInt(s[1]);
System.out.println(a-a/b*b);
}
}

Answer 18

Bash, 21 chars
echo $(($1-$1/$2*$2))

Answer 19

C, 226 chars
Late entry: I decided to go for the least number of characters while avoiding arithmetic operations altogether. Instead, I use the file system to compute the result:
#include <stdio.h>
#define z "%d"
#define y(x)x=fopen(#x,"w");
#define g(x)ftell(x)
#define i(x)fputs(" ",x);
main(a,b){FILE*c,*d;scanf(z z,&a,&b);y(c)y(d)while(g(c)!=a){i(c)i(d)if(g(d)==b)fseek(d,0,0);}printf(z,g(d));}

Answer 20

Java: 127 Chars
import java.util.*;enum M{M;M(){Scanner s=new Scanner(System.in);int a=s.nextInt(),b=s.nextInt();System.out.println(a-a/b*b);}}
Note the program does work, but it also throws
Exception in thread "main" java.lang.NoSuchMethodError: main
after the inputs are entered and after the output is outputted.

Answer 21

Common Lisp, 170 chars (including indentation):
(defun mod-divide()
(flet((g(p)(format t"Input of ~a number:"p)(read)))
(let*((a(g"first"))(b(g"second")))
(format t "Result:~d~%"(- a(* b(truncate a b)))))))
Old version (187 characters):
(defun mod-divide()
(flet((g(p)(format t"Input of ~a number:"p)(read)))
(let*((a(g"first"))(b(g"second")))
(multiple-value-bind(a b)(truncate a b)(format t "Result:~d~%"b)))))

Answer 22

DC: 8 chars
odO/O*-p
$ echo '17 3 odO/O*-p' | dc
2

Answer 23

Java, 110 chars
class C{public static void main(String[]a){Long x=new Long(a[0]),y=x.decode(a[1]);System.out.print(x-x/y*y);}}

Answer 24

Rebmu: 10 chars (no I/O) and 15 chars (with I/O)
If I/O is not required as part of the program source and you're willing to pass in named arguments then we can get 10 characters:
>> rebmu/args [sbJmpDVjKk] [j: 20 k: 7]
== 6
If I/O is required then that takes it to 15:
>> rebmu [rJrKwSBjMPdvJkK]
Input Integer: 42
Input Integer: 13
3
But using multiplication and division isn't as interesting (or inefficient) as this 17-character solution:
rJrKwWGEjK[JsbJk]
Which under the hood is turned into the equivalent:
r j r k w wge j k [j: sb j k]
Documented:
r j ; read j from user
r k ; read k from user
; write out the result of...
w (
; while j is greater than or equal to k
wge j k [
; assign to j the result of subtracting k from j
j: sb j k
]
; when a while loop exits the expression of the while will have the
; value of the last calculation inside the loop body. In this case,
; that last calculation was an assignment to j, and will have the
; value of j
)

Answer 25

Perl 25 characters
<>=~/ /;say$`-$'*int$`/$'
usage:
echo 15 6 | perl modulo.pl
3

Answer 26

Haskell, 30 chars
m a b=a-last((a-b):[b,2*b..a])
This is my first code golf, be free to comment on code and post improvements. ;-)
I know I won't win, but I just wanted to share my solution using lists.

Answer 27

In ruby with 38 chars
p (a=gets.to_i)-((b=gets.to_i)*(a/b))
Not a winner :(

Answer 28

DC: 7 Chars (maybe 5 ;)
??37axp
Used as follows:
echo "X\nY" | dc -e "??37axp"
[And, referencing some other examples above, if input is allowed to be inserted into the code, it can be 5 chars:
37axp
as in:
dc -e "17 3 37axp"
Just thought it worth a mention]

Answer 29

F#, 268 chars
Did I win?
printf "Input of first number:"
let x = stdin.ReadLine() |> int
printf "Input of second number:"
let y = stdin.ReadLine() |> int
let mutable z = x
while z >= 0 do
z <- z - y
// whoops, overshot
z <- z + y
// I'm not drunk, really
printfn "Result:%d" z