BitCycle, (削除) 146 (削除ここまで) (削除) 79 (削除ここまで) 64 bytes

Just realized a whole section of my original code was unneccessary. Huge reduction!

v <>!
A\B^^=
? D^>^<
 >\v^~ D@
 >/ C/
 > C ^
A/B v
^ <
? D^

The program takes input in unary from the command line, with the divisor first. It outputs the quotient and remainder in unary, separated by a 0. For example, here's a=11, b=4, a/b=2, a%b=3:

C:\>python bitcycle.py divmod.btc 1111 11111111111
110111

Ungolfed, in action

Here's my ungolfed version computing a=3, b=5 with animation turned on (sorry about the glitchiness):

Divmod running in BitCycle

Attempt at an explanation

The explanation applies to the ungolfed version. Before you tackle it, I highly recommend you read the Esolangs page to get a feel for how the language works.

The algorithm goes like this:

• Run an outer loop until the program is terminated.
  • Run an inner loop over the bits of the divisor, pairing them off with bits from the dividend.
    • If all bits of the divisor have matching dividend bits, output a single bit.
    • If not all bits of the divisor have matching dividend bits, output the separator 0 followed by what dividend bits there were, then terminate.

The heart of the code is the relationships among the collectors (the uppercase letters). Since there are multiple separate collectors with each letter, let's refer to them as A1, A2, B1, B2, etc., numbering from top to bottom.

• A1 and A2 hold the divisor and dividend, respectively, at the beginning of the main loop.
• The inner loop peels off one bit at a time from the divisor and the dividend.
  • The rest of the divisor, if any, always goes into B1.
  • If both the divisor and dividend were nonempty, one bit goes into C1 and one into C3. The rest of the dividend goes into B2.
  • If only the divisor was nonempty, we've reached the end of the dividend, and it's time to print the remainder. The bit from the divisor goes into C2.
  • If only the dividend was nonempty, we've reached the end of the divisor; it's time to process the bits in C3 or C2 for output. The rest of the dividend goes into C4.
• If there are any bits in the B collectors, they cycle their contents back around to the A collectors and continue in the inner loop.
• Once the A and B collectors are all empty, the C collectors open and we proceed to the processing stage:
  • C1 and C4 dump their contents (the divisor and the remaining dividend, respectively) into D1 and D3.
  • If C2 is empty, we're still printing the quotient.
    • The contents of C3 go up to the top right = switch. The first 1 bit passes straight through to ! and is output.
    • When the 1 bit passes through, it activates the switch to point rightward, which sends all the subsequent bits off the board.
  • If C2 is not empty, we're printing the remainder.
    • The first bit of C2 is negated to a 0 and passed through the switch. The 0 goes on to ! and is output.
    • When the 0 bit passes through, it activates the switch to point leftward. Now all the bits from C3 go leftward from the switch and are redirected around into the !, outputting the entire remainder.
    • A copy of the first bit from C2 is also sent into D2.
• Now the D collectors open.
  • If there is anything in D2, that means we just printed the remainder. The bit from D2 hits the @, which terminates the program.
  • Otherwise, the contents of D1 and D3 loop back into A1 and A2 respectively, and the main loop starts over.

• \$\begingroup\$ that is awesome \$\endgroup\$

  Evan Carslake

  – Evan Carslake

  2017年03月30日 17:33:20 +00:00

  Commented Mar 30, 2017 at 17:33
• \$\begingroup\$ "The program takes input in unary from the command line": That looks like binary to me? \$\endgroup\$

  2xsaiko

  – 2xsaiko

  2017年03月30日 18:36:28 +00:00

  Commented Mar 30, 2017 at 18:36
• \$\begingroup\$ Whoops. Because the output looked like binary, I though the input should be too. Then I read the text. Nevermind. :P \$\endgroup\$

  2xsaiko

  – 2xsaiko

  2017年03月31日 15:37:02 +00:00

  Commented Mar 31, 2017 at 15:37

Funciton, (削除) 224 (削除ここまで) 108 bytes

Byte count assumes UTF-16 encoding with BOM.

 ┌──┬───┐
┌┴╖╓┴╖ ┌┴╖
│%╟║f╟┐│÷╟┘
╘╤╝╙─╜│╘╤╝
 └────┴─┘

Try it online!

The above defines a function f, which takes two integers and returns both their division and their product (functions in Funciton can have multiple outputs as long as the sum of inputs and outputs doesn't exceed 4).

Using two input values for multiple purposes is actually quite trivial: you simply split off the connector with a T-junction at the value will be duplicated along both branches, which we can then feed separately to the built-ins for division and modulo.

It actually took me twice as long to figure out how to display the result to the user than just to implement the solution.

Also, Funciton has a built-in divmod, ÷%, and amusingly the built-ins ÷ and % that my solution uses are implemented in terms of ÷%. However, my function f above isn't quite identical to ÷%: I had to swap the order of the inputs and although it seems like it should be easy to change that, so far I haven't been able to do so without increasing the byte count.

brainfuck, (削除) 43 (削除ここまで) 41 bytes

,<,[>->+<[>]>>>>+<<<[<+>-]<<[<]>-]>>.>>>.

This uses a modified version of my destructive modulus algorithm on Esolangs.

The program reads two bytes – d and n, in that order – from STDIN and prints two bytes – n%d and n/d, in that order – to STDOUT. It requires a brainfuck interpreter with a doubly infinite or circular tape, such as the one on TIO.

Try it online!

How it works

Before the program starts, all cells hold the value 0. After reading d from STDIN (,), moving one step left (<) and reading n from STDIN (,), the tape looks as follows.

 v
A B C D E F G H J
0 n d 0 0 0 0 0 0

Next, assuming that n> 0, we enter the while loop

[>->+<[>]>>>>+<<<[<+>-]<<[<]>-]

which transforms the tape as follows.

First, >->+< advances to cell C and decrements it, then advances to cell D and increments it, and finally goes back to cell C. What happens next depends on whether the value of cell C is zero or not.

• If cell C hold a positive value, [>] (go right while the cell is non-zero) will advance to cell E.

  >>>>+<<< advances to cell J to increment it, then goes back to cell F.

  Since cell F will always hold 0, the while loop [<+>-] is skipped entirely, and << goes back to cell D.

  Finally, since neither D nor C hold 0, [<] (go left while the cell is non-zero) will retrocede to cell A.

• If cell C holds 0, the loop [>] is skipped entirely; >>>>+<<< advances to cell G to increment it, then goes back to cell D.

  At this point, D will hold d (in fact, the sum of the values in C and D will always be d), so [<+>-] (while D is positive, increment C and decrement D) will set C to d and D to 0.

  Finally, << retrocedes to cell B, [<] (go left while the cell is non-zero) further left to cell A.

In both cases, >- advances to cell B and decrements it, and the loop starts over unless this zeroes it out.

After k iterations, the tape looks as follows.

 v
A B C D E F G H J
0 n-k d-k%d k%d 0 0 k/d 0 k-k/d

After n iterations B is zeroed out and we break out of the loop. The desired values (n%d and n/d) will be stored in cells D and G, so >>.>>>. prints them.

Regex (ECMAScript / Python), 57 bytes

(x(x*)),(x*?)(?=1円*$)(x?(x*))(?=4円*$)((?=2円+$)2円5円*$|$4円)

Try it online! - ECMAScript
Try it online! - Python

Takes its arguments in unary, as two strings of x characters whose lengths represent the numbers. The divisor comes first, followed by a , delimiter, followed by the dividend. The quotient and remainder are returned in the capture groups 4円 and 3円, respectively.

I developed the basic form of this on 2014年04月03日 while working on my abundant numbers regex. It had been a month earlier that teukon and I had independently come up with the multiplication algorithm described here, but up until this point we hadn't written or golfed any regex to find the unknown quotient of a known dividend and divisor. And it wasn't until dividing by \$\sqrt 2\$ that I adapted it to handle a dividend of zero correctly.

It is used, in its various forms (shown above or below), by these other regexes:

• Abundant numbers
• Fibonacci numbers - the version that returns the index
• OEIS A033286 (\$np_n)\$
• Factorial numbers
• Proth numbers
• Consecutive-prime/constant-exponent numbers
• Is the number binary-heavy?
• Euler's totient function by Grimmy
• Shift right by half a bit
• Decompose a number!
• Decide symmetry of fractions

Commented and indented:

(x(x*)), # 1円 = divisor; 2円 = 1円-1; tail = dividend
(x*?)(?=1円*$) # 3円 = remainder of division; tail -= 3円
(x?(x*)) # 4円 = conjectured quotient - find the largest one that matches the
 # following assertions; 5円 = 4円-1, or 0 if 4円==0; tail -= 4円
(?=4円*$) # assert tail is divisible by quotient
(
 (?=2円+$) # assert tail is positive and divisible by divisor-1
 2円5円*$ # assert tail-(divisor-1) is divisible by quotient-1
|
 $4円 # if dividend == 0, assert that quotient == dividend
)

To show what's going on, here is a Python function that does division using the regex's algorithm:

def modulo_is_zero(n, modulus):
 return n==0 if modulus==0 else n>=0 and n % modulus == 0
def divide(dividend, divisor):
 if divisor==0:
 return
 remainder = dividend % divisor
 dividend -= remainder
 quotient = dividend
 if dividend != 0:
 while 1:
 if modulo_is_zero(dividend - quotient, quotient) and \
 modulo_is_zero(dividend - quotient, divisor-1) and \
 modulo_is_zero(dividend - quotient - (divisor-1), quotient-1):
 break
 quotient -= 1
 if quotient == 0:
 return
 return [quotient, remainder]

Try it online!

Notice that in regex, \0ドル \equiv 0 \pmod 0\$. This is very convenient, as it allows a quotient of \1ドル\$ to be returned with no special case. The only special case we actually need to handle is when the dividend is zero.

First, to demonstrate that the correct answer satisfies the assertions used – assume the remainder has already been subtracted away, and we are dividing \$B=C/A\$:

\$\begin{aligned} C &= AB \\ C-B &= (A-1)B \\ C-B-(A-1) &= (A-1)(B-1) \end{aligned}\$

Thus,

\$\begin{aligned} C &\equiv 0 \pmod A \\ C &\equiv 0 \pmod B \\ C-B &\equiv 0 \pmod {A-1} \\ C-B-(A-1) &\equiv 0 \pmod {B-1} \end{aligned}\$

The last two can be simplified to:

\$\begin{aligned} C &\equiv B \pmod {A-1} \\ C &\equiv A \pmod {B-1} \end{aligned}\$

At this point it becomes apparent why neither of these two alone would be enough. If \$A<B\$, it looks like the first one could yield false positives, and if \$B<A\$, it looks like the second one could. It turns out that only one of them is needed if we're guaranteed the input meets certain constraints, but until recently I simply accepted this intuitively and from testing the algorithm up to large numbers.

Unlike the multiplication algorithm, this division algorithm can't simply be proved correct from the Chinese remainder theorem, because \$C\$ is now constant while \$B\$, which defines one of the moduli, is the unknown. But thanks to H.PWiz, we finally have a rigorous proof for why the generalized form of division always works, and rigorously defined thresholds for when the division works in its two shortened forms.

Now let's look at ways to shorten the regex. First, if we assume \$C\ge A\$, it becomes 50 bytes:

(x(x*)),(x*?)(?=1円*$)(x(x*))(?=4円*$)(?=2円+$)2円5円*$

Try it online! - ECMAScript
Try it online! - Python
Try it online! - Python equivalent

If we know that all divisors will be in the required range, the division regex can be shortened further.

With \$C \equiv A \pmod {B-1}\$ and \$C \equiv 0 \pmod B\$, we need to find the largest matching quotient.
With \$C \equiv B \pmod {A-1}\$, we need to find the smallest matching quotient.

Coming in at 42 bytes, here is the case of using only \$C \equiv A \pmod {B-1}\$ and \$C \equiv 0 \pmod B\$:

(x(x*)),(x*?)(?=1円*$)(x(x*))(?=4円*$)2円5円*$

Try it online! - ECMAScript

This can be shortened further, thanks to a trick found by Grimmy (and used in this post). I've used it to shorten factorial, Proth, \$np_n\$, and consecutive-prime/constant-exponent regexes. As a standalone division regex, it comes in at 39 bytes:

(x+),(x*?)(?=1円*$)((x*)(?=1円4円*$)x)3円*$

Try it online! - ECMAScript
Try it online! - Python
Try it online! - Python non-regex equivalent

(x+), # 1円 = divisor
(x*?)(?=1円*$) # 2円 = remainder of division; tail -= 2円
( # 3円 = conjectured quotient - find the largest one that matches
 (x*) # 4円 = 3円-1; tail -= 4円
 (?=1円4円*$) # assert tail-(quotient-1)-divisor is divisible by quotient-1
 x # tail -= 1
)
3円*$ # assert both tail and dividend are divisible by quotient

The above algorithm is guaranteed to return the correct quotient if at least one of the following constraints is met (along with \$C \equiv 0 \pmod A\$):

• \$A^2+2A < 4C\$ or equivalently \$A+2 < 4B\$
• \$A=C\$ or equivalently \$B=1\$
  • This works because the regex subtracts \$C-(B-1)-A\$ in order to assert that it is \$\equiv 0 \pmod {B-1}\$, and that subtraction will always result in a non-match (as it can't have a negative result) if \$A=C\$, unless \$B=1\$.
• \$A\$ is a prime power
  • One of the cases in which this must be true, is when \$C\$ is semi-prime and neither of the two other conditions are met. It also must be true if \$C\$ is a prime power.

The proof of the first constraint follows. We start from the assertions made by the regex:

\$\begin{aligned} \qquad C &\equiv 0 \pmod B \\ C &\equiv A \pmod {B-1} \end{aligned}\$

Suppose the algorithm finds a \$B'\$ satisfying these moduli, with \$C=A'B'\$. Since the search for \$B\$ is done from largest to smallest, the only way for it to first find a \$B'\ne B\$ is with \$B<B'\$. Then:

\$\begin{aligned} \qquad C &\equiv 0 \pmod {B'} \\ C &\equiv A \pmod {B'-1} \\ A'B' &\equiv A \pmod {B'-1} \\ B' &\equiv 1 \pmod {B'-1} \end{aligned}\$

So, by modular division of \$A'B'/B'\$,

\$\qquad A' \equiv A \pmod {B'-1}\$

Due to \$AB=A'B'\$ and \$B<B'\$,

\$\begin{aligned} \qquad A &> A' \\ A &\ge A'+B'-1 \end{aligned}\$

Due to the AM-GM inequality, we have:

\$\begin{aligned} A'+B' &\ge 2\sqrt C \\ A'+B'-1 &\ge 2\sqrt C-1 \\ A &\ge 2\sqrt C-1 \\ {(A+1)}^2 &\ge {(2\sqrt C)}^2 \\ A^2+2A+1 &\ge 4C \end{aligned}\$

So for a \$B'\$ to be found, \$A^2+2A+1 \ge 4C\$ must be true. Therefore, if \$A^2+2A+1 < 4C\$, no such \$B'\$ will be found. But \$A^2+2A+1 = 4AB\$ has no solutions:

\$\begin{aligned} A^2+2A+1 &\stackrel?= 4AB \\ A &> 0 \\ A+2+{1\over A} &\ne 4B \end{aligned}\$

So we can simplify the inequality that guarantees no \$B'\$ will be found:

\$\begin{aligned} A^2+2A+1 &\le 4C \\ A^2+2A &< 4C \end{aligned}\$

And now the proof of the prime power constraint. Suppose that \$A\$ is a prime power and the algorithm has found a \$B'\$:

\$\begin{aligned} \qquad A&=p^n \\ A'B'&=C\\ B'&>B \\ A'&<A \\ A' &= p^{n-k},\ k>0 \end{aligned}\$

So \$B'\$ must have gained the prime power factor that \$A'\$ lost:

\$\begin{aligned} \qquad B' &= xp^k\\ B' &\equiv 0 \pmod p \\ A &\equiv 0 \pmod p \end{aligned}\$

So \$A\$ and \$B'-1\$ are coprime, and we can do modular division by \$A\$, since it is not \$\equiv 0 \pmod {B'-1}\$:

\$\begin{aligned} \qquad C &\equiv A \pmod {B'-1} \\ AB &= A\pmod {B'-1} \\ A &\equiv A \pmod {B'-1} \\ B &\equiv 1 \pmod {B'-1} \\ B &< B' \\ B &\le B'-1 \\ B &= 1 \end{aligned}\$

So for a \$B'\$ to be found when \$A=p^n\$, it must be the case that \$B=1\$, meaning \$A=C\$. But we already know that the algorithm works properly when \$A=C\$.

Coming in at (削除) 38 (削除ここまで) 30 bytes, here is the case of using only \$C \equiv B \pmod {A-1}\$:

(x(x*)),(x*?)(?=1円*$)(x+?)2円+$

Try it online! - ECMAScript
Try it online! - Python
Try it online! - Python non-regex equivalent

(x(x*)), # 1円 = divisor; 2円 = 1円-1; tail = dividend
(x*?)(?=1円*$) # 3円 = remainder of division; tail -= 3円
(x+?) # 4円 = conjectured quotient - find the smallest one that matches
2円+$ # assert tail is positive and divisible by divisor-1

The above algorithm is guaranteed to return the correct quotient if at least one of the following constraints is met:

• \$A^2 > C \ge A\$ or equivalently \$A > B \ge 1\$
• \$A=1\$ or equivalently \$B=C\$

The proof of the first constraint follows. Starting from the assertion made by the regex:

\$\qquad C \equiv B \pmod {A-1}\$

Suppose the algorithm finds a \$B'\$ satisfying this modulus. Since the search for \$B\$ is done from smallest to largest, the only way for it to first find a \$B'\ne B\$ is with \$B>B'>0\$. Then:

\$\begin{aligned} \qquad C &\equiv B' \pmod {A-1} \\ AB &\equiv B' \pmod {A-1} \\ A &\equiv 1 \pmod {A-1} \end{aligned}\$

So, by modular division of \$AB/A\$,

\$\begin{aligned} B &\equiv B' \pmod {A-1} \\ B &> B' \\ B &\ge B' + A-1 \\ B-(A-1) &\ge B' > 0 \\ B-(A-1) &> 0 \\ B &> A-1 \\ B &\ge A \end{aligned}\$

So for a \$B'\$ to be found, \$B \ge A\$ must be true. Therefore, if \$B < A\$, no such \$B'\$ will be found.

This regex was actually discovered just for the sake of this post, to complete the symmetry. Then the math above proved it could be shortened from 38 to just 30 bytes by dropping the assertion of \$C \equiv 0 \pmod B\$, which was not referenced anywhere in the proof; I hadn't even tried shortening it that much beforehand.

At the time of posting it had never yet been used in a larger regex. Since then I found the perfect use for it in Shift right by half a bit, shortening that regex by 141 bytes.

So now let's prove that the generalized form of division works for all inputs. We have:

\$\begin{aligned} \qquad C &\equiv 0 \pmod {B} \\ C &\equiv B \pmod {A-1} \\ C &\equiv A \pmod {B-1} \end{aligned}\$

Suppose the algorithm finds a \$B'\$ satisfying these moduli, with \$C=A'B'\$. Since the search for \$B\$ is done from largest to smallest, the only way for it to first find a \$B'\ne B\$ is with \0ドル<B<B'\$. So now, doing the same steps we did for the two shortened forms of division:

\$\begin{aligned} C &\equiv 0 \pmod {B'} \\ C &\equiv A \pmod {B'-1} \\ A'B' &\equiv A \pmod {B'-1} \\ B' &\equiv 1 \pmod {B'-1} \\ A' &\equiv A \pmod {B'-1} \\ A' &< A \\ A'+B'-1 &\le A \end{aligned}\$ \$\begin{aligned} \ \\ C &\equiv B' \pmod {A-1} \\ AB &\equiv B' \pmod {A-1} \\ A &\equiv 1 \pmod {A-1} \\ B &\equiv B' \pmod {A-1} \\ B &< B' \\ B+A-1 &\le B' \end{aligned}\$

Putting it together,

\$\begin{aligned} B+A-1 &\le B' \\ B+(A'+B'-1)-1 &\le B' \\ B+A'+B'-2 &\le B' \\ B+A' &\le B'-B'+2 \\ B+A' &\le 2 \\ B &> 0 \\ A' &> 0 \\ B = A' &= 1 \\ B' &= C \\ A &= C \end{aligned}\$

So for a \$B'\$ to be found, \$B'=C\$ and \$A=C\$ must be true. But the regex tries to assert the modulus

\$C-B'-(A-1) \equiv 0 \pmod {B'-1}\$

But if \$B'=C\$ and \$A=C\$, this becomes

\$\begin{aligned} C-C-(C-1) \equiv 0 \pmod {B'-1} \\ 0-(C-1) \equiv 0 \pmod {B'-1} \end{aligned}\$

If \$C>1\$, the regex will fail to subtract \$C-1\$ from \0ドル\$ before it can even try to assert the modulus, thus the search for \$B\$ will continue with other values, and the correct one will be found.

If \$C=A=1\$, the answer \$B'=1\$ is correct, as \$B=1\$ in this case.

Therefore the regex will always give the correct answer for \$C>0\$, and since \$C=0\$ is handled as a special case, it will give the correct answer for all inputs.

APL (Dyalog), 5 bytes

-2 bytes thanks to @ngn

⌊÷,|⍨

This is an atop (2-train) of a fork (3-train), where the atop's right tine is a derived function (the result of an operator applied to a function):

 result 
 ↑┌──────────┐
 ││ ┌────┐│┌──────┐ (derived function)
 │↓ │ ↓│↓ │╱
 ┌───┐ ┌───┐ ┌───┐ ╔═══╤═══╗
 │ ⌊ │ │ ÷ │ │ , │ ║ | │ ⍨ ║ 
 └───┘ └───┘ └───┘ ╚═══╧═══╝
 ↑ ↑ ↑ ↑ ╲
left argument ┴─────────────────┘ (operator)
 └─────────┴ right argument

⌊ floor of

÷ division

, catenated to

| division remainder
⍨ with swapped arguments (APL modulus is "backwards")

Try it online!

• \$\begingroup\$ How did you make that cool diagram? \$\endgroup\$

  emiflake

  – emiflake

  2017年03月27日 20:19:32 +00:00

  Commented Mar 27, 2017 at 20:19
• 3
  \$\begingroup\$ @WolfgangTS Painstakingly. Dyalog APL comes with ability to make basic tree diagrams of tacit functions. Try it online! I started with that... \$\endgroup\$

  Adám

  – Adám

  2017年03月27日 20:23:13 +00:00

  Commented Mar 27, 2017 at 20:23
• \$\begingroup\$ Ouch, looks very difficult. I don't have the patience for that I'm afraid, haha \$\endgroup\$

  emiflake

  – emiflake

  2017年03月27日 21:48:05 +00:00

  Commented Mar 27, 2017 at 21:48
• \$\begingroup\$ shorter: ⌊÷,|⍨ \$\endgroup\$

  ngn

  – ngn

  2017年03月28日 11:05:10 +00:00

  Commented Mar 28, 2017 at 11:05
• \$\begingroup\$ @ngn Ouch, you got me. Happy to see that you're still here. \$\endgroup\$

  Adám

  – Adám

  2017年03月28日 11:05:37 +00:00

  Commented Mar 28, 2017 at 11:05

JavaScript (ES6), 17 bytes

Thanks to @Arnauld for golfing off one byte

x=>y=>[x/y|0,x%y]

Receives input in format (x)(y)

Gets floor of x/y by performing bitwise or
Gets remainder by x%y
Puts both values in an array so that they can both be returned

Try it online!

Brachylog, 6 bytes

{÷|%}f

Try it online!

Explanation

We abuse the metapredicate f findall to apply two different predicates to the Input list of two arguments:

{ }f Findall for the Input [A,B] :
 ÷ Integer division
 | Or...
 % Modulo

Python 2, 20 bytes

lambda a,b:(a/b,a%b)

Try it online!

Built-in, 6 bytes

divmod

Try it online!

Java 8, 18 Bytes

(a,b)->a/b+","+a%b

This is a lambda expression of the type BiFunction<Integer, Integer, String>.

I'm surprised... this is actually a fairly concise solution for Java. Go lambda expressions!

• 4
  \$\begingroup\$ You can golf a byte by using a currying lambda a->b-> instead of (a,b)->: Try it online. \$\endgroup\$

  Kevin Cruijssen

  – Kevin Cruijssen

  2020年01月21日 11:17:02 +00:00

  Commented Jan 21, 2020 at 11:17

MATL, (削除) 12 (削除ここまで) 10 bytes

Qt:ie=&fhq

Input is a, then b. Output is remainder, then quotient.

Try it online!

Explanation

This avoids both modulo and division. Instead it uses array reshaping:

1. Build an array of a+1 nonzero elements.
2. Reshape as a 2D array of b rows. This automatically pads with zeros if needed.
3. The row and column indices of the last nonzero entry, minus 1, are respectively the remainder and quotient.

Consider for example a=7, b=3.

Q % Input a implicitly. Push a+1
 % STACK: 8
t: % Duplicate. Range from 1 to that
 % STACK: 8, [1 2 3 4 5 6 7 8]
ie % Input b. Reshape as a matrix with b rows (in column major order)
 % STACK: 8, [1 4 7;
 2 5 8]
 3 6 0]
= % Compare for equality
 % STACK: [0 0 0;
 0 0 1;
 0 0 0]
&f % Row and column indices (1-based) of nonzero element
 % STACK: 2, 3
hq % Concatenate. Subtract 1. Implicitly display
 % STACK: [1 2]

• 2
  \$\begingroup\$ This is a fine example of lateral thinking, nice work! \$\endgroup\$

  Leo

  – Leo

  2017年03月27日 14:34:12 +00:00

  Commented Mar 27, 2017 at 14:34

Regex (Perl / PCRE), (削除) 36 (削除ここまで) 35 bytes

x(x*),((x(?=((?(4)4円)1円)))*)4円?(x*)

Try it online!

Takes its arguments in unary, as two strings of x characters whose lengths represent the numbers. The divisor comes first, followed by a , delimiter, followed by the dividend. The quotient and remainder are returned in the capture groups 2円 and 5円, respectively.

In contrast to the ECMAScript regex solution, this one doesn't have to do anything anywhere near as fancy or mathematically interesting. Just count the number of times \$divisor\$ fits into \$dividend\$ by splitting the divisor to keep two tandem running totals that are both subtracted from \$dividend\$, one that keeps subtracting \$divisor-1\$, and one that keeps subtracting \1ドル\$ and adding it to the total quotient. We must do a split like this, because regex refuses to repeat a zero-width group more than once (this, along with the limited space to work in, is exactly what prevents it from being Turing-complete).

I never wrote a division algorithm in any regex flavor besides ECMAScript before. So it's interesting to now know how they compare in golfed size.

x(x*), # 1円 = divisor-1; tail = dividend
( # 2円 = what will be the quotient
 (
 x # tail -= 1
 (?=
 ( # 4円 = running total
 (?(4)4円) # recall the previous contents of 4,円 if any
 1円 # 4円 += divisor-1
 )
 )
 )* # Loop the above as many times as possible (zero or more); if
 # it loops zero times, 4円 will be unset (we'll treat that as 0)
)
4円? # tail -= 4,円 or leave tail unchanged if 4円 is unset
(x*) # 5円 = remainder

Regex (Java), (削除) 41 (削除ここまで) (削除) 40 (削除ここまで) 38 bytes

x(x*),((x(?=(4円1円|(?!3円)1円)))*)4円?(x*)

Try it online!

This is a port of the Perl/PCRE regex to a flavor that has no conditionals. Emulating a conditional costs (削除) 5 (削除ここまで) 3 bytes here. The quotient and remainder are returned in the capture groups 2円 and 5円, respectively.

x(x*), # 1円 = divisor-1; tail = dividend
( # 2円 = what will be the quotient
 ( # 3円 = the following (after the first iteration has finished),
 # which is always 1
 x # tail -= 1
 (?=
 ( # 4円 = running total
 4円 # recall the previous contents of 4円 (only if it is set)
 1円 # 4円 += divisor-1
 |
 (?!3円) # Match this alternative only if this is the first iteration of
 # the loop, meaning 4円 is unset; 3円 can never not match here if
 # 3円 is set (because 3円==1) unless 1円==0, in which case it
 # doesn't matter if this alternative is taken instead of the
 # above – in that case, the value of 4円 isn't changed anyway.
 1円 # 4円 = divisor-1
 )
 )
 )* # Loop the above as many times as possible (zero or more); if
 # it loops zero times, 4円 will be unset (we'll treat that as 0)
)
4円? # tail -= 4,円 or leave tail unchanged if 4円 is unset
(x*) # 5円 = remainder

Regex (Python^regex / Ruby), 43 bytes

x(x*),((x(?=((?(5)5円)1円))(?=(4円)))*)4円?(x*)

Try it online! - Python _{^{import regex}}
Try it online! - Ruby

This is a port of the Perl/PCRE regex to flavors that have no support for nested backreferences. Python's built-in re module does not even support forward backreferences, so for Python this requires regex.

Emulating a nested backreference by copying the group back and forth costs 8 bytes here. The quotient and remainder are returned in the capture groups 2円 and 6円, respectively.

x(x*), # 1円 = divisor-1; tail = dividend
( # 2円 = what will be the quotient
 (
 x # tail -= 1
 (?=
 ( # 4円 = running total
 (?(5)5円) # recall the previous contents of 4円 (as copied into 5円) if any
 1円 # 4円 += divisor-1
 )
 )
 (?=(4円)) # 5円 = 4,円 to make up for Python's lack of nested backreferences
 )* # Loop the above as many times as possible (zero or more); if
 # it loops zero times, 4円 will be unset (we'll treat that as 0)
)
4円? # tail -= 4,円 or leave tail unchanged if 4円 is unset
(x*) # 6円 = remainder

Regex (.NET), 29 bytes

(x+),(?=(1円)*(x*))((?<-2>x)*)

Try it online!

This uses .NET's Balanced Groups feature. It returns the quotient and remainder in the lengths of 4円 and 3円, respectively.

(x+), # 1円 = divisor; assert 1円 > 0; tail = dividend
(?=
 (1円)* # push 2円 onto the stack for each time 1円 fits into dividend
 (x*) # 3円 = remainder
)
((?<-2>x)*) # 4円 = quotient: pop all 2円 from stack, doing 4円 += 1 for each

Regex 🐘 (.NET), 14 bytes

(x+),(1円)*(x*)

Try it online!

This uses .NET's Balanced Groups feature. It returns the quotient in the capture count of 2円, and remainder in the length of 3円.

Jelly, 3 bytes

:,%

Try it online!

• \$\begingroup\$ Darn, I was just going to post this, I tested it a few extra times and got ninja'd :( good job! \$\endgroup\$

  hyperneutrino

  – hyperneutrino ♦

  2017年03月27日 13:06:23 +00:00

  Commented Mar 27, 2017 at 13:06

Mathematica, (削除) 20 (削除ここまで) 18 bytes

⌊#/#2⌋@Mod@##&

Minor abuse of the flexible output rules: the result is given as div[mod], which will remain unevaluated. The individual numbers can be extracted with result[[0]] and result[[1]].

And hey, it's only one byte longer than the ridiculously named built-in QuotientRemainder.

Mathematica, actually has a neat way to apply multiple functions to the same input, but it's three bytes longer:

Through@*{Quotient,Mod}

• 2
  \$\begingroup\$ You know it's bad when you language creates built-ins that simply combines built-ins... \$\endgroup\$

  Fatalize

  – Fatalize

  2017年03月27日 13:08:06 +00:00

  Commented Mar 27, 2017 at 13:08
• 1
  \$\begingroup\$ @Fatalize Is it? I find divmod built-ins quite useful, and Mathematica is by far not the only language to have one. \$\endgroup\$

  Martin Ender

  – Martin Ender

  2017年03月27日 13:11:29 +00:00

  Commented Mar 27, 2017 at 13:11
• 10
  \$\begingroup\$ @Fatalize, a lot of the same work is required to compute quotients as is required to compute remainders. If both results are to be used, a properly-engineered quotRem builtin can save significant time over calling quot and rem separately. \$\endgroup\$

  Julian Wolf

  – Julian Wolf

  2017年03月27日 14:26:17 +00:00

  Commented Mar 27, 2017 at 14:26

05AB1E, 5 bytes

÷21%‚

Try it online!

(削除) 05AB1E has a bug, so implicit input doesn't work :( (削除ここまで) Emigna noted that inputs are often pushed in reverse.

• \$\begingroup\$ You could do ÷²¹%) for 5 bytes. \$\endgroup\$

  Emigna

  – Emigna

  2017年03月27日 13:23:28 +00:00

  Commented Mar 27, 2017 at 13:23
• \$\begingroup\$ @Emigna I dunno if it's valid though. Wait, how did that work? \$\endgroup\$

  Erik the Outgolfer

  – Erik the Outgolfer

  2017年03月27日 13:26:15 +00:00

  Commented Mar 27, 2017 at 13:26
• 1
  \$\begingroup\$ I don't see why it wouldn't be valid. It works because implicit inputs are pushed to the stack in the reverse order of what you'd assume in cases like this. \$\endgroup\$

  Emigna

  – Emigna

  2017年03月27日 13:27:29 +00:00

  Commented Mar 27, 2017 at 13:27
• 2
  \$\begingroup\$ I take You may choose any format you like for both input and output, as long as the two numbers are clearly distinguishable to mean that you can decide that the inputs are taken as divisor, dividend. You could just specify "Input are taken as divisor, dividend" in the answer and they'll be clearly distinguishable :) \$\endgroup\$

  Emigna

  – Emigna

  2017年03月27日 13:34:02 +00:00

  Commented Mar 27, 2017 at 13:34
• 1
  \$\begingroup\$ s‰ also works, retaining args order, or ‰ with swapped arguments. 2-byte and 1-byte each respectively. \$\endgroup\$

  Magic Octopus Urn

  – Magic Octopus Urn

  2017年04月17日 16:26:18 +00:00

  Commented Apr 17, 2017 at 16:26

Jellyfish, 14 bytes

p
m
,|S
% i
Ei

Try it online!

Explanation

Jellyfish is a beautiful language when it comes to applying multiple functions to the same input. The language is 2D and all binary functions look south for one input and east for another. So by approaching one value from the west and from the north, we can feed it to two functions without having to duplicate it in the code.

The two is in the program are replaced with the two input values when the program starts. Now % is division. It takes one input directly from the east, and when going south it hits the E which redirects that search east as well. So both inputs get fed to % as arguments.

| is the built-in for modulo, which basically does the same thing, but ends up looking south for both in puts.

We concatenate both results into a pair with ,. Then m is the floor function (which we need because % is floating-point division) and finally we print the result with p.

sed, 36 bytes

35 bytes of code, +1 for the -r flag.

:a;s/^(1+)( 1*)1円/1円2円x/;ta;s/.* //

Takes input in unary, space-separated, with the smaller number first. Outputs as unary, with the quotient first in 1s and the remainder second in xs. (If this isn't acceptable, let me know and I'll change it to space-separated 1s like the input.)

Explanation

:a; Define label a
 s/ / /; Perform this substitution:
 ^(1+) Match the first unary number...
 ( 1*) ... followed by a space and 0 or more 1s...
 1円 ... followed by the the first group again
 1円2円x Keep the first two parts unchanged; replace the third
 with an x
 ta; If the substitution succeeded, goto a
 s/.* // After the loop is over, remove the first number

Cubix, 12 (削除) 13 (削除ここまで) bytes

;W@o,I|\S%;O

Which maps onto the following cube

 ; W
 @ o
, I | \ S % ; O
. . . . . . . .
 . .
 . .

Try it here

Explanation with steps as executed
,I|I, - starts with an superflous integer divide, gets the first integer from input, reflects back and gets the next integer from input, then divides again
O; - Output the result of the integer division and pop it
% - do the mod. This could be done later, but ended up here
S\o - Add space character to stack, redirect up and output space
W; - Shift left and pop the space from the stack
O|@ - Output the mod previously calculated, pass through the horizontal reflector and halt.

• \$\begingroup\$ Beat me by two minutes. Nice answer! \$\endgroup\$

  Luke

  – Luke

  2017年03月27日 19:17:33 +00:00

  Commented Mar 27, 2017 at 19:17
• \$\begingroup\$ @Luke Thanks, thought I could get another one off, but proving elusive \$\endgroup\$

  MickyT

  – MickyT

  2017年03月27日 19:22:39 +00:00

  Commented Mar 27, 2017 at 19:22

Brain-Flak, (削除) 56 (削除ここまで) 54 bytes

({}<>)<>([()]{()<(({})){({}[()])<>}{}>}<><([{}()]{})>)

Try it online!

-2 bytes thanks to Wheat Wizard

Explanation

The current best known integer division and modulo in Brain-Flak are very similar (in fact the currently used integer division is just a modification I made on feersum's modulo).

Modulo: ({}(<>))<> { (({})){({}[()])<>}{} }{}<> ([{}()]{})
Division: ({}(<>))<>([()]{()<(({})){({}[()])<>}{}>}{}<>< {} {} >)

Conveniently, the integer division program uses only the third stack for storing data while the modulo program uses only the normal two stacks for storing data. Thus by simply running them both at the same time they do not collide at each other.

Modulo: ({}(<>))<> { (({})){({}[()])<>}{} }{}<> ([{}()]{})
Division: ({}(<>))<>([()]{()<(({})){({}[()])<>}{}>}{}<>< {} {} >)
Combined: ({}(<>))<>([()]{()<(({})){({}[()])<>}{}>}{}<><([{}()]{})>)

Finally, both the integer division and modulo programs used in this combination were designed to be stack clean (not leave garbage on the stacks/not depend on the (non)existence of values on the stacks other than their input) but that is not necessary for this problem. Thus we can save two bytes by not bothering to pop the zero at the end of the main loop and another two bytes by not pushing zero at the start, instead relying on the zero padding on the bottom of the stacks.

({}<>)<>([()]{()<(({})){({}[()])<>}{}>}<><([{}()]{})>)

For the explanation for the integer division program see feersum's answer

Integer Division Explanation Coming Soon...

Brain-Flak, (削除) 168 148 (削除ここまで) 110 bytes

I guess I should have checked the Wiki first

(({})(<({}(<(({})<>)>))<>([()]{()<(({})){({}[()])<>}{}>}{}<><{}{}>)>))<>{(({})){({}[()])<>}{}}{}<>([{}()]{}<>)

Format:

Input: Output:
A (18) remainder (2)
B (4) division (4)

Try it online!

(({})(< # Copy A
({}(< # Pick up A
(({})<>) # Copy B to the other stack
>)) # Put A on top of a 0 on the second stack
 # At this point the stacks look like this: A
 0
 B B
 ^
<>([()]{()<(({})){({}[()])<>}{}>}{}<><{}{}>) # Positive division from the wiki
>)) # Put down A on top of a 0
 # The stack now: A
 0
 Div B
 ^
<>{(({})){({}[()])<>}{}}{}<>([{}()]{}<>) # Modulo from the wiki

• \$\begingroup\$ Golf \$\endgroup\$

  Wheat Wizard

  – Wheat Wizard ♦

  2017年03月27日 14:52:45 +00:00

  Commented Mar 27, 2017 at 14:52

OIL, (削除) 134 (削除ここまで) (削除) 106 (削除ここまで) (削除) 103 (削除ここまで) 102 bytes

Takes the input from stdin, the two numbers seperated by a newline. Outputs the result of the integer division, then a newline, and then the remainder.

This is one of the most complicated OIL programs I've ever written, as OIL lacks builtins for division, remainder, addition, substraction, and so on. It works with the primitive way of doing division: repeated nested decrementation.

I present the code in an annotated format, with comments in the style of scripting languages. Before executing, the comments have to be removed.

5 # read input into lines 0 and 2
5
2
0 # the zero to compare to (nops)
1 # make a backup of the second input at line 3
2
3
10 # check if the second input is 0. %
4
2
24 # if so, jump to 24 (marked with §)
13 # else, go on
10 # check if the first input is zero &
4
31 # if so, jump to 31 (marked with $)
18 # else, go on
9 # decrement both numbers
9
2
6 # jump to line 8 (marked with %)
8
8 # increment the value in line 1 (initially a zero) §
1
1 # "restore the backup"
3
2
6 # jump to line 13 (marked with &)
13
10 # is the second number zero? $
4
2
42 # if so, jump to 42 (marked with +)
36 # else go on
9 # decrement both the second number and the backup
2
9
3
6 # jump to 31 (marked with $)
31
4 # print the division +
1
11 # a newline
4
3 # and the remainder (from the backup)

edit: Shaved off 3 more bytes by moving a "constant" to a one-digit location (less bytes to reference), and then implicit-ing 2 zero-locations (By using an empty line instead. One of them I could have done before).

edit: And another byte by making the initial zero implicit. We really only need a single literal zero.

• \$\begingroup\$ Great work! That's exactly the kind of answer I hoped this challenge would receive :) Just a note: you are guaranteed that the divisor will always be strictly positive, so you don't need to check for a division by 0 ;) \$\endgroup\$

  Leo

  – Leo

  2017年03月27日 14:12:39 +00:00

  Commented Mar 27, 2017 at 14:12
• \$\begingroup\$ @Leo I'm guaranteed that the divisor will always be strictly positive at the beginning. It won't work if I take the division by zero part out, this case can happen even when the "actual" division is a normal one. If I remember correctly this occurs when the remainder is zero. \$\endgroup\$

  L3viathan

  – L3viathan

  2017年03月27日 14:38:47 +00:00

  Commented Mar 27, 2017 at 14:38
• \$\begingroup\$ I'm talking about the check on line 4, not the one on line 12... Doesn't it execute just once at the start of the program? \$\endgroup\$

  Leo

  – Leo

  2017年03月27日 14:41:30 +00:00

  Commented Mar 27, 2017 at 14:41

Retina, 14 bytes

Let's abuse the input/output formats!

(.*)¶(1円)*
$#2

Takes input as b\na, in unary, using for unary digit any single non-digit, non-newline character. Outputs the quotient in decimal, immediately followed by the remainder in unary, using the same character as the input.

Try it online!

(.*) ¶(1円)* matches the first number , then a newline (¶ is Retina's shorthand for \n), then the first number again as many times as possible. The number of matches of the second group will be the result of the division, and the part not matched will be the remainder.

With $#2, we replace everything that was matched in the previous line with the number of captures of the second group, and get then our result.

• \$\begingroup\$ Haha, quite right, I clearly shouldn't be writing programs that late in the evening. \$\endgroup\$

  FryAmTheEggman

  – FryAmTheEggman

  2017年03月30日 18:37:29 +00:00

  Commented Mar 30, 2017 at 18:37

Excel 2013, (削除) 31 30 (削除ここまで) 26 bytes

=INT(A1/B1)&","&MOD(A1;B1)

Explanation

Input is in cell A1 and B1. This simply returns the return values of the FLOOR and MOD function, which are for flooring the division and for the remainder. These values are separated by a comma.

• \$\begingroup\$ I think you mean cell A1 and B1 not A1 and A2 \$\endgroup\$

  fəˈnɛtɪk

  – fəˈnɛtɪk

  2017年03月27日 13:57:24 +00:00

  Commented Mar 27, 2017 at 13:57
• \$\begingroup\$ Save 1 byte with FLOOR(A1/B1;1) instead of QUOTIENT(A1;B1) \$\endgroup\$

  Engineer Toast

  – Engineer Toast

  2017年03月30日 12:36:11 +00:00

  Commented Mar 30, 2017 at 12:36
• \$\begingroup\$ Because the input is always a Natural number, I think you can replace FLOOR(A1/B1;1) with `INT(A1/B1)' to save 4 more bytes \$\endgroup\$

  Wernisch

  – Wernisch

  2017年09月20日 13:39:42 +00:00

  Commented Sep 20, 2017 at 13:39
• \$\begingroup\$ Could this be split into two cells and still qualify as a valid answer? If so, it could be 22 bytes for the two cells combined. \$\endgroup\$

  Axuary

  – Axuary

  2021年04月06日 14:37:36 +00:00

  Commented Apr 6, 2021 at 14:37

Labyrinth, 11 bytes

?:?:}/!\{%!

Try it online!

Explanation

?: Read a and duplicate.
?: Read b and duplicate.
} Move a copy of b over to the auxiliary stage.
/ Compute a/b.
! Print it.
\ Print a linefeed.
{ Get back the other copy of b.
% Compute a%b.
! Print it.

The IP then hits a dead end, turns around and the program terminates due to the attempted division by zero when % is executed again.

C, 32 bytes

f(a,b){printf("%d %d",a/b,a%b);}

Try it online!

ArnoldC, (削除) 286 (削除ここまで) 283 bytes

HEY CHRISTMAS TREE c
YOU SET US UP 0
HEY CHRISTMAS TREE d
YOU SET US UP 0 
GET TO THE CHOPPER c
HERE IS MY INVITATION a
HE HAD TO SPLIT b
ENOUGH TALK
GET TO THE CHOPPER d
HERE IS MY INVITATION a
I LET HIM GO b
ENOUGH TALK
TALK TO THE HAND c
TALK TO THE HAND d
YOU HAVE BEEN TERMINATED

Try it online!

How It Works

HEY CHRISTMAS TREE c //DECLARE VARIABLE c = 0
YOU SET US UP 0
HEY CHRISTMAS TREE d //DECLARE VARIABLE d = 0
YOU SET US UP 0
GET TO THE CHOPPER c /*
HERE IS MY INVITATION a SET c = a/b
HE HAD TO SPLIT b 
ENOUGH TALK */
GET TO THE CHOPPER d /*
HERE IS MY INVITATION a SET d = a mod b
I LET HIM GO b
ENOUGH TALK */
TALK TO THE HAND c // PRINT c
TALK TO THE HAND d // PRINT d
YOU HAVE BEEN TERMINATED //END

Output Format

a/b
a mod b

Quipu, 54 bytes

\/\/[]\n[]
 1&/1円&
 [] []
 // %%
 /\ /\

Try it online!

Explanation

Each pair of columns constitutes a "thread." Threads are executed one at a time, top to bottom, left to right. Each thread also stores a value (initially 0).

Each of threads 0 and 1 uses the \/ knot to read a number from stdin and stores that number as its value.

Thread 2 does the following:

• []: Get the value of thread N, where N is the previous value of this thread. Here, there is no previous value, so we get the value of thread 0.
• 1&: Push a 1.
• []: Get the value of thread N, where N is the 1 we just pushed.
• //: Integer-divide the last two values.
• /\: Output the result.

Thread 3 sets its value to a string containing a newline (\n) and then outputs it (/\).

Thread 4 does the same thing as thread 2, but with modulo (%%) instead of division.

TypeScript's Type System, 78 bytes

//@ts-ignore
type M<A,B,T=[]>=A extends[...B,...infer I]?M<I,B,[...T,1]>:[T,A]

Try it at the TypeScript Playground!

This submission is a generic type taking in two numbers in unary and outputting a tuple of two unary numbers, the first being the division and the second the remainder.

Explanation:

//@ts-ignore // ignore any compiler errors in this type
type M<
 A, // A is the numerator
 B, // B is the denominator
 T = [] // T is the # of times recursed (division result)
> =
 A extends [...B, // if B fits in A (A >= B):
 ...infer I // set I to the difference (A - B)
 ] ? M< // recurse with:
 I, // I as the numerator
 B, // B as the denominator
 [...T, 1] // T with 1 appended (i.e. increment)
 >
 : [T, A] // otherwise return [T, A]

This uses tail recursion to increase the maximum recursion depth from 50 to 999.

• \$\begingroup\$ TypeScript's type checker has tail recursion now? Dang. \$\endgroup\$

  Bbrk24

  – Bbrk24

  2023年07月27日 17:21:26 +00:00

  Commented Jul 27, 2023 at 17:21
• \$\begingroup\$ @Bbrk24 Yep, been around since late 2021. typescriptlang.org/docs/handbook/release-notes/… \$\endgroup\$

  noodle person

  – noodle person

  2023年07月27日 19:35:08 +00:00

  Commented Jul 27, 2023 at 19:35
• \$\begingroup\$ Ah, neat. So, in the spirit of these challenges, we don't count the Unary converter? What about stuff like Equals from Type Challenges and such? \$\endgroup\$

  General Grievance

  – General Grievance

  2025年08月20日 14:51:16 +00:00

  Commented Aug 20 at 14:51
• \$\begingroup\$ @GeneralGrievance Idk tbh, I kinda just go with whatever feels right for the challenge, like if it's something simple I might just expect the input in unary, if it's something more complex I would include the converter. Don't think there were ever any hard rules abt it \$\endgroup\$

  noodle person

  – noodle person

  2025年08月21日 02:54:33 +00:00

  Commented Aug 21 at 2:54
• \$\begingroup\$ No problem. Just trying to see what your perspective was. \$\endgroup\$

  General Grievance

  – General Grievance

  2025年08月21日 12:25:04 +00:00

  Commented Aug 21 at 12:25

><>, (削除) 27 26 (削除ここまで) 16 + 1 = 17 bytes

:r:{%:n','o-,ドルn;

Note

• Input using the -v flag, see TIO for an example.
• This outputs the remainder first, then a comma and lastly the integer division.

Try it online!

Explanation

Note that the stack starts as A, B, where A and B represent the first and second input, because of the -v flag used.

:r:{%:n','o-,ドルn; # Explanation
:r:{ # Do some stack modifications to prepare it for
 # next part
 # (Stack: B, A, A, B)
 % # Take the modulo of the top two items
 # (Stack: B, A, A%B)
 : # Duplicate it
 # (Stack: B, A, A%B, A%B)
 n # Pop once and output as number
 # (Stack: B, A, A%B)
 ','o # Print separator
 # (Stack: B, A, A%B)
 - # Subtract the modulo from the first input
 # (Stack: B, A-A%B)
 ,ドル # Swap the two inputs so they are back in order
 # and divide, so we get an integer
 # (Stack: floor(A/B))
 n; # Output that as a number and finish.

• \$\begingroup\$ How can you provide input values up to 255? \$\endgroup\$

  Leo

  – Leo

  2017年03月27日 14:43:52 +00:00

  Commented Mar 27, 2017 at 14:43
• \$\begingroup\$ Just use higher ASCII/Unicode values. That way, į becomes 255. \$\endgroup\$

  Luke

  – Luke

  2017年03月27日 14:46:14 +00:00

  Commented Mar 27, 2017 at 14:46
• \$\begingroup\$ Ok, nice :) By the way, wouldn't it be shorter to take input numbers from command line directly with the -v flag? \$\endgroup\$

  Leo

  – Leo

  2017年03月27日 15:18:57 +00:00

  Commented Mar 27, 2017 at 15:18
• \$\begingroup\$ It would, but I couldn't get that to work on TIO, so I settled with this solution. It would save 8 bytes - 1 (for the -v flag). \$\endgroup\$

  Luke

  – Luke

  2017年03月27日 15:23:30 +00:00

  Commented Mar 27, 2017 at 15:23
• \$\begingroup\$ Here you are :) \$\endgroup\$

  Leo

  – Leo

  2017年03月27日 15:42:19 +00:00

  Commented Mar 27, 2017 at 15:42

C, 21 bytes

#define f(a,b)a/b,a%b

A macro that replaces f(a,b) with the 2 terms comma separated. Though you'd better be passing it to a function or else there's no way to pick the 2 apart.

Try It Online

Haskell, 21 bytes

a#b=(div a b,mod a b)

Try it online! Example usage: 13#2 returns (6,1). Yes, this is pretty boring, however slightly more interesting than the divMod build-in which works the same.

While we are at it, there is also quot, rem and quotRem which behave the same on natural numbers as div, mod and divMod. However, for negative inputs the result of mod has the same sign as the divisor, while the result of rem has the same sign as the dividend. Or, as it is put in the Prelude documentation, quot is integer division truncated toward zero and div is integer division truncated toward negative infinity.

How about no div or mod build-ins?

No build-ins, (削除) 36 32 (削除ここまで) 31 bytes

a#b|a<b=(a,0)|m<-a-b=(+1)<$>m#b

Try it online! Example usage: 13#2 returns (1,6), that is the mod result is first and the div result second. If a is smaller b, then a mod b is a and a div b is 0, so (a,0) is returned. Otherwise recursively compute mod and div of a-b and b, add 1 to the division result and keep the remainder.

Adding 1 to the division result is achieved by using <$>, which is commonly used as map to map functions over lists, but works on tuples too, however the function is applied to the second tuple element only.

Edit: Saved one byte thanks to xnor!

• 2
  \$\begingroup\$ Your second solution can shave a byte using <$> on a tuple to act on its second element :a#b|a<b=(a,0)|m<-a-b=(+1)<$>m#b. \$\endgroup\$

  xnor

  – xnor

  2017年03月28日 00:20:28 +00:00

  Commented Mar 28, 2017 at 0:20

• code-golf
• arithmetic
• division