# How long should my microwave run?

I'm hungry. Let's microwave something. Given a numerical input of between 1 and 4 digits, output the number of seconds that the microwave should run.

## Details

The trick is figuring out if the user is inputting seconds or a combination of seconds and minutes. The ones and the tens places should be interpreted as seconds and the hundreds and thousands places should be minutes. For example, the value 1234 should be interpreted as 12 minutes, 34 seconds and 9876 should be 98 minutes, 76 seconds. Typing 130 and 90 should both result in a cook time of 90 seconds.

Here are a few other inputs and outputs:

• 1 = 1
• 11 = 11
• 111 = 71
• 1111 = 671
• 9 = 9
• 99 = 99
• 999 = 639
• 9999 = 6039

## Rules

This is , so the shortest program in bytes wins. Standard loopholes are not allowed. The winning entry must return the right answer when given any integer input from 1 to 9999.

• @WheatWizard, I'm happy to edit the question. Do you have a suggestion for what I should say in the Details section? Maybe I could make this sentence clearer: "The ones and the tens places should be interpreted as seconds and the hundreds and thousands places should be minutes." Nov 11, 2017 at 4:34
• @WheatWizard I just added more detail, let me know if you think I should add more. Nov 11, 2017 at 4:35
• Does that work with an input of 9876? Nov 11, 2017 at 4:37
• Strange, I was about to sandbox this exact challenge haha Nov 11, 2017 at 10:46
• What would the output be for 190? Nov 12, 2017 at 17:49

# Python 2, 19 bytes

lambda t:t-t/100*40


Try it online!

• im confused about the order of operations... Apr 13 at 13:41
• @Hippopotomonstrosesquipedalian I'm unsure about the source of the confusion, but maybe a fully parenthesized version helps: t-((t/100)*40). The / is integer division in Python 2.
– ovs
Apr 14 at 11:21
• this also works in python 3 Apr 14 at 11:23
• No in Python 3 it would have to be t-t//100*40
– ovs
Apr 14 at 11:24

# Japt, 6 bytes

ìL ì60


Test it online! ìL converts to base-100, and ì60 converts back to base 60, resulting in floor(n/100)*60 + n%100. Also works with hours (10000 -> 3600, the number of seconds in an hour).

• Exactly what I had :) Nov 11, 2017 at 22:49

# Pyth - 9 8 bytes

Converts input to base 100, then interprets that as a base 60 number.

ijQ*TT60


# TI-Basic (83 series), 8 bytes

Ans-40int(sub(Ans


Requires OS version 1.15 or higher.

## C, C++, Java, C#, D : 36 bytes

### C : 28 bytes

First time i have an answer that short !

int r(int i){return i/100*60+i%100;}


D can have a special optimization because of the golfy template system :

T r(T)(T i){return i/100*60+i%100;}


C has a special optimization with the implicit int :

r(i){return i/100*60+i%100;}


Code to test

In C ( have to include stdio.h ) :

int main() {
int testArr[] = {1,11,111,1111,9,99,999,9999};
for(int i=0;i<8; ++i) {
printf("%d = %d\n",testArr[i],r(testArr[i]));
}
return 0;
}


In C++ ( have to include iostream ) :

int main() {
std::initializer_list<int> testList{
1,11,111,1111,9,99,999,9999
};

for (auto x : testList) {
std::cout << r(x) << '\n';
}
}


Try it online!

In Java :

public class MainApp {

int r(int i){return i/100*60+i%100;}

public static void main(String[]a) {
MainApp m = new MainApp();
int testArr[] = new int[]{
1,11,111,1111,9,99,999,9999
};

for (int v : testArr) {
System.out.println(v + " = " + m.r(v));
}
}
}


Try it online!

In C#

class Program {
int r(int i){return i/100*60+i%100;}
static void Main(string[] args) {
var p = new Program();
int[] testArr = new int
{
1,11,111,1111,9,99,999,9999
};
foreach(int a in testArr) {
Console.WriteLine(a + " = " + p.r(a));
}
}
}


In D ( have to import std.stdio ) ( exactly, i have no idea how to use arrays in D ) :

void main() {
int[] arr = [1,11,111,1111,9,9,999,9999];
for(int i = 0; i < arr.length; i++)
writeln(arr[i]," = ",r(arr[i]));
}


• The D test code be the footer of this TIO: tio.run/…, and I see you've learned the template system :). (There is a foreach in D, I just forgot how to use it sadly) Nov 13, 2017 at 12:53
• C can be golfed to 28 bytes using C89 implicit-int. Nov 13, 2017 at 16:23
• You should post all these as separate answers. Nov 25, 2017 at 18:53

## dc, 10 bytes

?9A~r60*+p


Try it online!

Explanation: in dc when you push sth. on the stack it goes on top

?         # read and push the input number on the stack
9A        # push 100: 9 * 10^1 + A * 10^0 :D
~         # divide 2nd nr. by the top nr., push quotient, then remainder
r60*      # swap top 2 nr., then multiply the top by 60
+p        # add top 2 nr., then print result


# Bash bc + sed, 30 28 bytes

-2 bytes thanks to @seshoumara.

bc<<<0sed 's/..\?$/*60+&/'  Try it online! Takes input from stdin. Went for a more creative approach: inserts *60+ before the last 1 or 2 digits, and prepends a 0 to the beginning to account for inputs with only 1 or 2 digits. The result is then passed to bc. • If you remove -r and use \?, you can loose 2 bytes. Nov 11, 2017 at 7:08 # Perl 5, 15+1(-p) bytes /..$/;$_-=40*$

• -l switch not counted because for tests readability

Try it online

# C (gcc), 50 bytes

t;f(){scanf("%d",&t);printf("%d",t%100+t/100*60);}


Try it online!

# Jelly, 5 bytes

b³ḅ60


Try it online!

... Or as a full program:

bȷ2ḅ60


This can be ported easily to 05AB1E, so:

# 05AB1E, 5 bytes

тв60β


Try it online!

Simply converts the input integer to base 100 and then converts the result from base 60 to integer. Hence, it is equivalent to Input % 100 + 60 * ⌊Input / 100⌋

• As a function, 5 bytes Nov 11, 2017 at 12:40
• @cairdcoinheringaahing I think you meant this, but thanks for the heads-up anyway Nov 11, 2017 at 12:56

# Java 8, 13 bytes

n->n-n/100*40


Port of @ovs' amazing Python 2 formula.

Try it here.

# PowerShell, 38 33 bytes

-5 bytes thanks to mazzy

$args|%{$_-40*($_-replace'..?$')}


Try it online!

Uses the same formula popularizes by ovs. Truncating is still expensive in PowerShell. Now uses a better regex to instead rip the last one or two characters off to division.

• Oct 21, 2020 at 18:58
• @mazzy Brilliant! Oct 21, 2020 at 19:34

# JavaScript, 21 bytes

a=>(a/100^0)*60+a%100


Try it online!

• Save 4 bytes by using ovs's trick - a-(a/100^0)*40 Nov 11, 2017 at 8:33
• @IanF1. Thanks, but I think it would be literally stealing their idea.
– user72349
Nov 11, 2017 at 11:48
• yeah you're right. Too enthusiastic, sorry. Nov 11, 2017 at 11:56
• @ThePirateBay You don't really live up to your name then ;) Nov 11, 2017 at 17:06

# J, 12 bytes

-40*&<.%&100


It's ovs' Python 2 solution expressed in J. It consist of a hook and a fork:

┌─┬───────────────────────┐
│-│┌──┬────────┬─────────┐│
│ ││40│┌─┬─┬──┐│┌─┬─┬───┐││
│ ││  ││*│&│<.│││%│&│100│││
│ ││  │└─┴─┴──┘│└─┴─┴───┘││
│ │└──┴────────┴─────────┘│
└─┴───────────────────────┘

%&100  - divides the number by 100
*&<.       - finds the floor of the left argument and multiplies it to the left arg.
40           -
-             - subtracts the result of the above fork from the input


Try it online!

• Same byte count as 60#.0 100#:]. Nov 11, 2017 at 9:37
• @FrownyFrog - your solution looks prettier, cheers! Nov 11, 2017 at 9:43
• Galen, I think it might be because Frowny's has a smiley :]
– RGS
Oct 21, 2020 at 22:47
• @RGS Yes, indeed! Oct 22, 2020 at 4:37

## Batch, 23 bytes

@cmd/cset/a%1-%1/100*40


# bash, 20 bytes

echo $[$1-$1/100*40]  Try it online # Haskell, 18 bytes f t=t-tdiv100*40  Try it online! Anotha port. ## Pointfree solution, 21 bytes (-)<*>(*40).(div40)  ## Retina, 11 bytes .{100} 60$*


Try it online!

Input and output in unary. The test suite converts from and to decimal for convenience.

Doing this kind of base conversion for only up to two digits is surprisingly simple to do in unary. We just match runs of 100 1s and replace them with 60 1s. Anything that's left over would correspond to the last two digits in the decimal representation and remains unchanged.

## Alice, 19 bytes

/o
\i@/.'d%~'d:'<*+


Try it online!

### Explanation

Too bad I removed divmod from the language, I guess...

/o
\i@/...


This is just the usual framework for linear programs with decimal I/O operating purely in Cardinal (arithmetic) mode.

.     Duplicate input.
'd%   Mod 100.
~     Swap with other copy.
'd:   Divide by 100.
'<*   Multiply by 60.


## Labyrinth, 19 bytes

?:_100%}#00/_60*{+!


Try it online!

### Explanation

?      Read input.
:      Duplicate.
_100%  Mod 100.
}      Move off to auxiliary stack.
#00/   Divide by 100, using the stack depth to get a 1, instead of _1.
_60*   Multiply by 60.
{+     Retrieve the earlier result and add it.
!      Print.


The IP then hits a dead end and starts moving backward. When it reaches the / it attempts a division by zero which terminates the program.

# Excel VBA, 29 Bytes

Anonymous VBE immediate window function that takes input from range [A1] and outputs to the VBE immediate window.

?[A1]Mod 1E2+60*[Int(A1/100)]


# APL (Dyalog), 11 10 bytes

60⊥0 100⊤⊢


Try it online!

How?

0 100⊤ - encode in base 100, stopping at the second LSB, effectively producing n ÷ 100, n % 100.

60⊥ - decode in base 60

# PARI/GP, 16 bytes

Straightforward:

n->n\100*60+n%100


Unfortunately this nice method is simply too long to use:

n->[60,1]*divrem(n,100)


# R, 21 bytes

x=scan();x-x%/%100*40


Try it online!

• I created my own solution in R and it was far less elegant than this. Nov 13, 2017 at 13:25
• You can cut the scan since most of the answers just assume the variable is already defined. Nov 28, 2017 at 13:58
• The rules are so inconsistent with regards to that. One most other challenges you have to have a function or scan to capture the value.
– Mark
Nov 29, 2017 at 13:30

# Common Lisp, 34 bytes

(lambda(n)(- n(*(floor n 100)40)))


Try it online!

Another port of @ovs' formula.

# Pushy, 10 9 bytes

Kevin outgolfed me in my own language... (using the approach from ovs' answer)

2dH/40*-#


Try it online!

# 10 bytes

sjvj60*^+#


Try it online!

s             \ Split input into digits
jvj          \ Join the first two and the last two
60*       \ Multiply the first by 60
#    \ Print


# 11 bytes

For one byte more we can use the Input % 100 + 60 * ⌊Input / 100⌋ approach:

H2d%}/60*+#


Try it online!

• 9 bytes by creating a port of @ovs' Python 2 answer: 2dH/40*-#. Never programmed in Pushy before, but it seems like a pretty cool language. :) Nov 13, 2017 at 13:33
• @KevinCruijssen it's quite a generic stack based language, I guess the only slightly different thing it brings to the table is the double stack... but thank you, and thanks for the golf :) Nov 13, 2017 at 17:37

# Husk, 7 bytes

B60B100


Try it online! Not very exciting, just converts to base 100 and back from base 60.

• This is neat! Husk and other golf-oriented languages make my brain hurt, but in a good way. Oct 20, 2020 at 22:55

# Symbolic Python, 6648 46 bytes

Thanks to Jo King for -16 bytes!

__=-~-~_-_
_-=_/(__<<__^__)**__*(__-~__<<-~__)


Try it online!

## Explanation

                      # input is initially in _
__=-~-~_-_            # set __ to (2+input-input) = 2
_-=                   # subtract from _:
_/                 # _ divided by:
(__<<__^__)**__  # (2<<2 ^ 2) ** 2 = (8^2)**2 = 100
*                 # multiplied by
(__-~__<<-~__)   # (2 - ~2 << -~2) = 5 << 3 = 40
# value in _ is the implicit output

• You can use *= and -= for 59 bytes. Or change to generating 5 on the second line for -1 more
– Jo King
Oct 21, 2020 at 20:34
• Or generate 10 for 49 bytes
– Jo King
Oct 21, 2020 at 21:25
• @JoKing Thanks for the suggestions! I was able to get it one byte shorter by not defining ___.
– ovs
Oct 21, 2020 at 22:03

# GAP, 45 bytes

Try it online!

f:=function(t) return t-QuoInt(t,100)*40;end;


# Nibbles, 4 bytes (8 nibbles)

@60@@

    @      # convert input in command-line arg to base
@     # 100 (default value for @ when no input from STDIN)
@          # and convert from base
60        # 60
` 