How long should my microwave run?

Question

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 code-golf, 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.

Answer 1

Python 2, 19 bytes

lambda t:t-t/100*40

Try it online!

Answer 2

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).

Answer 3

Pyth - 9 8 bytes

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

ijQ*TT60

Test Suite.

Answer 4

TI-Basic (83 series), 8 bytes

Ans-40int(sub(Ans

Requires OS version 1.15 or higher.

Answer 5

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

D : 35 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;
}

TIO Link

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[8]
        {
            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]));
} 

TIO Link

Answer 6

dc, 10 bytes

?9A~r60*+p

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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] * 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

Answer 7

Bash bc + sed, 30 28 bytes

-2 bytes thanks to @seshoumara.

bc<<<0`sed 's/..\?$/*60+&/'`

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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.

Answer 8

Perl 5, 15+1(-p) bytes

/..$/;$_-=40*$`

• -l switch not counted because for tests readability

Try it online

Answer 9

C (gcc), 50 bytes

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

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Answer 10

Jelly, 5 bytes

As a monadic link (thanks for the heads-up, caird!):

b³ḅ60

Try it online!

... Or as a full program:

bȷ2ḅ60

This can be ported easily to 05AB1E, so:

05AB1E, 5 bytes

тв60β

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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⌋

Answer 11

Java 8, 13 bytes

n->n-n/100*40

Port of @ovs' amazing Python 2 formula.

Try it here.

Answer 12

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.

Answer 13

JavaScript, 21 bytes

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

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Answer 14

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!

Answer 15

Batch, 23 bytes

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

Answer 16

bash, 20 bytes

echo $[$1-$1/100*40]

Try it online

Answer 17

Haskell, 18 bytes

f t=t-t`div`100*40

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Anotha port.

Pointfree solution, 21 bytes

(-)<*>(*40).(`div`40)

Answer 18

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.

Answer 19

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.
+     Add.

Answer 20

Labyrinth, 19 bytes

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

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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.

Answer 21

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)]

Answer 22

APL (Dyalog), 11 10 bytes

60⊥0 100⊤⊢

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How?

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

60⊥ - decode in base 60

Answer 23

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)

Answer 24

R, 21 bytes

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

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Answer 25

Common Lisp, 34 bytes

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

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Another port of @ovs' formula.

Answer 26

Pushy, 10 9 bytes

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

2dH/40*-#

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10 bytes

sjvj60*^+#

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

11 bytes

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

H2d%}/60*+#

Try it online!

Answer 27

Husk, 7 bytes

B60B100

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

Answer 28

Symbolic Python, 66 48 46 bytes

Thanks to Jo King for -16 bytes!

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

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---

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

Answer 29

GAP, 45 bytes

Try it online!

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

Answer 30

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