What is the standard scratch?

Question

In golf, the standard scratch of a course is calculated using this formula:

(3*num_of_3_par_holes + 4*num_of_4_par_holes + 5*num_of_5_par_holes) - difficulty_rating

Your task is to calculate the standard scratch of a golf course, given these 4 inputs.

You should take input in any standard form in the format

[num_of_3_pars, num_of_4_pars, num_of_5_pars], difficulty rating

but if it saves bytes, you many take input in a different way.

You should output the final result by any accepted method on meta, such as returning from a function.

Shortest code wins because this is code-golf!

Answer 1

05AB1E, 7 bytes

3LÌ*O²-

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Answer 2

GolfScript, 23 bytes

~[.;3,]zip{~3+*}%{+}*\-

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Takes input as difficulty rating [num_of_3_pars num_of_4_pars num_of_5_pars].

Answer 3

PHP>=7.1, 43 bytes

[,$x,$y,$z,$d]=$argv;echo$x*3+$y*4+$z*5-$d;

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Answer 4

Pyth, 9 bytes

-s*V}3 5E

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Takes input as difficulty rating\n[num_of_3_pars, num_of_4_pars, num_of_5_pars].

Answer 5

CJam, 12 bytes

{6,3>.*:+\-}

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-1 thanks to Challenger5.

Takes input as difficulty rating [num_of_3_pars num_of_4_pars num_of_5_pars].

Answer 6

Batch, 28 bytes

@cmd/cset/a%1*3+%2*4+%3*5-%4

Can't believe I'm tying with Python!

Answer 7

QBIC, 14 bytes

?:*3+:*4+:*5-:

Functionally equivalent to this.

Answer 8

MATL, 7 bytes

3:5*si-

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Input vector times range 3:5 minus the second input. Contrary to my Octave answer, it's actually shorter to have the inputs as two separate inputs, and shorter to element-wise multiply, then sum, than to do a direct dot product.

Answer 9

cQuents, 14 bytes

#|1:3A+4B+5C-D

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Input is 3par 4par 5par diff. cQuents is not at all built for this type of problem, but it still did fairly well.

Explanation

#|1               Append 1 to the end of the user input - call it n
   :              Mode: sequence
    3A+4B+5C-D    Each item in the sequence equals the first input times three plus
                  the second input times four plus the third input times five minus
                  the fourth input. The last input, n, which came from the program's
                  parameters, prints the nth item in the sequece.
                  

Answer 10

Pyt, 13 8 bytes

35Ř←*Ʃ←-

Takes as input from stdin inputs separately in the following order: [num_of_3_pars,num_of_4_pars,num_of_5_pars], difficulty_rating

Answer 11

K (ngn/k), 12 bytes

{-y-/x*3+!3}

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Called like f[1 2 3;4]. Takes the number of par 3/4/5 holes as one list of three values, and the difficulty rating as the second arg.

• x*3+!3 generate 3 4 5, multiplying by the first input
• -y-/ equivalent to (-y)+/x, or (+/x)-y. In this instance, the leading - negates the entire y-/... expression.

Answer 12

MMIX, 24 bytes (6 instrs)

00000000: 2a020202 28000000 26020203 2a000100  *£££(¡¡¡&££¤*¡¢¡
00000010: 22000002 f8010000                    "¡¡£ẏ¢¡¡

scratch 4ADDU $2,$2,$2      // p5 *= 5
        2ADDU $0,$0,$0      // p3 *= 3
        SUBU  $2,$2,$3      // p5 -= diff
        4ADDU $0,$1,$0      // p3 += 4 * p4
        ADDU  $0,$0,$2      // p3 += p5
        POP   1,0           // return p3

I ordered the instructions to avoid as many short-term data dependencies as possible.

Answer 13

Factor, 23 22 bytes

[ { 3 4 5 } v. - neg ]

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-1 thanks to @Bubbler

The quotation takes the input as difficulty list.

• { 3 4 5 } v. Take the dot product between the input list and { 3 4 5 }.
• - neg Subtract the difficulty rating from the dot product.

Answer 14

BRASCA, 18 bytes

Takes input as <3-par>,<4-par>,<5-par>,<difficulty>

Ci',G,3*$4*+$5*+$-

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Explanation

C                   - Set implicit output to number mode
 i',G               - Use numbers instead of charcodes for 0-9. Then split by commas.
     ,              - Reverse stack so the 3par is at the top.
      3*            - 3par * 3
        $4*+        - 4par * 4, then add to 3par
            $5*+    - 5par * 5, then add to total
                $-  - total - difficulty
<implicit>          - Print it

Answer 15

Vyxal as, 8 7 bytes

-1 thanks to Aaron Miller

2ʀ3+uJ*

Takes input as par3 \n par4 \n par5 \n difficulty. Try it Online!

2ʀ3+uJ*
2ʀ        # Push the range [0, 1, 2]
  3+      # Add 3: [3, 4, 5]
    uJ    # Append -1: [3, 4, 5, -1]
      *   # Multiply the list by its counterpart in...
          # the (implicit) input, and...
          # Print the sum of the result (s flag) 

Answer 16

TI-Basic, 19 bytes

Prompt A,B,C,D:3A+4B+5C-D

Alternatively (21 bytes):

3Ans(1)+4Ans(2)+5Ans(3)-Ans(4

Answer 17

Add++, 27 bytes

D,g,@@@@!,3 4 5 1ECBcB*0$_s

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How it works

D,g,        - Create a function called g...
    @@@@    - ...that takes 4 arguments...
    !,      - ...as a list. e.g.     [4 3 2 1]
    3 4 5 1 - Push 3 4 5 1; STACK = [[4 3 2 1] 3 4 5 1]
    EC      - Collect;      STACK = [[4 3 2 1] [3 4 5 1]]
    Bc      - Zip;          STACK = [[4 3] [3 4] [2 5] [1 1]]
    B*      - Products;     STACK = [12 12 10 1]
    0$_     - Negate;       STACK = [12 12 10 -1]
    s       - Sum;          STACK = [33]

Answer 18

Ly, 13 bytes

3n*4n*5n*&+n-

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Answer 19

Desmos, 21 bytes

f(a,b,c,d)=3a+4b+5c-d

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Input is a function where \$a=3\ \text{par holes},b=4\ \text{par holes},c=5\ \text{par holes},d=\text{difficulty rating}\$

Example Test Case:

Answer 20

Stax, 13 bytes

╨E⌡u╥Diüaë¬~▼

Run and debug it

If I can find a way to read in the array and the number as is instead of evaluating them from string, this program will be shorter.

Answer 21

PHP, 66 bytes

list($a,$b,$c,$d,$e)=$argv;echo eval("return 3*$b+4*$c+5*$d-$e;");

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first attempt:

PHP, 145 bytes

<?php $a=$_SERVER['argv'];$b=[0,3,4,5,-1];$r=array_map(function($x,$y){return$x*$y;},$a,$b);echo array_reduce($r,function($i,$c){return$i+=$c;});

Ungolfed:

<?php
$a = $_SERVER['argv'];
$b = [0,3,4,5,-1]; // the first element always contains the filename, and is hereby disregarded
$r = array_map(function($x,$y){return $x*$y;},$a,$b);
echo array_reduce($r, function ($i,$c) {return $i+=$c;});

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With 10, 5 and 3 holes respectivly, and a difficulty of 5, it's to be called with $ php codegolf.php 10 5 3 5

Note: it throws a warning, as we multiply a string with zero. Add 1 byte for the error-supression-operator to get rid if that.

Answer 22

Julia 1.0, 13 bytes

v->[3:5;-1]'v

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improvement of Dennis's Julia 0.5 answer