# Sum the array times n, except the last

I've been posting relatively hard challenges recently, so here goes an easy one.

Given an array $$\A\$$ and a number $$\n\$$, calculate the sum of all numbers of $$\A\$$ multiplied by $$\n\$$, except the last one. All numbers (the elements of $$\A\$$ and the value of $$\n\$$) are positive integers, and $$\A\$$ is non-empty. Shortest code in bytes wins.

I have a 3-byte J solution. Can you find it (or beat it in a different language)?

## Test cases

A           N   Ans   Explanation
3 1 4 1 5   10  95    (3+1+4+1)*10+5
3 1 4 1 5   1   14    (3+1+4+1)*1+5
1           999 1     1

• Can we take the list in reverse? – user92069 Jul 11 '20 at 8:20
• @Third-party'Chef' No. – Bubbler Jul 11 '20 at 8:26
• I wonder if your J solution used mixed base conversion – xnor Jul 11 '20 at 9:37
• Can we take the input as numbers instead of a single array? – null Jul 11 '20 at 9:54
• @HighlyRadioactive Yes, that's fine. – Bubbler Jul 11 '20 at 12:19

# J, 3 bytes

That was fun to find.

&+/


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### How it works

10 (&+/) 3 1 4 1 5 will bind 10 as an argument of + as 10&+, one verb that gets inserted between the elements of the list by /. So we have: 3 (10&+) 1 (10&+) 4 (10&+) 1 (10&+) 5. Now x n&v y means that y gets applied to n&v for x times. With J's right to left evaluation we get: to 5 add 1 times 10, add 4 times 10, add 1 times 10, add 3 times 10. A challenge made for J's stranger parts. :-) And because + is commutative, +&/ would also be a valid solution.

• Perfect, you nailed it! – Bubbler Jul 11 '20 at 12:17

# JavaScript (ES6),  28  23 bytes

Saved 3 bytes thanks to @Mukundan314

Expects (A)(n).

A=>n=>eval(A.join*n+)


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### How?

We simply join the input array with "*n+", so that [1,2,3] is turned into "1*n+2*n+3" and evaluate the resulting string.

• I'm simultaneously amazed and disgusted. Good work! – Jhal Jul 11 '20 at 22:02

foldr1.((+).).(*)


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It turns out this this was close to a port of the intended J solution. The pointfree function ((+).).(*) takes the argument n to the map \a b->a*n+b, that is, to add n times the left value to the right value. This creates the same "verb" as J used, and the foldr1 does the same a J's automatic right to left evaluation. It starts with the rightmost value in the list, which never gets multiplied by n, and applies it right-to-left, effectively increasing the sum so far with n times to the new element.

# Python 3, 27 bytes

lambda a,n:a.pop()+sum(a)*n


Port of my Japt solution to python

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• Snap! Just wrote the exact same code without looking at any answers! – Noodle9 Jul 11 '20 at 10:31
• +1 for 2 min earlier – ZaMoC Jul 11 '20 at 11:29

# Wolfram Language (Mathematica), 19 bytes

#2Tr@Most@#+Last@#&


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# Python 3, 27 bytes

lambda a,n:a.pop()+sum(a)*n


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## Clojure 41 bytes

#(+(last %1)(* %2(apply +(butlast %1))))

Unfortunately, + does have to be applyed.

Try It Online

• Welcome to the community! Why not add a Try-It-Online link for your code so users can try it out :) – mkst Jul 14 '20 at 8:45
• You can save 2 bytes by writing just % instead of %1, it is always recognized as the first argument, even when there are more. – Kirill L. Jul 15 '20 at 12:42
• And another one by swapping the order of operands: tio.run/… – Kirill L. Jul 15 '20 at 12:49

# R, 3736 35 bytes

-2 bytes with help from Giuseppe

function(l,n)rev(l)%*%n^(seq(!l)>1)


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Reverse the vector, and perform dot product with the vector $$\(n^0, n^1, n^1, \ldots,n^1) = (1, n, n,\ldots, n)\$$.

I just discovered this behaviour of seq, which gains 1 byte on item 4 of this tip: seq(!l) is equivalent to seq(along.with = l) (giving the vector 1 2 3 ... length(l)) in all situations, even if l is of length 1. That is because !l is a logical, not an integer, and so we avoid the call to seq.int when l is a (length 1) integer.

• If the list were guaranteed to be at least length 2, this would be a byte shorter, but as it is, it's longer by one instead. – Giuseppe Jul 13 '20 at 18:09
• @Giuseppe Good old seq_along! I parlayed it into a 36-byter. Thanks! – Robin Ryder Jul 14 '20 at 8:36
• @Giuseppe seq(!l) works and is equivalent to seq(a=l), even if l is of length 1! – Robin Ryder Jul 15 '20 at 10:59
• Wow, it's been a while since I saw such a neat and applicable golfing trick! That's probably more due to my lack of participation here than anything else. – Giuseppe Jul 15 '20 at 18:15

# Perl 5 + -pa -MList::Util+sum, 19 bytes

$_=pop(@F)+<>*sum@F  Try it online! • Did you mean -MList::Util=sum? – msh210 Jul 14 '20 at 12:12 • Interesting, I didn't know = would work there. I've always used + for places where spaces can't be used without quotes or to help precedence (e.g. /\d+\D/,say+($1)x$&). Looks like a bunch of things work though. Good to know, thanks! :) – Dom Hastings Jul 14 '20 at 12:56 # Japt, 7 bytes o +V*Ux  Try it online! ## Explanation o +V*Ux o // Pop and return last element of first input + // plus V* // second input times Ux // Sum of first input  # Pyth, 7 bytes +*sPQEe  Try it online! ## Explanation +*sPQEe Q # First input P # Remove the last element s # Sum elements * E # Multiply by the second input + e # Add the last element of the first input  # 05AB1E, 5 bytes -2 bytes thanks to @KevinCruijssen. *²÷O  Try it online! ## Explanation * Multiply list by second operand  Dump ÷ Divide the last item by ² the second operand O Sum the stack  # 05AB1E, 7 bytes „²*ý.VO  Try it online! ## Explanation „ 2-char string ²* (Which does when evaluated) Multiply by the second input ý Join the input list by this .V Evaluate O Sum the resulting stack  # APL (Dyalog Extended), 9 bytes (SBCS) Anonymous tacit infix function. Takes $$\A\$$ as left argument and $$\n\$$ as right argument. ⊢/+.×+×∘~  Try it online! ×∘~$$\A×(1-n)\$$ +.×+$$\\big(\sum_{i=1}^N A_i×n\big)+\$$ ⊢/ rightmost element (lit. right-argument reduction) So this effectively implements: $$\Bigg(\bigg(\sum_{i=1}^N A_i×n\bigg)+A×(1-n)\Bigg)_N\\ \bigg(\sum_{i=1}^N A_i×n\bigg)+A_N×(1-n)\\ \bigg(\sum_{i=1}^N A_i×n\bigg)+A_N-n×A_N\\ \bigg(\sum_{i=1}^{N-1} A_i×n\bigg)+A_N$$ • @Bubbler Ugh, thanks, that'll be much harder to explain. Also, it isn't really related to my solution at all. You should self-answer with that, I guess. – Adám Jul 13 '20 at 7:55 # APL (Dyalog Unicode), 5 bytes +⍣⎕/⎕  Try it online! A full program, which pretty much works like the 3-byte J solution. Takes two lines of input, $$\A\$$ first and $$\n\$$ second. ### How it works +⍣⎕/⎕ ⎕ ⍝ Take the input A / ⍝ Reduce by... + ⍝ Add the left argument ⍣⎕ ⍝ n times For n=10 and A = 3 1 4 1 5, this becomes: +⍣10/3 1 4 1 5 3 (+⍣10) 1 (+⍣10) 4 (+⍣10) 1 (+⍣10) 5 3 added 10 times to 1 added 10 times to 4 added 10 times to 1 added 10 times to 5  # APL (Dyalog Extended), 8 bytes 1¨⍛,⊥0,⊣  Try it online! A longer but more interesting one. A tacit dyadic function that takes $$\A\$$ on its left and $$\n\$$ on the right. Uses mixed base conversion ⊥, which does the following: Base: 1 1 1 ... 1 n Digit value: n n n ... n 1 Array value: 0 a1 a2 ... ax-1 ax Total: a1n + a2n + ... + ax-1n + ax  ### How the code works 1¨⍛,⊥0,⊣ ⍝ Input: left=A, right=n 1¨ ⍝ An array of ones as long as A ⍛, ⍝ Append n, which becomes the base 0,⊣ ⍝ A prepended with single zero, which becomes the values ⊥ ⍝ Mixed base conversion as described above  • Another interesting method could be (⎕⊥,)/⎕, but I don't think you can get rid of the brackets – Jo King Jul 16 '20 at 1:25 # x86-16 machine code, 18 bytes 33 DB XOR BX, BX ; clear running sum 49 DEC CX ; decrement array length 74 09 JZ ADD_LAST ; handle array length of 1 case LOOP_SUM: AD LODSW ; load next value into AX 03 D8 ADD BX, AX ; BX = BX + AX E2 FB LOOP LOOP_SUM ; keep looping 93 XCHG AX, BX ; move sum into AX F7 E2 MUL DX ; DX:AX = AX * DX 93 XCHG AX, BX ; move result back to BX ADD_LAST: AD LODSW ; load last value into AX 03 C3 ADD AX, BX ; AX = AX + BX C3 RET ; return to caller  As a callable function: [SI] to input array, CX array length, DX = N. Output to AX. Rather than make an elaborate test program, here's it being run using DOS DEBUG, entering the input array into memory and setting registers as they would be called: Explanation of above: Enter input array into memory address DS:200 as 16-bit, little-endian words: -e 200 3 0 1 0 4 0 1 0 5 0  Point SI to this input array: -r SI :200  Set CX to array's length: -r CX :5  Set N to 10 (0xA in hex): -r DX :A  Execute and stop before last instruction (RET will "return to DOS" and clobber registers): -g 111  Result is AX=005F or 95 in decimal. # Golfscript, 13 bytes ~:i;-1%{i*+}*  Try it online! Explanation: ~ to convert string input to array and integer on stack. :i; assigns $$\n\$$ to i and pops value. -1% reverses the array and {i*+}* folds the array with (a, b) -> a*n + b • 11 bytes, if you take input in a different format – user92069 Jul 29 '20 at 13:47 # Emacs Lisp with dash library: 38 51 bytes (lambda(n A)(+(car(last A))(* n(-sum(butlast A)))))  (38 bytes was the function body' size only.) • You forgot to add the last back in. – Sandra Jul 15 '20 at 16:06 • My bad, unfortunately the answer is now much longer. – Daanturo Jul 16 '20 at 0:55 # Pyramid Scheme, 407 bytes  ^ ^ /l\ /+\ /oop\ ^---^ ^-----^ - /x\ / \ / \ --- /arg\ /set\ -----^-----^ /2\ /+\ --- ^---^ ^- /1\ ^- --- ^- /]\ ^---^ / \ /2\ /set\--- ^-----^ /x\ ^- --- /]\ ^---^ ^- /#\ / \ ---^ /set\ / \ ^-----^ /arg\ - /+\-----^ ^---^ /2\ /*\ - --- ^---^ ^- /#\ /x\ ^--- ---/ \ /arg\ ^----- /1\ ---  Try it online! Takes input through command arguments, with n as the first argument. This basically implements the algorithm: i = 2 x = 0 o = 0 while args[i]: o += x*args x = args[i] i += 1 print(o + x)  But with more nesting and some shortcuts, like using the variable 2. ## MAWP, 26 bytes %@_2A<\:.>2M3A[1A~M~]%\WM:  Now it works properly on the testcases. Works on MAWP 1.1's integer input. Try it! • The new program works now. Only took 6 days! – Razetime Aug 31 '20 at 4:16 • +1 for persistence! – Dingus Aug 31 '20 at 4:28 # K (oK), 14 13 bytes Solution: {*|x+/y*-1_x}  Try it online! Explanation: Couldn't figure out a smart way of solving this. {*|x+/y*-1_x} / the solution { } / lambda taking implicity x, y -1_x / drop (_) 1 element from end of x y* / multiply by y x+/ / sum up with x as accumulator *| / take last (reverse, first)  Notes: • -1 byte thanks to coltim - thanks! • You can trim a byte by doing {*|x+/y*-1_x} – coltim Nov 19 '20 at 14:04 # Retina 0.8.2, 31 bytes \d+$*
1(?=.*,1*;(1*)|1*$)$1
1


Try it online! Link includes test cases. Explanation:

\d+
$*  Convert to unary. 1(?=.*,1*;(1*)|1*$)
$1  Multiply all but the last element of A by n and delete A. 1  Take the sum and convert to decimal. # Raku, 20 bytes {@^a.pop+$^b*@a.sum}


# Zsh-P, 24 bytes

a=(0 \*<&0+$@) <<<$[a]


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Alternate solution using the -P flag, which enables RC_EXPAND_PARAM to do the same thing.

# Vimscript 36 Bytes

Disgusted to report that Arnauld's solution also works for vimscript.

let F={a,n->eval(join(a,"*".n."+"))}


# PHP, 41 bytes

fn($a,$n)=>array_pop($a)+array_sum($a)*\$n


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Just trying to use all the built-ins!

# T-SQL, 40 bytes

I am using a table instead of an array, sql doesn't have arrays

The test uses a temporary table instead of a real table, because of lack of permissions to create a table.

SELECT sum(a*@-i/@@rowcount*a*~-@)FROM t


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