Hampir setara dengan pertanyaan pertama Project Euler:

Jika kita mendaftar semua bilangan alami di bawah 10 yang merupakan kelipatan 3 atau 5, kita mendapatkan 3, 5, 6 dan 9. Jumlah kelipatan ini adalah 23.

Temukan jumlah semua kelipatan 3 atau 5 di bawah 1000.

Tantangan:

Dengan bilangan bulat positif Ndan sekurang-kurangnya satu bilangan bulat positif A, hasilkan jumlah dari semua bilangan bulat positif kurang dari Nitu adalah kelipatan dari setidaknya satu anggota A.

Misalnya, untuk kasus Project Euler, inputnya adalah:

1000
3
5

Kasus uji:

Input : 50, [2]
Output: 600

Input : 10, [3, 5]
Output: 23

Input : 28, [4, 2]
Output: 182

Input : 19, [7, 5]
Output: 51

Input : 50, [2, 3, 5]
Output: 857

Jelly, 6 bytes

ḍþṖḅTS

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How it works

ḍþṖḅTS  Main link. Left argument: D (array). Right argument: n (integer)

ḍþ       Divisible table; test each k in [1, ..., n] for divisibility by all
        integers d in D.
  Ṗ     Pop; discard the last Boolean array, which corresponds to n.
   ḅ    Unbase; convert the Boolean arrays of base n to integer. This yields a 
        non-zero value (truthy) and and only if the corresponding integer k is 
        divisible by at least one d in D.
    T   Truth; yield the array of all indices of truthy elements.
     S  Compute their sum.

Python, 59 55 bytes

lambda n,l:sum(v*any(v%m<1for m in l)for v in range(n))

repl.it

Unnamed function taking an integer, n and a list of integers l. Traverses a range of the Natural numbers (plus zero) up to but not including n and sums (sum(...)) those that have a remainder after division of zero (v%m<1) for any of the integers m in the list l. Uses multiplication rather than a conditional to save 3 bytes.

Octave, 38 36 33 bytes

@(x,y)(1:--x)*~all(mod(1:x,y),1)'

Take input as: f(10, [3;5]). This would be 2 bytes shorter if the input could be f(9,[3;5]) for the same test case.

Verify all test cases here.

Explanation:

@(x,y)        % Anonymous function that takes two inputs, x and y
              % x is a scalar and y is a vertical vector with the set of numbers
(1:--x)*      % Pre-decrement x and create a vector 1 2 ... x-1    

Octave can pre-decrement, so using 1:--x instead of 1:x-1 (two times) saves two bytes.

mod(a,b) gives 1 2 0 1 2 0 1 2 0 for mod(1:9,3). If the second argument is a vertical vector, it will replicate the first input vertically and take the modulus for each of the values in the second input argument. So, for input mod(1:9, [3;5]) this gives:

1 2 0 1 2 0 1 2 0
1 2 3 4 0 1 2 3 4

Taking ~all(_,1) on this gives true for the columns where at least one value is zero, and false where all values are non-zero:

~all(mod(1:x,y),1)
0 0 1 0 1 1 0 0 1

The ,1 is needed in case there is only one number in y. Otherwise it would act on the entire vector instead of number-by-number.

Transposing this to a vertical matrix and use matrix multiplication, will give us the correct answer, without the need for explicit summing:

JavaScript (ES6), 40 39 36 bytes

Input: integer n and array of integer(s) a with currying syntax (n)(a)

n=>F=a=>n--&&!a.every(v=>n%v)*n+F(a)

Test cases

let f =

n=>F=a=>n--&&!a.every(v=>n%v)*n+F(a)

console.log(f(50)([2]));        // 600
console.log(f(10)([3, 5]));     // 23
console.log(f(28)([4, 2]));     // 182
console.log(f(19)([7, 5]));     // 51
console.log(f(50)([2, 3, 5]));  // 857

MATL, 9 bytes

q:ti\~*us

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Retina, 34 bytes

Byte count assumes ISO 8859-1 encoding.

\d+
$*
M&!`(.+)\1*(?=1¶.*\b\1\b)
1

Input format is

50
2,3,5

Try it online!

Python, 67 bytes

a,b,c=input()
x=y=0
exec("if x%c<1or 1>x%b:y+=x\nx+=1\n"*a)
print y

After writing this I noticed my code was similar to the existing python answer, however I came up with it independently and am posting it anyway.

Mathematica, 37 27 bytes

Thanks to Martin Ender for a shrewd observation that led to big byte savings!

Tr[Union@@Range[#,#2-1,#]]&

Unnamed function taking two arguments, a list # of integers (the desired divisors A) and an integer #2 (the upper bound N) , and returning an integer. Range[#,#2-1,#] gives, for each element d of the list #, all the multiples of d less than or equal to #-1 (hence less than #); the union of these lists is then computed and summed with Tr.

Previous version:

Tr[x=#;Union@@(Range[#,x-1,#]&/@#2)]&

Perl 6, 25 bytes

{sum grep *%%@_.any,^$^a}

A lambda that takes the input numbers as arguments. (One argument for N, and an arbitrary number of arguments for A).

(Try it online.)

Explanation:

• { ... }: A lambda.
• $^a: First argument of the lambda.
• @_: Remaining arguments of the lambda ("variadic parameter").
• ^$^a: Range from 0 to $^a - 1.
• * %% @_.any: Another lambda, which tests its argument * using the divisible-by operator %% against an any-Junction of the list @_.
• grep PREDICATE, RANGE: iterates the range of numbers and returns the ones for which the predicate is true.

R, 67 bytes

a=scan();x=c();for(i in a[-1])x=c(x,seq(i,a[1]-1,i));sum(unique(x))

Takes a vector to STDIN in the following format: [N, a_1, a_2, ...]. Supports any number of a. For each a, creates the sequence a to N-1 with stepsize a. Then takes the sum of all the unique entries in that vector.

Haskell, 42 39 bytes

a!b=sum[x|x<-[1..a-1],any((<1).mod x)b]

Usage:

Main> 50![2,3,5]
857

Thanks to @Zgarb for 3 bytes

05AB1E, 9 bytes

FND²%P_*O

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F         For N in [0, ..., input[0]-1]
 ND²%     Evaluate N%input[1]; yields an array of results
     P    Take the total product of the array. Yields 0 only if at least one of the value is 0, in other words if N is multiple of at least one of the specified values
      _   Boolean negation, yields 1 if the last value is 0 and yields 0 otherwise
       *  Multiply by N: yields N if the last value is 0 and yields 0 otherwise
        O Display the total sum

Octave, 49 37 bytes

@(A,N)sum(unique((z=(1:N)'.*A)(z<N)))

the function will be called as f([2 3 4],50)

Assume that A=[2 3 4]; we require to have sum of numbers as

sum(
2,4,6...,50-1 ,
3,6,9...,50-1,
4,8,12,...50-1)

we can multiply [2 3 4] by 1:50 to get matrix (1:N)'.*A

[2 4 6 ... 2*50
3 6 9 ... 3*50
4 8 12 ...4*50]

then extract from the matrix those that are smaller than 50 : z(z<N)

Since there are repeated elements in the matrix we extract unique values and sum them.

previous answer: (this solution will fail if N==1)

@(A,N)sum((k=uint64(1:N-1))(any(k==(k./A').*A')))

function should be called as f(unit64([2 3 4]),uint64(50))

Pyth, 10 bytes

s{sm:0hQdt

Explanation

s{sm:0hQdtQ   Implicit input
    :0hQd     Get multiples of d below the bound
   m     tQ   ... for each d given
  s           Concatenate results
 {            Remove repeats
s             Take the sum

T-SQL, 87 bytes

This will work as long as @i has a value of 2048 or lower

USE master--needed for databases not using master as default
DECLARE @i INT=50
DECLARE @ table(a int)
INSERT @ values(2),(3),(5)

SELECT sum(distinct number)FROM spt_values,@ WHERE number%a=0and abs(number)<@i

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APL (Dyalog Unicode), 12 bytes

+/⊢∘⍳∩∘∊×∘⍳¨

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Anonymous tacit function. Thanks to @Adám for helping me shave 3 bytes off of this. Uses ⎕IO←0.

How:

+/⊢∘⍳∩∘∊×∘⍳¨ ⍝ Tacit function. Left and right arguments will be called ⍺ and ⍵ respectively.

        ×∘⍳¨ ⍝ Multiply ⍺ with each element of [0..⍵-1]
       ∊     ⍝ Enlist (flattens the vector)
     ∩∘      ⍝ Then, get the intersection of that vector with
  ⊢∘⍳        ⍝ The vector [0..⍵-1].
+/           ⍝ Then sum

Pip, 43 41 39 35 bytes

b^:sFc,a{f:0Fdb{f?0c%d?0(f:i+:c)}}i

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Explanation:

Takes inputs like so:

    arg1 1000
    arg2 3 5

b^:s                      ;read rest of inputs as array
                          ;(s is " " and ^ is split into array on char)
F c ,a{                   ;for(c in range(0,a))
  f:0                     ;flag to prevent double counting 15,30,etc.
  F d b {                 ;forEach(d in b)
    f? 0 c%d? 0 (f:i+:c)  ;if flag {continue}elif c%d {f=i+=c}
                          ;      (i will always be truthy so why not)     
  }
}
i                         ;print sum

Python 2, 80 Bytes

This is very long. Can definitely be shortened. Taking the 3 numbers as separate inputs is definitely hurting the score.

i=input
x=i();y=i();z=i();s=c=0
exec("if c%z<1 or c%y<1:s+=c\nc+=1\n"*x)
print s

Common Lisp, 77

(lambda(n x)(loop for i below n when(some(lambda(u)(zerop(mod i u)))x)sum i))

Ungolfed

(lambda (limit seeds)
  (loop for i below limit
        when (some (lambda (u) (zerop (mod i u))) seeds)
          sum i))

PowerShell, 57 bytes

param($a,$b)(1..--$a|?{$i=$_;$b|?{!($i%$_)}})-join'+'|iex

Try it online!

Iterative solution. Takes input as a number $a and as a literal array $b. Loops from 1 up to one below $a (via --$a), using a Where-Object operator |?{...} with a clause to select certain numbers.

The clause sets $i to be the current number before sending input array $b into another |?{...}, here picking out those items where the current number is evenly divided by at least one of the numbers in $b. Those elements of $b that do divide evenly are left on the pipeline.

Thus, if there is at least one element from $b, the pipeline contains an element, so the outer Where is $true and the current number is left on the pipeline. Otherwise, with no elements from $b on the pipeline, the outer Where is $false, so the current number is not placed on the pipeline.

Those numbers are all gathered up in parens, -joined together with + signs, and piped to |iex (short for Invoke-Expression and similar to eval). The summation result is left on the pipeline, and output is implicit.

PHP, 78 76 74 bytes

for(;++$i<$argv[$f=$k=1];$s+=$i*!$f)for(;$v=$argv[++$k];)$f*=$i%$v;echo$s;

The outer loop runs $i from 1 to below first argument and adds $i to $s if $f is not set.
The inner loop multiplies $f with ($i modulo argument) for all subsequent arguments, setting $f to 0 if $i is the multiple of any of them.

Run with -r.

Scala, 47 bytes

n=>1.to(n(0)-1).filter(i=>n.exists(i%_==0)).sum

n is a List which contains of a first argument N, the rest are elements of A

Works by filtering out numbers where there doesn't exist at least one A of which i is a multiple, then summing. Strictly speaking we should use n.tail.exists inside the closure, but as i is always less than N and therefore never a multiple of N the solution is still complete without this.

Java 8, 75 bytes

(N,A)->IntStream.range(1,N).filter(x->A.stream().anyMatch(y->x%y==0)).sum()

The method signature for this is int f(int N, List<Integer> A)

Ruby, 52 48 46 bytes

->b{b[s=0].times{|x|b.find{|y|x%y<1&&s+=x}};s}

C11, 177 bytes

#include"object.h"
#define S size_t
S g(S m,array_t*d){S s,i,l,j;for(s=i=0;i<m;i++){for(l=1,j=0;j<d->idx+1;l*=i%(S)(*array_get_ref(d,j++,NULL))->fwi->value);s+=l?0:i;}return s;}

Requires this set of headers in the same folder, and the fnv-hash library found there as well. Compile like gcc 1.c ../fnv-hash/libfnv.a -o 1 -DNODEBUG

Test Program:

#include "../calc/object/object.h"
#include <stdio.h>

size_t f (const size_t max, const size_t a, const size_t b);
size_t f2 (const size_t max, const array_t* const divs);
size_t g (size_t max, array_t* divs);

define_array_new_fromctype(size_t);

int main(void) {
  printf("%zu\n", f(10, 3, 5));
  static const size_t a[] = {
    3, 5
  };
  array_t* b = array_new_from_size_t_lit(a, 2, t_realuint);
  printf("%zu\n", f2(10, b));
  printf("%zu\n", g(10, b));
  array_destruct(b);
  return 0;
}

size_t f (const size_t max, const size_t a, const size_t b) {
  size_t sum = 0;
  for (size_t i = 0; i < max; i++) {
    sum += (i % a * i % b) ? 0 : i;
  }
  return sum;
}

size_t f2 (const size_t max, const array_t* const divs) {
  size_t sum = 0;
  const size_t len = array_length(divs);

  for (size_t i = 0; i < max; i++) {
    size_t mul = 1;
    for (size_t j = 0; j < len; j++) {
      object_t** this = array_get_ref(divs, j, NULL);

      fixwid_t*   num = (*this)->fwi;

      mul *= i % (size_t) num->value;
    }
    sum += mul ? 0 : i;
  }
  return sum;
}

#define S size_t
S g(S m,array_t*d){S s,i,l,j;for(s=i=0;i<m;i++){for(l=1,j=0;j<d->idx+1;l*=i%(S)(*array_get_ref(d,j++,NULL))->fwi->value);s+=l?0:i;}return s;}

outputs

23
23
23

Japt -x, 9 7 6 bytes

Ç*VøZâ

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           :Implicit input of integer U and array V
Ç          :Map each Z in the range [0,U)
 *         :  Multiply by
  Vø       :  Does V contain
    Zâ     :   Any of the divisors of Z
           :Implicit output of sum of resulting array

Whispers v2, 178 bytes

> Input
> Input
> ℕ
>> (1)
>> ∤L
>> {L}
>> L∩2
>> #L
>> L∈3
>> L⋅R
>> Each 5 4
>> Each 6 11
>> Each 7 12
>> Each 8 13
>> Each 9 14
>> Each 10 15 4
>> ∑16
>> Output 17

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Structure tree:

How it works

Very simply, we put each number (the lines with Each on them) through a series of functions (the lines with L on them), then, based off the results of those functions, we discard some numbers and keep the rest, before finally summing them. In fact, we can define those functions, where $\alpha$ denotes the set of numbers given as input:

This is what lines 5 through to 10 represent. Lines 11 through 16 are simply the application of those three functions. Once we've defined all the functions, we then mutate $\alpha$ to $\beta$ according to the following rule:

where $\alpha_i$ denotes the $i$th element of $\alpha$, and the same for $\beta$. Finally, we can simply take the sum of $\beta$, as the $0$ elements do not affect the sum.

K (oK), 15 14 bytes

Solution:

{+/&|/~y!\:!x}

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Examples:

{+/&|/~y!\:!x}[50;,2]
600
{+/&|/~y!\:!x}[10;3 5]
23

Explanation:

{+/&|/~y!\:!x} / the solution
{            } / lambda taking implicit x and y
           !x  / range 0..x-1
       y!\:    / modulo (!) x with each-left (\:) item in y
      ~        / not
    |/         / min-over to flatten into single list
   &           / indices where true
 +/            / sum up

Actually, 13 bytes

DR∙`i;)%Y*`MΣ

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Explanation:

DR∙`i;)%Y*`MΣ
DR             range(1, N)
  ∙            Cartesian product with A
   `i;)%Y*`M   for each pair:
    i;)          flatten, make a copy of the value from the range
       %Y        test if value from range divides value from A
         *       value from range if above is true else 0
            Σ  sum

Processing, 88 bytes

int q(int a,int[]b){int s=0,i=0;for(;++i<a;)for(int j:b)if(i%j<1){s+=i;break;}return s;}

Uses the simple for-loop approach, sums all the multiples up and returns it. Input is the format int, int[]array.