Biorąc pod uwagę liczbę naturalną, nnapisz program lub funkcję, aby uzyskać listę wszystkich możliwych mnożników dwóch czynników, które można wykorzystać do osiągnięcia n. Aby lepiej zrozumieć to, co udawał można przejść do http://factornumber.com/?page=16777216 aby zobaczyć, kiedy nto 16777216otrzymujemy następującą listę:

   2 × 8388608  
   4 × 4194304  
   8 × 2097152  
  16 × 1048576  
  32 ×  524288  
  64 ×  262144  
 128 ×  131072  
 256 ×   65536  
 512 ×   32768  
1024 ×   16384
2048 ×    8192
4096 ×    4096

Nie musisz drukować takich rzeczy jak tutaj. Wymagane jest, aby każdy wpis (para czynników) był dobrze odróżniony od siebie, a wewnątrz każdej pary pierwszy czynnik był również dobrze odróżniony od drugiego. Jeśli zdecydujesz się zwrócić listę / tablicę, element wewnętrzny może być listą / tablicą z dwoma elementami lub strukturą twojego języka, która obsługuje parę rzeczy takich jak C ++ std::pair.

Nie drukuj mnożenia przez 1 wpis, ani nie powtarzaj wpisów z pierwszym czynnikiem zamienionym na drugi, ponieważ są one dość bezużyteczne.

Brak zwycięzcy; będzie to golf oparty na języku.

Java (OpenJDK 8) , 81 66 65 bajtów

• -15 Bajtów dzięki Olivier Grégoire.
• -1 bajt: ++j<=i/j-> j++<i/j.

i->{for(int j=1;j++<i/j;)if(i%j<1)System.out.println(j+" "+i/j);}

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Stary (w celach informacyjnych)

Java (OpenJDK 8) , 126 bajtów

i->{java.util.stream.IntStream.range(2,i).filter(d->d<=i/d&&i%d==0).forEach(e->System.out.println(""+e+"x"+i/e));return null;}

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First codegolf submit and first lambda usage. Future self, please forgive me for the code.

05AB1E, 8 bytes

Ñ‚ø2äн¦

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C (gcc), 58 54 53 bytes

• Saved a byte thanks to Steadybox.

f(N,j){for(j=1;j++*j<N;)N%j||printf("|%d,%d",j,N/j);}

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Python 2, 51 bytes

f=lambda n,k=2:n/k/k*[f]and[(k,n/k)][n%k:]+f(n,k+1)

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51 bytes (thanks to Luis Mendo for a byte)

lambda n:[(n/k,k)for k in range(1,n)if(k*k<=n)>n%k]

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51 bytes

lambda n:[(n/k,k)for k in range(1,n)if n/k/k>n%k*n]

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Haskell, 38 bytes

f x=[(a,b)|a<-[2..x],b<-[2..a],a*b==x]

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APL (Dyalog), 28 bytes

{(⊢,⍵÷⊢)¨o/⍨0=⍵|⍨o←1↓⍳⌊⍵*.5}

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Perl 6, 38 bytes

{map {$^a,$_/$a},grep $_%%*,2.. .sqrt}

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Expanded:

{   # bare block lambda with implicit parameter ｢$_｣

  map
    { $^a, $_ / $a },  # map the number with the other factor

    grep
      $_ %% *,         # is the input divisible by *
      2 .. .sqrt       # from 2 to the square root of the input
}

Brachylog, 8 bytes

{~×≜Ċo}ᵘ

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Explanation

{~×≜Ċo}ᵘ
{     }ᵘ  List the unique outputs of this predicate.
 ~×       Pick a list of integers whose product is the input.
   ≜      Force concrete values for its elements.
    Ċ     Force its length to be 2.
     o    Sort it and output the result.

The ~× part does not include 1s in its output, so for input N it gives [N] instead of [1,N], which is later culled by Ċ. I'm not entirely sure why ≜ is needed...

Japt, 9 bytes

â¬Å£[XZo]

Test it online! Returns an array of arrays, with some nulls at the end; -R flag added to show output more clearly.

Jelly, 8 bytes

½ḊpP⁼¥Ðf

A monadic link taking a number and returning a list of lists (pairs) of numbers.

Try it online! (times out on TIO for the 16777216 example since it would create a list of 68.7 billion pairs and filter down to those with the correct product!)

How?

½ḊpP⁼¥Ðf - Link: number, n     e.g. 144
½        - square root of n          12
 Ḋ       - dequeue*                 [2,3,4,5,6,7,8,9,10,11,12]
  p      - Cartesian product**      [[2,1],[2,2],...[2,144],[3,1],...,[3,144],...,[12,144]
      Ðf - filter keep if:
     ¥   -   last two links as a dyad (n is on the right):
   P     -     product
    ⁼    -     equals
         -                          [[2,72],[3,48],[4,36],[6,24],[8,18],[9,16],[12,12]]

* Ḋ, dequeue, implicitly makes a range of a numeric input prior to acting, and the range function implicitly floors its input, so with, say, n=24 the result of ½ is 4.898...; the range becomes [1,2,3,4]; and the dequeued result is [2,3,4]

** Similarly to above, p, Cartesian product, makes ranges for numeric input - here the right argument is n hence the right argument becomes [1,2,3,...,n] prior to the actual Cartisian product taking place.

Husk, 8 bytes

tüOSze↔Ḋ

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Explanation

tüOSze↔Ḋ  Implicit input, say n=30.
       Ḋ  List of divisors: [1,2,3,5,6,10,15,30]
      ↔   Reverse: [30,15,10,6,5,3,2,1]
   Sze    Zip with original: [[1,30],[2,15],[3,10],[5,6],[6,5],[10,3],[15,2],[30,1]]
 üO       Deduplicate by sort: [[1,30],[2,15],[3,10],[5,6]]
t         Drop first pair: [[2,15],[3,10],[5,6]]

JavaScript (ES6), 55 bytes

n=>eval('for(k=1,a=[];k*++k<n;n%k||a.push([k,n/k]));a')

Demo

let f =

n=>eval('for(k=1,a=[];k*++k<n;n%k||a.push([k,n/k]));a')

console.log(JSON.stringify(f(6)))
console.log(JSON.stringify(f(7)))
console.log(JSON.stringify(f(16777216)))

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Python 2, 59 bytes

lambda N:{(n,N/n,n)[n>N/n:][:2]for n in range(2,N)if N%n<1}

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Jelly, 9 bytes

ÆḌḊµżUṢ€Q

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Octave, 42 bytes

@(n)[y=find(~mod(n,x=2:n)&x.^2<=n)+1;n./y]

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PHP, 70 bytes

As string (70 bytes):

$i++;while($i++<sqrt($a=$argv[1])){echo !($a%$i)?" {$i}x".($a/$i):'';}

As array dump (71 bytes):

$i++;while($i++<sqrt($a=$argv[1]))!($a%$i)?$b[$i]=$a/$i:'';print_r($b);

(im not sure if i can use return $b; instead of print_r since it no longer outputs the array, otherwise i can save 2 bytes here. )

The array gives the results like:

Array
(
    [2] => 8388608
    [4] => 4194304
    [8] => 2097152
    [16] => 1048576

Jelly, 12 bytes

ÆDµżUḣLHĊ$$Ḋ

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How it works

ÆDµżUḣLHĊ$$Ḋ - Main monadic link;
             - Argument: n (integer) e.g. 30
ÆD           - Divisors                   [1, 2, 3, 5, 6, 10, 15, 30]
    U        - Reverse                    [30, 15, 10, 6, 5, 3, 2, 1]
   ż         - Interleave                 [[1, 30], [2, 15], [3, 10], [5, 6], [6, 5], [10, 3], [15, 2], [30, 1]]
         $$  - Last 3 links as a monad
      L      -   Length                   8
       H     -   Halve                    4
        Ċ    -   Ceiling                  4
     ḣ       - Take first elements        [[1, 30], [2, 15], [3, 10], [5, 6]]
           Ḋ - Dequeue                    [[2, 15], [3, 10], [5, 6]]

Wolfram Language (Mathematica), 41 bytes

nRest@Union[Sort@{#,n/#}&/@Divisors@n]

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 is the Function operator, which introduces an unnamed function with named parameter n.

Factor, 58

Well, there has to be some Factor in this question!

[ divisors dup reverse zip dup length 1 + 2 /i head rest ]

It's a quotation. call it with the number on the stack, leaves an assoc (an array of pairs) on the stack.

I'm never sure if all the imports count or not, as they're part of the language. This one uses:

USING: math.prime.factors sequences assocs math ;

(If they count, I should look for a longer solution with shorter imports, which is kind of silly)

As a word:

: 2-factors ( x -- a ) divisors dup reverse zip dup length 1 + 2 /i head rest ;

50 2-factors .
 --> { { 2 25 } { 5 10 } }

Ruby, 43 bytes

->n{(2..n**0.5).map{|x|[[x,n/x]][n%x]}-[p]}

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How it works:

For every number up to sqrt(n), generate the pair [[x, n/x]], then take the n%xth element of this array. If n%x==0 this is [x, n/x], otherwise it's nil. when done, remove all nil from the list.

Pari/GP, 49 34 38 bytes

n->[[d,n/d]|d<-divisors(n),d>1&d<=n/d]

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Set builder notation for all pairs [d, n/d] where d runs through all divisors d of n subject to d > 1 and d <= n/d.

Huge improvement by alephalpha.

Husk, 14 12 bytes

tumoOSe`/⁰Ḋ⁰

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Explanation

tum(OSe`/⁰)Ḋ⁰  -- input ⁰, eg. 30
           Ḋ⁰  -- divisors [1..⁰]: [1,2,3,5,6,10,15,30]
  m(      )    -- map the following function (example on 10):
     Se        --   create list with 10 and ..
       `/⁰     --   .. flipped division by ⁰ (30/10): [10,3]
    O          --   sort: [3,10]
               -- [[1,30],[2,15],[3,10],[5,6],[5,6],[3,10],[2,15],[1,30]]
 u             -- remove duplicates: [[1,30],[2,15],[3,10],[5,6]]
t              -- tail: [[2,15],[3,10],[5,6]]

APL+WIN, 32 bytes

m,[.1]n÷m←(0=m|n)/m←1↓⍳⌊(n←⎕)*.5

Explanation:

(n←⎕) Prompts for screen input

m←(0=m|n)/m←1↓⍳⌊(n←⎕)*.5 Calculates the factors dropping the first

m,[.1]n÷ Identifies the pairs and concatenates into a list.

Add++, 18 15 bytes

L,F@pB]dBRBcE#S

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How it works

L,   - Create a lambda function
     - Example argument:     30
  F  - Factors;     STACK = [1 2 3 5 6 10 15]
  @  - Reverse;     STACK = [15 10 6 5 3 2 1]
  p  - Pop;         STACK = [15 10 6 5 3 2]
  B] - Wrap;        STACK = [[15 10 6 5 3 2]]
  d  - Duplicate;   STACK = [[15 10 6 5 3 2] [15 10 6 5 3 2]]
  BR - Reverse;     STACK = [[15 10 6 5 3 2] [2 3 5 6 10 15]]
  Bc - Zip;         STACK = [[15 2] [10 3] [6 5] [5 6] [3 10] [2 15]]
  E# - Sort each;   STACK = [[2 15] [3 10] [5 6] [5 6] [3 10] [2 15]]
  S  - Deduplicate; STACK = [[[2 15] [3 10] [5 6]]]

Mathematica, 53 bytes

Array[s[[{#,-#}]]&,⌈Length[s=Divisors@#]/2⌉-1,2]&

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Befunge-93, 56 bytes

&1vg00,+55./.:   <
+1< v`\g00/g00:p00
_ ^@_::00g%!00g\#v

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Julia 0.6, 41 bytes

~x=[(y,div(x,y))for y=2:x if x%y<1>y^2-x]

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Redefines the inbuild unary operator ~ and uses an array comprehension to build the output.

• div(x,y) is neccessary for integer division. x/y saves 5 bytes but the output is ~4=(2,2.0).
• Julia allows chaining the comparisons, saving one byte.
• Looping all the way to x avoids Int(floor(√x)).

APL NARS 99 chars

r←f w;i;h
r←⍬⋄i←1⋄→0×⍳0≠⍴⍴w⋄→0×⍳''≡0↑w⋄→0×⍳w≠⌊w⋄→0×⍳w≠+w
A:i+←1⋄→A×⍳∼0=i∣w⋄→0×⍳i>h←w÷i⋄r←r,⊂i h⋄→A

9+46+41+3=99 Test: (where not print nothing, it return something it return ⍬ the list null one has to consider as "no solution")

  f 101    

  f 1 2 3

  f '1'

  f '123'

  f 33 1.23

  f 1.23

  ⎕←⊃f 16777216      
   2 8388608
   4 4194304
   8 2097152
  16 1048576
  32  524288
  64  262144
 128  131072
 256   65536
 512   32768
1024   16384
2048    8192
4096    4096
  f 123
3 41 

Pyt, 67 65 bytes

←ĐðĐ0↔/⅟ƖŽĐŁ₂20`ŕ3ȘĐ05Ș↔ŕ↔Đ4Ș⇹3Ș⦋ƥ⇹⁺Ɩ3ȘĐ05Ș↔ŕ↔Đ4Ș⇹3Ș⦋ƤĐ3Ș⁺ƖĐ3Ș<łĉ

I'm pretty sure this can be golfed.

Basically, the algorithm generates a list of all of the divisors of the input (let's call it n), makes the same list, but flipped, interleaves the two (e.g., if n=24, then, at this point, it has [1,24,2,12,3,8,4,6,6,4,8,3,12,2,24,1]), and prints out the elements from index 2 until half the array length, printing each number on a new line, and with an extra new line in between every pair.

Most of the work is done in actually managing the stack.

Saved 2 bytes by using increment function.

Perl 5, 50 bytes

sub{map[$_,$_[0]/$_],grep!($_[0]%$_),2..sqrt$_[0]}

Ungolfed:

sub {
    return map  { [$_, $_[0] / $_] }
           grep { !($_[0] % $_) }
           (2 .. sqrt($_[0]));
}

Try it online.