给定两个正整数a和b，输出两个正整数c，d使得：

• c 分界 a
• d 分界 b
• c并且d是互质的
• 的最小公倍数的c和d等于最小公倍数a和b。

如果可能的答案不只一个，则只能输出其中一个或全部。

测试用例：

 a  b  c  d
12 18  4  9
18 12  9  4
 5  7  5  7
 3  6  1  6 or 3 2
 9  9  9  1 or 1 9
 6 15  2 15 or 6 5
 1  1  1  1

这是代码高尔夫球。以字节为单位的最短答案将获胜。

果冻，21 13字节

ÆEz®0iṂ$¦€ZÆẸ

在线尝试！

如果a = 2 ^A ·3 ^B ·5 ^C ·…并且b = ^2α · ^3β · ^5γ ·…，则我们计算

• c = 2 ^A>α？A：0 ·3 ^B>β？B：0 ·5 ^C>γ？C：0 ·…

• d = 2 ^{A>α？0：α} ·3 ^{B>β？0：β} ·5 ^{C>γ？0：γ} ·…

现在lcm（c，d）= 2 ^{max（A>α？A：0，A>α？0：α）} ·…= 2 ^{max（A，α）} ·3 ^{max（B，β）} ·…= lcm（ a，b）

并且gcd（c，d）= 2 ^{分钟（A>α？A：0，A>α？0：α）} ·…= 2 ⁰ ·3 ⁰ ·5 ⁰ ·…= 1。

换句话说：从（c，d）=（a，b）开始。然后，对于每个素数，将其完全除以c或d的因式分解：以那个素数最小的那个为准。（在此实现中，如果出现平局，则c失去其指数。）

因此，如果a = 2250 = 2 ¹ ·3 ² ·5 ³和b = 360 = 2 ³ ·3 ² ·5 ¹，

那么c = 2 ⁰ ·3 ⁰ ·5 ³ = 125，d = 2 ³ ·3 ² ·5 ⁰ = 72。

乔纳森·艾伦（Jonathan Allan）击倒了8个字节！谢谢〜

R，143个 139 123字节

f=function(a,b,q=1:(a*b))for(i in 1:a)for(j in 1:b)if(!a%%i+b%%j&max(q[!i%%q+j%%q])<2&i*j==min(q[!q%%a+q%%b]))cat(i,j,"\n")

（感谢@Giuseppe节省了这19个字节！）

带有缩进，换行符和一些说明：

f=function(a,b,
           q=1:(a*b)) #Defined as function arguments defaults to avoid having to use curly brackets
    for(i in 1:a)
        for(j in 1:b)
            if(!a%%i + b%%j & #Is a divided by c and b divided by d
               max(q[!i%%q+j%%q])<2 & #Are c and d coprimes
               i*j==min(q[!q%%a+q%%b])) #Is this the same lcm
                   cat(i,j,"\n") #Then print

测试用例：

> f=function(a,b,q=1:(a*b))for(i in 1:a)for(j in 1:b)if(!a%%i+b%%j&max(q[!i%%q+j%%q])<2&i*j==min(q[!q%%a+q%%b]))cat(i,j,"\n")
> f(5,7)
5 7 
> f(12,18)
4 9 
> f(6,15)
2 15 
6 5 
> f(1,1)
1 1 

外壳，10个字节

→ÖF§-⌋⌉ΠmḊ

蛮力。获取并返回列表，并且也适用于两个以上的数字。在线尝试！

说明

→ÖF§-⌋⌉ΠmḊ  Implicit input, say [6,15]
        mḊ  Map divisors: [[1,2,3,6],[1,3,5,15]]
       Π    Cartesian product:[[1,1],[2,1],[1,3],[2,3],[3,1],[1,5],[3,3],[6,1],[1,15],[2,5],[3,5],[6,3],[2,15],[6,5],[3,15],[6,15]]
 Ö          Sort by
  F         reduce by
     ⌉      lcm
   -⌋       minus gcd: [[1,1],[3,3],[2,1],[1,3],[3,1],[6,3],[1,5],[2,3],[6,1],[2,5],[3,15],[1,15],[3,5],[6,15],[2,15],[6,5]]
→           Get last element: [6,5]

Mathematica，82个字节

#&@@Select[Subsets[Flatten@Divisors[{t=#,r=#2}],{2}],GCD@@#==1&&LCM@@#==t~LCM~r&]&

JavaScript（ES6），90 84 80字节

以currying语法获取输入，(a)(b)并返回2个整数的数组。

a=>g=(b,c=1)=>(G=(a,b)=>b?G(b,a%b):a)(c,d=a*b/G(a,b)/c)-1|a%c|b%d?g(b,c+1):[c,d]

测试用例

let f =

a=>g=(b,c=1)=>(G=(a,b)=>b?G(b,a%b):a)(c,d=a*b/G(a,b)/c)-1|a%c|b%d?g(b,c+1):[c,d]

console.log(JSON.stringify(f(12)(18))) // [ 4,  9 ]
console.log(JSON.stringify(f(18)(12))) // [ 9,  4 ]
console.log(JSON.stringify(f( 5)( 7))) // [ 5,  7 ]
console.log(JSON.stringify(f( 3)( 6))) // [ 1,  6 ] or [ 3, 2 ]
console.log(JSON.stringify(f( 9)( 9))) // [ 9,  1 ] or [ 1, 9 ]
console.log(JSON.stringify(f( 6)(15))) // [ 2, 15 ] or [ 6, 5 ]
console.log(JSON.stringify(f( 1)( 1))) // [ 1,  1 ]

怎么样？

a =>                            // a = first input
  g = (                         // g = recursive function that takes:
    b,                          //   b = second input
    c = 1                       //   c = first output divisor, initially set to 1
  ) =>                          //
    (G = (a, b) =>              // G = function that takes a and b
      b ? G(b, a % b) : a       //     and returns the greatest common divisor
    )(                          // we call it with:
      c,                        //   - c
      d = a * b / G(a, b) / c   //   - d = LCM(a, b) / c = a * b / GCD(a, b) / c
    ) - 1 |                     // if the result is not 1 (i.e. c and d are not coprime)
    a % c |                     // or c does not divide a
    b % d ?                     // or d does not divide b:
      g(b, c + 1)               //   do a recursive call with c + 1
    :                           // else:
      [c, d]                    //   return [c, d], a valid factorization of the LCM

MATL，17 16字节

&YFt&X>2:!=*^!Xp

在线尝试！

与Lynn Jelly解决方案相同的方法

自从我使用任何MATL（或者使用matlab）以来已经有一段时间了，所以可能会有很多改进。

哈斯克尔，50 48 47 45 42字节

(?)=gcd;a!b|c<-div a$a?b=(c*c?b,div b$c?b)

想法：我注意到了c*d = a*b/gcd(a,b)。因此，该算法执行两个步骤：

1. 以c' = a/gcd(a,b)和开头d' = b。这满足不同的是所有要求c'，并d'必须互质。
2. 为了使它们互质，我计算e = gcd(c',d')然后设置c = c'*e和d = d'/e。这使所有属性（因为综合因素保持不变），但因为我删除所有共享的因素d，我做c和d互素。

在我的实现中，c'称为c。

在线尝试！

-3字节感谢Laikoni

05AB1E，12个字节

Ñ`âʒ.¿I.¿Q}н

在线尝试！ 或作为测试套件

R，126个字节

function(a,b,g=function(x,y)ifelse(o<-x%%y,g(y,o),y),l=a*b/g(a,b))matrix(c(C<-(1:l)[!l%%1:l],D<-l/C),,2)[g(C,D)<2&!a%%C+b%%D,]

在线尝试！

与其他R答案相比，这采用了一种不同的方法（并且似乎少了些高尔夫球运动）来找到值。

说明：

function(a,b){
 G <- function(x,y)ifelse(o<-x%%y,G(y,o),y) #gcd function, vectorized for x,y
 l <- a*b/g(a,b)                            #lcm of a,b
 C <- (1:l)[!l%%1:l]                        #divisors of l
 D <- l/C                                   #l/C is the other half of the pair
 rel_prime <- G(C, D) < 2                   #pairs where C,D are relatively prime, lol, GCD
 a_div <- !a%%C                             #divisors of a
 b_div <- !b%%D                             #divisors of b
 C <- C[rel_prime & a_div & b_div]
 D <- D[rel_prime & a_div & b_div]          #filter out the bad pairs
 matrix(c(C,D),,ncol = 2)                   #matrix of pairs, returned
}

除了我将所有定义作为默认参数外，我还考虑了一条线的所有计算。

J，19个字节

(*/:"1)&.|:&.(_&q:)

在线尝试！

基于@Lynn的解决方案。

说明

(*/:"1)&.|:&.(_&q:)  Input: [a, b]
              _&q:   Get exponenets of each prime
         |:&         Transpose
  /:"1 &             Grade each row
 *                   Multiply elementwise
       &.|:          Transpose
           &. _&q:   Convert exponents back to numbers

Haskell，91 74字节

a!b=[(x,y)|x<-[1..a],y<-[1..b],rem a x+rem b y+gcd x y<2,lcm a b==lcm x y]

在线尝试！

Saved 17 bytes thanks to Laikoni

Python 2, 75 bytes

def f(x):n=1;exec'n+=1;j=k=1\nwhile x[j]%k<1:k*=n**j;j^=1\nx[j]/=k/n;'*x[0]

Input is taken as a list, which the function modifies in place.

Try it online!

Python 3, 129 bytes

lambda a,b:[[c,d]for c in range(1,-~a)for d in range(1,-~b)if((gcd(c,d)<2)*a*b/gcd(a,b)==c*d/gcd(c,d))>a%c+b%d]
from math import*

Try it online! or Try the test suite.

Outputs all possible combinations in the form of a nested list.

Jelly,  19 15  14 bytes

-4 with pointer from Leaky Nun (use divisor built-in)

^{I am almost 100% certain this is not the way to actually do this one, but here is a first attempt.
Let's see who outgolfs it with a seven or eight byter!
Yep... see Lynn's answer with explanation!}

g/־l/
ÆDp/ÇÐṂ

A monadic link taking a list of the two numbers and returning a list of lists of the possibilities.

Try it online!

How?

g/־l/  - Link: gcd divided by lcm: list [x, y]
g/      - reduce by gcd = gcd(x, y)
   æl/  - reduce by lcm = lcm(x,y)
  ÷     - divide

ÆDp/ÇÐṂ - Main link: list [a, b]    e.g. [160, 90]
ÆD      - divisors (vectorises)          [[1,2,4,5,8,10,16,20,32,40,80,160],[1,2,3,5,6,9,10,15,18,30,45,90]]
  p/    - reduce by Cartesian product    [[1,1],[1,2],...,[1,90],[2,1],[2,2],...,[2,90],....,[160,90]]
     ÐṂ - entries for which this is minimal:
    Ç   -   call the last link (1) as a monad

Perl 6, 72 bytes

{([X] map {grep $_%%*,1..$_},@^a).grep:{([lcm] @a)==([lcm] $_)==[*] $_}}

Try it online!

Takes a list (a, b). Returns a list of all possible lists (c, d).

Explanation:

-> @ab {
    # Generate all pairs (c, d)
    ([X]
         # where c divides a and d divides b.
         map { grep $_%%*, 1..$_ }, @ab)
    # Only keep pairs with lcm(a, b) = lcm(c, d) and lcm(c, d) = c * d.
    # The latter implies gcd(c, d) = 1.
    .grep: { ([lcm] @ab) == ([lcm] $_) == [*] $_ }
}

Python 2, 126 121 bytes

def f(a,b):
 c=[1,1];p=2
 while p<=a*b:
	t=m=1
	while(a*b)%p<1:m*=p;t=b%p<1;a/=p**(a%p<1);b/=p**t
	p+=1;c[t]*=m
 return c

Try it online!

Python 2 + sympy, 148 bytes

from sympy import*
a,b=input()
c=d=z=1
while(a/c*c+b/d*d<a+b)+gcd(c,d)-1+(lcm(c,d)!=lcm(a,b)):E=c==d==z;Q=c==z;d=+E or Q+d;c=+Q or-~c;z+=E
print c,d

Try it online!

-1 thanks to Jonathan Frech.

This answer works in Python 2 (not Python 3), using sympy.gcd and sympy.lcm instead of math.gcd and math.lcm which are only available in Python 3. And yes, this is brute force :)

Jelly, 13 bytes

Ụ€’×
ÆEz0ÇZÆẸ

Try it online! My first Jelly answer! Edit: ÆEz0µỤ€’×µZÆẸ also works for 13 bytes. Explanation:

ÆE              Get prime factor exponents of both values (vectorises)
  z0            Zip but fill the shorter array with 0
    µ           New monadic link
     Ụ€         Grade up each pair (1-indexed)
       ’        Convert to 0-indexing (vectorises)
        ×       Multiply each pair by its grade (vectorises)
         µ      New monadic link
          Z     Zip back into separate lists of prime factor exponents
           ÆẸ   Turn prime exponent lists back into values (vectorises)

PARI/GP, 86 bytes

This just does what Lynn says in her answer:

f(a,b)=forprime(p=2,a*b,v=valuation(a,p);w=valuation(b,p);if(w<v,b/=p^w,a/=p^v));[a,b]

If I do not count the f(a,b)= part, it is 79 bytes.

05AB1E, 32 26 24 22 20 19 bytes

Ó0ζεD`›0sǝ}øεā<ØsmP

Try it online! I still have no idea how to write in this language, but at least it's not a brute-force algorithm. Explanation:

Ó                       Get exponents of prime factors (vectorised)
 0ζ                     Zip, filling with 0
   ε      }             For each prime
    D`                  Extract the pair of exponents
      ›0sǝ              Overwrite the smaller with 0
           ø            Zip back into two lists of prime exponents
            ε           For each list (} implied)
             ā<Ø        Get a list of primes
                sm      Raise each prime to the exponent
                  P     Take the product