Programming-Idioms

# 75 Compute LCM
New implementation

Be concise.

Be useful.

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Please try to avoid dependencies to third-party libraries and frameworks.

Other implementations 
#include <gmp.h>
mpz_t _a, _b, _x;
mpz_init_set_str(_a, "123456789", 10);
mpz_init_set_str(_b, "987654321", 10);
mpz_init(_x);

mpz_lcm(_x, _a, _b);
gmp_printf("%Zd\n", _x);
#include <numeric>
auto x = std::lcm(a, b);
int gcd(int a, int b)
{
  while (b != 0)
  {
    int t = b;
    b = a % t;
    a = t;
  }
  return a;
}

int lcm(int a, int b)
{
  if (a == 0 || b == 0)
    return 0;
  return (a * b) / gcd(a, b);
}

int x = lcm(140, 72);
import std.numeric: gcd
uint x = (a * b) / gcd(a, b);
x = lcm(a, b);

int lcm(int a, int b) => (a * b) ~/ gcd(a, b);

int gcd(int a, int b) {
  while (b != 0) {
    var t = b;
    b = a % t;
    a = t;
  }
  return a;
}
defmodule BasicMath do
	def gcd(a, 0), do: a
	def gcd(0, b), do: b
	def gcd(a, b), do: gcd(b, rem(a,b))
	
	def lcm(0, 0), do: 0
	def lcm(a, b), do: (a*b)/gcd(a,b)
end
gcd(A,B) when A == 0; B == 0 -> 0;
gcd(A,B) when A == B -> A;
gcd(A,B) when A > B -> gcd(A-B, B);
gcd(A,B) -> gcd(A, B-A).

lcm(A,B) -> (A*B) div gcd(A, B).
import "math/big"
gcd.GCD(nil, nil, a, b)
x.Div(a, gcd).Mul(x, b)
x = lcm a b

const gcd = (a, b) => b === 0 ? a : gcd (b, a % b)
let x = (a * b) / gcd(a, b)
import java.math.BigInteger;
BigInteger a = new BigInteger("123456789");
BigInteger b = new BigInteger("987654321");
BigInteger x = a.multiply(b).divide(a.gcd(b));
(setf x (lcm a b))
extension=gmp
$gcd = gmp_lcm($a, $b);
echo gmp_strval($gcd);
sub lcm {
	use integer;
	my ($x, $y) = @_;
	my ($f, $s) = @_;
	while ($f != $s) {
		($f, $s, $x, $y) = ($s, $f, $y, $x) if $f > $s;
		$f = $s / $x * $x;
		$f += $x if $f < $s;
	}
	$f
}
sub gcd {
	my ($x, $y) = @_;
	while ($x) { ($x, $y) = ($y % $x, $x) }
	$y
}
 
sub lcm {
	my ($x, $y) = @_;
	($x && $y) and $x / gcd($x, $y) * $y or 0
}
from math import gcd
x = (a*b)//gcd(a, b)
x = a.lcm(b)
extern crate num;

use num::Integer;
use num::bigint::BigInt;
let x = a.lcm(&b);

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