Question

如何计算多个数字的最小公倍数？

到目前为止，我只能在两个数字之间进行计算。但不知道如何扩展它来计算3个或更多的数字。

到目前为止，我就是这样做的

LCM = num1 * num2 /  gcd ( num1 , num2 )

使用gcd是计算数字的最大公约数的函数。使用欧几里德算法

但我无法弄清楚如何计算3个或更多数字。

Answer 1

您可以通过迭代计算两个数字的最小公倍数来计算两个以上数字的最小公倍数，即

lcm(a,b,c) = lcm(a,lcm(b,c))

Answer 2

在Python中（修改后的primes.py）：

def gcd(a, b):
    """Return greatest common divisor using Euclid's Algorithm."""
    while b:      
        a, b = b, a % b
    return a

def lcm(a, b):
    """Return lowest common multiple."""
    return a * b // gcd(a, b)

def lcmm(*args):
    """Return lcm of args."""   
    return reduce(lcm, args)

用法：

>>> lcmm(100, 23, 98)
112700
>>> lcmm(*range(1, 20))
232792560

reduce()的作品类似于that：

>>> f = lambda a,b: "f(%s,%s)" % (a,b)
>>> print reduce(f, "abcd")
f(f(f(a,b),c),d)

Answer 3

这是一个ECMA风格的实现：

function gcd(a, b){
    // Euclidean algorithm
    var t;
    while (b != 0){
        t = b;
        b = a % b;
        a = t;
    }
    return a;
}

function lcm(a, b){
    return (a * b / gcd(a, b));
}

function lcmm(args){
    // Recursively iterate through pairs of arguments
    // i.e. lcm(args[0], lcm(args[1], lcm(args[2], args[3])))

    if(args.length == 2){
        return lcm(args[0], args[1]);
    } else {
        var arg0 = args[0];
        args.shift();
        return lcm(arg0, lcmm(args));
    }
}

Answer 4

我会选择这个（C＃）：

static long LCM(long[] numbers)
{
    return numbers.Aggregate(lcm);
}
static long lcm(long a, long b)
{
    return Math.Abs(a * b) / GCD(a, b);
}
static long GCD(long a, long b)
{
    return b == 0 ? a : GCD(b, a % b);
}

只是一些澄清，因为乍一看它没有接缝这么清楚这个代码在做什么：

Aggregate是一个Linq扩展方法，所以你不能忘记使用System.Linq添加到你的引用。

Aggregate获取累积函数，因此我们可以在IEnumerable上使用属性lcm（a，b，c）= lcm（a，lcm（b，c））。 More on Aggregate

GCD计算使用Euclidean algorithm。

lcm计算使用Abs（a * b）/ gcd（a，b），参考Reduction by the greatest common divisor。

希望这有帮助，

Answer 5

我只是在Haskell中想到了这一点：

lcm' :: Integral a => a -> a -> a
lcm' a b = a`div`(gcd a b) * b
lcm :: Integral a => [a] -> a
lcm (n:ns) = foldr lcm' n ns

我甚至花时间编写自己的gcd函数，只在Prelude中找到它！今天对我的学习很多：D

Answer 6

一些不需要gcd函数的Python代码：

from sys import argv 

def lcm(x,y):
    tmp=x
    while (tmp%y)!=0:
        tmp+=x
    return tmp

def lcmm(*args):
    return reduce(lcm,args)

args=map(int,argv[1:])
print lcmm(*args)

以下是终端的样子：

$ python lcm.py 10 15 17
510

Answer 7

这是一个Python单行（不计入导入），用于返回从1到20的整数的LCM：

Python 3.5+导入：

from functools import reduce
from math import gcd

Python 2.7导入：

from fractions import gcd

通用逻辑：

lcm = reduce(lambda x,y: x*y//gcd(x, y), range(1, 21))

在https://intense-bayou-64974.herokuapp.com/和Python 2中，运算符优先级规则规定*和//运算符具有相同的优先级，因此它们从左到右应用。因此，x*y//z表示(x*y)//z而非x*(y//z)。这两者通常产生不同的结果。这对于浮点除法并不重要，但它对Python 3有意义。

Answer 8

查找任何数字列表的lcm的函数：

 def function(l):
     s = 1
     for i in l:
        s = lcm(i, s)
     return s

Answer 9

这是Virgil Disgr4ce的C＃端口实现：

public class MathUtils
{
    /// <summary>
    /// Calculates the least common multiple of 2+ numbers.
    /// </summary>
    /// <remarks>
    /// Uses recursion based on lcm(a,b,c) = lcm(a,lcm(b,c)).
    /// Ported from http://stackoverflow.com/a/2641293/420175.
    /// </remarks>
    public static Int64 LCM(IList<Int64> numbers)
    {
        if (numbers.Count < 2)
            throw new ArgumentException("you must pass two or more numbers");
        return LCM(numbers, 0);
    }

    public static Int64 LCM(params Int64[] numbers)
    {
        return LCM((IList<Int64>)numbers);
    }

    private static Int64 LCM(IList<Int64> numbers, int i)
    {
        // Recursively iterate through pairs of arguments
        // i.e. lcm(args[0], lcm(args[1], lcm(args[2], args[3])))

        if (i + 2 == numbers.Count)
        {
            return LCM(numbers[i], numbers[i+1]);
        }
        else
        {
            return LCM(numbers[i], LCM(numbers, i+1));
        }
    }

    public static Int64 LCM(Int64 a, Int64 b)
    {
        return (a * b / GCD(a, b));
    }

    /// <summary>
    /// Finds the greatest common denominator for 2 numbers.
    /// </summary>
    /// <remarks>
    /// Also from http://stackoverflow.com/a/2641293/420175.
    /// </remarks>
    public static Int64 GCD(Int64 a, Int64 b)
    {
        // Euclidean algorithm
        Int64 t;
        while (b != 0)
        {
            t = b;
            b = a % b;
            a = t;
        }
        return a;
    }
}'

Answer 10

使用LINQ你可以写：

static int LCM(int[] numbers)
{
    return numbers.Aggregate(LCM);
}

static int LCM(int a, int b)
{
    return a * b / GCD(a, b);
}

应添加using System.Linq;并且不要忘记处理异常......

Answer 11

这是 Swift 。

// Euclid's algorithm for finding the greatest common divisor
func gcd(_ a: Int, _ b: Int) -> Int {
  let r = a % b
  if r != 0 {
    return gcd(b, r)
  } else {
    return b
  }
}

// Returns the least common multiple of two numbers.
func lcm(_ m: Int, _ n: Int) -> Int {
  return m / gcd(m, n) * n
}

// Returns the least common multiple of multiple numbers.
func lcmm(_ numbers: [Int]) -> Int {
  return numbers.reduce(1) { lcm($0, $1) }
}

Answer 12

只是为了好玩，一个shell（几乎所有shell）实现：

#!/bin/sh
gcd() {   # Calculate $1 % $2 until $2 becomes zero.
      until [ "$2" -eq 0 ]; do set -- "$2" "$(($1%$2))"; done
      echo "$1"
      }

lcm() {   echo "$(( $1 / $(gcd "$1" "$2") * $2 ))";   }

while [ $# -gt 1 ]; do
    t="$(lcm "$1" "$2")"
    shift 2
    set -- "$t" "$@"
done
echo "$1"

尝试：

$ ./script 2 3 4 5 6

获取

最大输入和结果应小于(2^63)-1或shell数学将包装。

Answer 13

Scala版本：

def gcd(a: Int, b: Int): Int = if (b == 0) a else gcd(b, a % b)
def gcd(nums: Iterable[Int]): Int = nums.reduce(gcd)
def lcm(a: Int, b: Int): Int = if (a == 0 || b == 0) 0 else a * b / gcd(a, b)
def lcm(nums: Iterable[Int]): Int = nums.reduce(lcm)

Answer 14

我正在寻找gcd和lcm的数组元素，并在以下链接中找到了一个很好的解决方案。

https://www.hackerrank.com/challenges/between-two-sets/forum

包括以下代码。 gcd的算法使用欧几里得算法在下面的链接中解释得很好。

https://www.khanacademy.org/computing/computer-science/cryptography/modarithmetic/a/the-euclidean-algorithm

private static int gcd(int a, int b) {
    while (b > 0) {
        int temp = b;
        b = a % b; // % is remainder
        a = temp;
    }
    return a;
}

private static int gcd(int[] input) {
    int result = input[0];
    for (int i = 1; i < input.length; i++) {
        result = gcd(result, input[i]);
    }
    return result;
}

private static int lcm(int a, int b) {
    return a * (b / gcd(a, b));
}

private static int lcm(int[] input) {
    int result = input[0];
    for (int i = 1; i < input.length; i++) {
        result = lcm(result, input[i]);
    }
    return result;
}

Answer 15

在R中，我们可以使用数字包中的 mGCD （x）和 mLCM （x）函数来计算最大值公共向量x中的所有数字的公约数和最小公倍数：

    library(numbers)
    mGCD(c(4, 8, 12, 16, 20))
[1] 4
    mLCM(c(8,9,21))
[1] 504
    # Sequences
    mLCM(1:20)
[1] 232792560

Answer 16

以下是 PHP 实施：

    // https://stackoverflow.com/q/12412782/1066234
    function math_gcd($a,$b) 
    {
        $a = abs($a); 
        $b = abs($b);
        if($a < $b) 
        {
            list($b,$a) = array($a,$b); 
        }
        if($b == 0) 
        {
            return $a;      
        }
        $r = $a % $b;
        while($r > 0) 
        {
            $a = $b;
            $b = $r;
            $r = $a % $b;
        }
        return $b;
    }

    function math_lcm($a, $b)
    {
        return ($a * $b / math_gcd($a, $b));
    }

    // https://stackoverflow.com/a/2641293/1066234
    function math_lcmm($args)
    {
        // Recursively iterate through pairs of arguments
        // i.e. lcm(args[0], lcm(args[1], lcm(args[2], args[3])))

        if(count($args) == 2)
        {
            return math_lcm($args[0], $args[1]);
        }
        else 
        {
            $arg0 = $args[0];
            array_shift($args);
            return math_lcm($arg0, math_lcmm($args));
        }
    }

    // fraction bonus
    function math_fraction_simplify($num, $den) 
    {
        $g = math_gcd($num, $den);
        return array($num/$g, $den/$g);
    }


    var_dump( math_lcmm( array(4, 7) ) ); // 28
    var_dump( math_lcmm( array(5, 25) ) ); // 25
    var_dump( math_lcmm( array(3, 4, 12, 36) ) ); // 36
    var_dump( math_lcmm( array(3, 4, 7, 12, 36) ) ); // 252

积分用他的answer above (ECMA-style code)转到@ T3db0t。

Answer 17

ES6风格

function gcd(...numbers) {
  return numbers.reduce((a, b) => b === 0 ? a : gcd(b, a % b));
}

function lcm(...numbers) {
  return numbers.reduce((a, b) => Math.abs(a * b) / gcd(a, b));
}

Answer 18

你可以用另一种方式做到 - 设n个数字。取一对连续数字并将其lcm保存在另一个数组中。在第一次迭代程序中执行此操作会进行n / 2次迭代。然后接下来从0开始拾取对，如（0,1），（2,3）等等。计算它们的LCM并存储在另一个数组中。这样做直到你留下一个阵列。（如果n是奇数，则无法找到lcm）

Answer 19

这是我用的 -

def greater(n):

      a=num[0]

      for i in range(0,len(n),1):
       if(a<n[i]):
        a=n[i]
      return a

r=input('enter limit')

num=[]

for x in range (0,r,1):

    a=input('enter number ')
    num.append(a)
a= greater(num)

i=0

while True:

    while (a%num[i]==0):
        i=i+1
        if(i==len(num)):
               break
    if i==len(num):
        print 'L.C.M = ',a
        break
    else:
        a=a+1
        i=0

Answer 20

对于python 3：

from functools import reduce

gcd = lambda a,b: a if b==0 else gcd(b, a%b)
def lcm(lst):        
    return reduce(lambda x,y: x*y//gcd(x, y), lst)

Answer 21

在python中：

def lcm(*args):
    """Calculates lcm of args"""
    biggest = max(args) #find the largest of numbers
    rest = [n for n in args if n != biggest] #the list of the numbers without the largest
    factor = 1 #to multiply with the biggest as long as the result is not divisble by all of the numbers in the rest
    while True:
        #check if biggest is divisble by all in the rest:
        ans = False in [(biggest * factor) % n == 0 for n in rest]
        #if so the clm is found break the loop and return it, otherwise increment factor by 1 and try again
        if not ans:
            break
        factor += 1
    biggest *= factor
    return "lcm of {0} is {1}".format(args, biggest)

>>> lcm(100,23,98)
'lcm of (100, 23, 98) is 112700'
>>> lcm(*range(1, 20))
'lcm of (1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19) is 232792560'

Answer 22

在Ruby中，它很简单：

> [2, 3, 4, 6].reduce(:lcm)
=> 12

> [16, 32, 96].reduce(:gcd)
=> 16

（在Ruby 2.2.10和2.6.3上测试。）

Answer 23

Python 3.9 math模块的gcd和lcm支持数字列表。

import math

lst = [1,2,3,4,5,6,7,8,9]

print(math.lcm(*lst))

print(math.gcd(*lst))

Answer 24

int gcd(int a, int b) { if (b == 0) return a; return gcd(b, a%b); } int lcm(int[] a, int n) { int res = 1, i; for (i = 0; i < n; i++) { res = res*a[i]/gcd(res, a[i]); } return res; }

Answer 25

对于任何寻找快速工作代码的人，请尝试以下方法：

我写了一个函数 lcm_n(args, num) ，它计算并返回数组args中所有数字的lcm。第二个参数num是数组中的数字计数。

将所有这些数字放在数组args中，然后调用lcm_n(args,num);

之类的函数

此功能返回所有这些数字的lcm。

以下是函数lcm_n(args, num)的实现：

int lcm_n(int args[], int num) //lcm of more than 2 numbers
{
    int i, temp[num-1];

    if(num==2)
    {
        return lcm(args[0], args[1]);
    }
    else
    {
        for(i=0;i<num-1;i++)
        {
           temp[i] = args[i];   
        }

        temp[num-2] = lcm(args[num-2], args[num-1]);
        return lcm_n(temp,num-1);
    }
}

此功能需要以下两个功能才能工作。所以，只需将它们与它一起添加即可。

int lcm(int a, int b) //lcm of 2 numbers
{
    return (a*b)/gcd(a,b);
}


int gcd(int a, int b) //gcd of 2 numbers
{
    int numerator, denominator, remainder;

    //Euclid's algorithm for computing GCD of two numbers
    if(a > b)
    {
        numerator = a;
        denominator = b;
    }
    else
    {
        numerator = b;
        denominator = a;
    }
    remainder = numerator % denominator;

    while(remainder != 0)
    {
        numerator   = denominator;
        denominator = remainder;
        remainder   = numerator % denominator;
    }

    return denominator;
}

Answer 26

方法compLCM采用向量并返回LCM。所有数字都在向量in_numbers内。

int mathOps::compLCM(std::vector<int> &in_numbers)
 {
    int tmpNumbers = in_numbers.size();
    int tmpMax = *max_element(in_numbers.begin(), in_numbers.end());
    bool tmpNotDividable = false;

    while (true)
    {
        for (int i = 0; i < tmpNumbers && tmpNotDividable == false; i++)
        {
            if (tmpMax % in_numbers[i] != 0 )
                tmpNotDividable = true;
        }

        if (tmpNotDividable == false)
            return tmpMax;
        else
            tmpMax++;
    }
}

Answer 27

LCM既是关联的又是可交换的。

LCM（A，B，C）= LCM（LCM（A，B），C）= LCM（一，LCM（B，C））

这是C：中的示例代码：

int main()
{
  int a[20],i,n,result=1;  // assumption: count can't exceed 20
  printf("Enter number of numbers to calculate LCM(less than 20):");
  scanf("%d",&n);
  printf("Enter %d  numbers to calculate their LCM :",n);
  for(i=0;i<n;i++)
    scanf("%d",&a[i]);
 for(i=0;i<n;i++)
   result=lcm(result,a[i]);
 printf("LCM of given numbers = %d\n",result);
 return 0;
}

int lcm(int a,int b)
{
  int gcd=gcd_two_numbers(a,b);
  return (a*b)/gcd;
}

int gcd_two_numbers(int a,int b)
{
   int temp;
   if(a>b)
   {
     temp=a;
     a=b;
     b=temp;
   }
  if(b%a==0)
    return a;
  else
    return gcd_two_numbers(b%a,a);
}

Answer 28

我们有工作实施of Least Common Multiple on Calculla，适用于任何数量的输入，同时显示步骤。

我们做的是：

0: Assume we got inputs[] array, filled with integers. So, for example:
   inputsArray = [6, 15, 25, ...]
   lcm = 1

1: Find minimal prime factor for each input.
   Minimal means for 6 it's 2, for 25 it's 5, for 34 it's 17
   minFactorsArray = []

2: Find lowest from minFactors:
   minFactor = MIN(minFactorsArray)

3: lcm *= minFactor

4: Iterate minFactorsArray and if the factor for given input equals minFactor, then divide the input by it:
  for (inIdx in minFactorsArray)
    if minFactorsArray[inIdx] == minFactor
      inputsArray[inIdx] \= minFactor

5: repeat steps 1-4 until there is nothing to factorize anymore. 
   So, until inputsArray contains only 1-s.

就是这样 - 你得到了你的lcm。

Answer 29

clc;

data = [1 2 3 4 5]

LCM=1;

for i=1:1:length(data)

    LCM = lcm(LCM,data(i))

end

Answer 30

这个怎么样？

from operator import mul as MULTIPLY

def factors(n):
    f = {} # a dict is necessary to create 'factor : exponent' pairs 
    divisor = 2
    while n > 1:
        while (divisor <= n):
            if n % divisor == 0:
                n /= divisor
                f[divisor] = f.get(divisor, 0) + 1
            else:
                divisor += 1
    return f


def mcm(numbers):
    #numbers is a list of numbers so not restricted to two items
    high_factors = {}
    for n in numbers:
        fn = factors(n)
        for (key, value) in fn.iteritems():
            if high_factors.get(key, 0) < value: # if fact not in dict or < val
                high_factors[key] = value
    return reduce (MULTIPLY, ((k ** v) for k, v in high_factors.items()))

Answer 31

GCD需要对负数进行一点修正：

def gcd(x,y):
  while y:
    if y<0:
      x,y=-x,-y
    x,y=y,x % y
    return x

def gcdl(*list):
  return reduce(gcd, *list)

def lcm(x,y):
  return x*y / gcd(x,y)

def lcml(*list):
  return reduce(lcm, *list)

Answer 32

如果没有时间限制，这将非常简单直接：

def lcm(a,b,c):
    for i in range(max(a,b,c), (a*b*c)+1, max(a,b,c)):
        if i%a == 0 and i%b == 0 and i%c == 0:
            return i

3个或更多数字的最小公倍数

32 个答案: