如何计算多个号码的最小公倍数？
到目前为止，我只能够两个号码之间进行计算。 但不知道如何将它扩大到3个计算或多个数字。
到目前为止，这是我做到了
LCM = num1 * num2 / gcd ( num1 , num2 )
随着GCD是计算数字的最大公约数的功能。 利用欧氏算法
但我无法弄清楚如何计算的3个或更多的数字。
Answer 1:
您可以通过反复计算两个数字，即的LCM计算超过两个号码的LCM
lcm(a,b,c) = lcm(a,lcm(b,c))
Answer 2:
在Python（改性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()的工作原理是这样认为 ：
>>> f = lambda a,b: "f(%s,%s)" % (a,b)
>>> print reduce(f, "abcd")
f(f(f(a,b),c),d)
Answer 3:
下面是一个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:
我会去与这一个（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，因为乍一看它并没有接缝很清楚这是什么代码是这样做的：
总结是一个LINQ扩展方法，让你无法忘记使用System.Linq的你引用的补充。
聚合得到一个累积功能，所以我们可以利用属性LCM（A，B，C）= LCM的（一个，LCM（B，C））在一个IEnumerable。 更多关于骨料
GCD计算利用了的欧几里德算法 。
LCM计算使用ABS（A * B）/ GCD（A，B），是指由最大公约数还原 。
希望这可以帮助，
Answer 5:
我只是在Haskell想通了这一点：
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
我甚至花时间来写我自己的gcd的功能，才发现它的前奏！ 今天我学习的地段：d
Answer 6:
一些Python代码，不需要GCD的功能：
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)
下面是它看起来像在终端：
$ python lcm.py 10 15 17
510
Answer 7:
这是一个Python的一行（不包括进口），从1返回整数的LCM且20以下：
Python的3.5 +进口：
from functools import reduce
from math import gcd
Python 2.7版进口：
from fractions import gcd
常见的逻辑：
lcm = reduce(lambda x,y: x*y//gcd(x, y), range(1, 21))
在这两个的Python 2和Python 3中 ，运算符优先级规则规定的*和//运营商具有相同的优先级，所以他们从左至右适用。 这样， x*y//z装置(x*y)//z而不是x*(y//z) 两个通常产生不同的结果。 这也无妨尽可能多的浮法事业部，但它对于地板师 。
Answer 8:
下面是维吉尔Disgr4ce的实行的C＃端口：
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 9:
功能查找号码中的任何名单的LCM：
def function(l):
s = 1
for i in l:
s = lcm(i, s)
return s
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);
}
应该添加using System.Linq; 不要忘了处理异常...
Answer 11:
这是斯威夫特 。
// 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:
你可以做的另一种方式 - 让那里是n numbers.Take一对连续的数字，并保存其LCM在另一个数组。 在第一次迭代程序这样做的n / 2 iterations.Then下一个拾取一对从0开始像（0,1），（2,3）等on.Compute其LCM，并存储在另一个阵列。 这样做，直到你留下了一个阵列。 （这是不可能找到LCM如果n为奇数）
Answer 13:
在R，我们可以使用功能MGCD（x）和MLCM（X）从所述包号 ，来计算在整数向量x一起的所有数字的最大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 14:
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 15:
和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 16:
只是为了好玩，一个shell（几乎所有的壳）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"
与尝试：
$ ./script 2 3 4 5 6
要得到
60
最大的输入和结果应当小于(2^63)-1或壳数学将包装。
Answer 17:
我一直在寻找满足gcd和数组元素的LCM，发现在以下链接一个好的解决方案。
https://www.hackerrank.com/challenges/between-two-sets/forum
它包括以下代码。 对于GCD算法使用欧几里德算法下面的链接很好地解释。
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 18:
下面是PHP实现：
// 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
积分去@ T3db0t与他的回答以上（ECMA-样式代码） 。
Answer 19:
GCD需要负数的小修正：
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 20:
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 21:
我们工作实现上Calculla最小公倍数它适用于任何数量的输入也显示的步骤。
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.
这就是它 - 你有你的LCM。
Answer 22:
LCM既是联想和交换。
LCM（A，B，C）= LCM（LCM（A，B），C）= LCM（一，LCM（B，C））
这里是在C语言的示例代码：
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 23:
方法compLCM需要一个载体，并返回LCM。 所有的数字都是矢量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 24:
clc;
data = [1 2 3 4 5]
LCM=1;
for i=1:1:length(data)
LCM = lcm(LCM,data(i))
end
Answer 25:
对于任何人寻找快速的工作代码，试试这个：
我写了一个函数lcm_n(args, num)其计算并返回在数组中的所有数字的LCM args 。 第二个参数num是阵列中的数字的计数。
把阵列中的所有这些数字args ，然后调用如函数lcm_n(args,num);
此功能将所有这些数字的LCM。
下面是函数的实现lcm_n(args, num)
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一起工作。 所以，只要加入他们与它一起。
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:
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 27:
在蟒蛇：
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 28:
这是我用什么 -
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 29:
为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 30:
如果没有时间的约束，这是非常简单和直接的：
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
文章来源: Least common multiple for 3 or more numbers
