Algorithm 3个或更多数字的最小公倍数
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Algorithm 3个或更多数字的最小公倍数,algorithm,math,lcm,Algorithm,Math,Lcm,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 ) gcd是计算数字的最大公约数的函数。使用欧几里德算法 但是我不知道如何计算3个或更多数字的LCM。你可以通过迭代计算两个数字的LCM来计算两个以上数字的LCM,即 lcm(a,b,c) = lcm(a,lcm(b,c)) 在Python中(已修改): 用法: >&g
LCM = num1 * num2 / gcd ( num1 , num2 )
gcd是计算数字的最大公约数的函数。使用欧几里德算法
但是我不知道如何计算3个或更多数字的LCM。你可以通过迭代计算两个数字的LCM来计算两个以上数字的LCM,即
lcm(a,b,c) = lcm(a,lcm(b,c))
在Python中(已修改):
用法:
>>> lcmm(100, 23, 98)
112700
>>> lcmm(*range(1, 20))
232792560
reduce()
的工作原理如下:
我刚刚在Haskell中发现了这一点:
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
我甚至花时间编写了自己的
gcd
函数,结果却在序曲中找到了它!今天我学到了很多:D这里是一个ECMA风格的实现:
function gcd(a, b){
// Euclidean algorithm
while (b != 0){
var temp = b;
b = a % b;
a = temp;
}
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));
}
}
#!/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 "$1"
// https://stackoverflow.com/q/12412782/1066234
function math_gcd($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
你可以用另一种方法- 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。在第一次迭代时这样做,程序会进行n/2次迭代。然后,从0开始选取一对,如(0,1),(2,3)等等。计算它们的LCM并存储在另一个数组中。执行此操作直到只剩下一个数组。
(如果n为奇数,则不可能找到lcm)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)
以下是终端中的外观:
$ python lcm.py 10 15 17
510
这是维吉尔·迪斯格尔4CE的一个C#port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;
}
}'
公共类数学
{
///
///计算2+个数字的最小公倍数。
///
///
///基于lcm(a,b,c)=lcm(a,lcm(b,c))使用递归。
///从http://stackoverflow.com/a/2641293/420175.
///
公共静态Int64 LCM(ILST编号)
{
如果(数字计数<2)
抛出新ArgumentException(“必须传递两个或多个数字”);
返回LCM(数字,0);
}
公共静态Int64 LCM(参数Int64[]编号)
{
返回LCM((ILST)编号);
}
专用静态Int64 LCM(IList编号,int i)
{
//递归地遍历参数对
//即lcm(参数[0]、lcm(参数[1]、lcm(参数[2]、参数[3]))
如果(i+2==数字.计数)
{
返回LCM(编号[i],编号[i+1]);
}
其他的
{
返回LCM(编号[i],LCM(编号,i+1));
}
}
公共静态Int64 LCM(Int64 a、Int64 b)
{
报税表(a*b/GCD(a,b));
}
///
///查找2个数字的最大公分母。
///
///
///也来自http://stackoverflow.com/a/2641293/420175.
///
公共静态Int64 GCD(Int64 a、Int64 b)
{
//欧几里德算法
INT64T;
而(b!=0)
{
t=b;
b=a%b;
a=t;
}
返回a;
}
}'
GCD需要对负数进行一些修正:
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)
def gcd(x,y):
而y:
如果y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()))
从运算符导入mul作为乘法
def系数(n):
f={}#创建“因子:指数”对需要一个dict
除数=2
当n>1时:
而(除数ES6型)
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));
}
使用LINQ,您可以编写:
static int LCM(int[] numbers)
{
return numbers.Aggregate(LCM);
}
static int LCM(int a, int b)
{
return a * b / GCD(a, b);
}
应该使用System.Linq;
添加,不要忘记处理异常…我们有一个工作实现,它可以用于任何数量的输入,也可以显示步骤
clc;
data = [1 2 3 4 5]
LCM=1;
for i=1:1:length(data)
LCM = lcm(LCM,data(i))
end
我们所做的是:
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.
就这样-你得到了你的lcm。lcm是结合的和交换的
LCM(a,b,c)=LCM(LCM(a,b,c)=LCM(a,b,c))
以下是C语言中的示例代码:
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);
}
intmain()
{
int a[20],i,n,result=1;//假设:计数不能超过20
printf(“输入要计算LCM的数字数量(小于20):”;
scanf(“%d”和“&n”);
printf(“输入%d个数字以计算其LCM:,n”);
对于R中的(i=0;i,我们可以使用来自包编号的函数mGCD(x)和mLCM(x),一起计算整数向量x中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
我会选择这个(C#):
只是一些澄清,因为乍一看,这段代码所做的事情并不那么清楚:
Aggregate是一种Linq扩展方法,因此您不能忘记使用System.Linq将其添加到引用中
Aggregate获取一个累加函数,因此我们可以利用IEnumerable上的属性lcm(a,b,c)=lcm(a,lcm(b,c))
GCD计算使用了
lcm计算使用Abs(a*b)/gcd(a,b),请参阅
希望这有帮助,方法compLCM获取一个向量并返回LCM。所有数字都在向量in_数中
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++;
}
}
int-mathOps::compLCM(std::vector和in_数)
{
int tmpNumbers=in_numbers.size();
int tmpMax=*max_元素(in_numbers.begin(),in_numbers.end());
bool tmpNotDividable=false;
while(true)
{
for(int i=0;i
任何想要快速工作的代码的人,请尝试以下方法:
我编写了一个函数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
将所有这些数字放入数组args
,然后调用类似lcm\n(args,num);
此函数返回所有这些数字的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);
}
}
intgcd(inta,intb){
如果(b==0),则返回a;
返回gcd(b,a%b);
}
整数lcm(整数[]a,整数n){
int res=1,i;
对于(i=0;
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);
}
}
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;
}
from functools import reduce
from math import gcd
from fractions import gcd
lcm = reduce(lambda x,y: x*y // gcd(x, y), range(1, 21))
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)
def function(l):
s = 1
for i in l:
s = lcm(i, s)
return s
#!/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"
$ ./script 2 3 4 5 6
60
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'
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;
}
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
// 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
// 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) }
}
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)