查找两个数字的LCM的程序

两个数字的LCM(最小公倍数)是可以除以两个数字的最小数字。

例如，15和20的LCM为60，5和7的LCM为35。

一个简单的解决方案是找到两个数字的所有素因子，然后找到两个数字中存在的所有因子的并集。最后，返回并集元素的乘积。

一个有效的解决方案基于以下公式来计算两个数字“ a”和“ b”的LCM。

a x b = LCM(a, b) * GCD (a, b)

   LCM(a, b) = (a x b) / GCD(a, b)

我们已经讨论了找到两个数字的GCD的函数。使用GCD，我们可以找到LCM。

下面是上述想法的实现：

C++

// C++ program to find LCM of two numbers
#include 
using namespace std;
 
// Recursive function to return gcd of a and b
long long gcd(long long int a, long long int b)
{
  if (b == 0)
    return a;
  return gcd(b, a % b);
}
 
// Function to return LCM of two numbers
long long lcm(int a, int b)
{
    return (a / gcd(a, b)) * b;
}
  
// Driver program to test above function
int main()
{
    int a = 15, b = 20;
    cout <<"LCM of " << a << " and "
         << b << " is " << lcm(a, b);
    return 0;
}

C

// C program to find LCM of two numbers
#include 
 
// Recursive function to return gcd of a and b
int gcd(int a, int b)
{
    if (a == 0)
        return b;
    return gcd(b % a, a);
}
 
// Function to return LCM of two numbers
int lcm(int a, int b)
{
    return (a / gcd(a, b)) * b;
}
 
// Driver program to test above function
int main()
{
    int a = 15, b = 20;
    printf("LCM of %d and %d is %d ", a, b, lcm(a, b));
    return 0;
}

Java

// Java program to find LCM of two numbers.
class Test
{
    // Recursive method to return gcd of a and b
    static int gcd(int a, int b)
    {
        if (a == 0)
            return b;
        return gcd(b % a, a);
    }
     
    // method to return LCM of two numbers
    static int lcm(int a, int b)
    {
        return (a / gcd(a, b)) * b;
    }
     
    // Driver method
    public static void main(String[] args)
    {
        int a = 15, b = 20;
        System.out.println("LCM of " + a +
                           " and " + b +
                      " is " + lcm(a, b));
    }
}

Python3

# Python program to find LCM of two numbers
 
# Recursive function to return gcd of a and b
def gcd(a,b):
    if a == 0:
        return b
    return gcd(b % a, a)
 
# Function to return LCM of two numbers
def lcm(a,b):
    return (a / gcd(a,b))* b
 
# Driver program to test above function
a = 15
b = 20
print('LCM of', a, 'and', b, 'is', lcm(a, b))
 
# This code is contributed by Danish Raza

C#

// C# program to find LCM
// of two numbers.
using System;
class GFG {
     
    // Recursive method to
    // return gcd of a and b
    static int gcd(int a, int b)
    {
        if (a == 0)
            return b;
        return gcd(b % a, a);
    }
     
    // method to return
    // LCM of two numbers
    static int lcm(int a, int b)
    {
        return (a / gcd(a, b)) * b;
    }
     
    // Driver method
    public static void Main()
    {
        int a = 15, b = 20;
        Console.WriteLine("LCM of " + a +
         " and " + b + " is " + lcm(a, b));
    }
}
 
// This code is contributed by anuj_67.

Javascript

 
// Javascript program to find LCM of two numbers
 
// Recursive function to return gcd of a and b
function gcd(a, b)
{
if (b == 0)
    return a;
return gcd(b, a % b);
}
 
// Function to return LCM of two numbers
function lcm(a, b)
{
    return (a / gcd(a, b)) * b;
}
 
// Driver program to test above function
  
    let a = 15, b = 20;
    document.write("LCM of " + a + " and "
    + b + " is " + lcm(a, b));
     
 
// This code is contributed by Mayank Tyagi
 

输出

LCM of 15 and 20 is 60

https://youtu.be/anSfYgbo694