C++ 求两个数的最小公倍数

LCM是 最小公倍数 的缩写，用于获取两个数字（n1和n2）的最小公倍数，且最小公倍数必须能被给定的两个数字整除。最小公倍数是两个数字中共同的一个数。表示两个数字的最小公倍数的方式为LCM(a, b)或lcm(a, b)。

C++ 求两个数的最小公倍数

例如，两个正数的最小公倍数（LCM）如LCM（12, 24）为24。因为数字12和24都可以整除24而不留下任何余数。类似地，数字3和4的最小公倍数为12，因为两个数字的最小公倍数是12。

两个数字的最小公倍数算法

步骤1： 从用户那里获取两个输入n1和n2。

步骤2： 将n1和n2的最小公倍数存储在max变量中。

步骤3： 验证max变量是否可被n1和n2整除，并将max打印为两个数字的最小公倍数。

步骤4： 否则，max值在每次迭代时更新为1，并跳转到步骤3以检查max变量的整除性。

步骤5： 结束程序。

使用if语句和while循环获取两个数字的最小公倍数的程序

Program1.cpp

#include 
using namespace std;
int main()
{
    int n1, n2, max_num, flag = 1;
    cout << " Enter two numbers: \n";
    cin >> n1 >> n2;    

    // use ternary operator to get the large number
    max_num = (n1 > n2) ? n1 : n2;

    while (flag)  
    {
        // if statement checks max_num is completely divisible by n1 and n2.
        if(max_num % n1 == 0 && max_num % n2 == 0)
        {
cout << " The LCM of " <

输出

Enter two numbers:
30
50
 The LCM of 30 and 50 is 150

使用while循环计算两个数字的最小公倍数的程序

Program2.cpp

#include 
using namespace std;

int main()
{
    // declare variables
    int num1, num2, lcm, gcd, temp;
    cout <<" Enter the first number: ";
    cin >> num1;
    cout <<" Enter the second number: ";
    cin >> num2;

    // assign num1 and num2 values to int a and b
    int a = num1;
    int b = num2;   
    // use while loop to define the condition
    while (num2 != 0)
    {
        temp = num2;
        num2 = num1 % num2;
        num1 = temp;
    }

    // assign num1 to gcd variable
    gcd = num1;
    lcm = (a * b) / gcd;
    cout << "\n LCM of " << a << " and " << b << " = " << lcm;
    return 0;
}

输出

Enter the first number: 15
 Enter the second number: 10

 LCM of 15 and 10 = 30

使用最大公约数计算两个数的最小公倍数的程序

Program3.cpp

#include 
using namespace std;

// definition of getGCD() function
int getGCD( int n1, int n2)
{
// here if statement checks the given number is not equal to 0.
if ( n1 == 0 || n2 == 0)
return 0;

// if n1 equal to n2, return n1
if (n1 == n2)
return n1;

// if n1 is greater than n2, execute the followed statement
if ( n1 > n2)
return getGCD (n1 - n2, n2);

return getGCD (n1, n2 - n1);
}

// definition of getLCM() function
int getLCM (int n1, int n2)
{
/* divide the multiplication of n1 and n2 by getGCD() function to return LCM. */    
return (n1 * n2) / getGCD (n1,n2);
}

int main()
{
// declare local variables  
int n1, n2;
cout << " Enter two positive numbers: ";
cin >> n1 >> n2; // get numbers from user

// print the LCM(n1, n2)
cout << " \n LCM of " <

输出

Enter two positive numbers: 50
60

 LCM of 50 and 60 is 300

使用递归函数获取两个数的最小公倍数的程序

程序4.cpp

# include 
using namespace std;

// define the getGCD()  function and pass n1 and n2 parameter
int getGCD( int n1, int n2)
{
    if (n1 == 0)
    {
        return n2;      
    }

return getGCD(n2 % n1, n1);
}

// define the getLCM function to return the LCM
int getLCM( int n1, int n2)
{
    return (n1 / getGCD(n1, n2)) * n2;
}

int main()
{
    // declare local variable
    int num1, num2;
    cout << " Enter two numbers: " <> num1 >> num2;
cout << " LCM of two numbers " <

输出

Enter two numbers:
12
36
 LCM of two numbers 12 and 36 is 36

使用函数和while循环计算多个数组元素的最小公倍数的程序

Program5.cpp

#include 
using namespace std;
int lcm_num (int n1, int n2)
{
    int max;
    max = (n1 > n2) ? n1 : n2;
    while (true)
    {
        if (max % n1 == 0 && max % n2 ==0)
        return max;
        max++;
    }
}

// definition of lcm_array() function
int lcm_array (int arr[], int num)
{
    int i;
    int lcm = lcm_num( arr[0], arr[1]);

    // use for loop to get the lcm
    for (i = 2; i < num; i++)
    {
        lcm = lcm_num(lcm , arr[i]);
    }
    return lcm;

}

int main()
{   
    // declare integer array
    int arr[] = {10, 5, 15, 30};
    int num = sizeof(arr) / sizeof(arr[0]); // get the number of elements
    cout << " LCM of multiple array elements is: " << lcm_array(arr, num);
}

输出