计算包含 N 和 M 倍数的两个数组中的公共元素

给定两个数组，第一个数组包含小于或等于 k 的整数 n 的倍数，类似地，第二个数组包含小于或等于 k 的整数 m 的倍数。
任务是找到数组之间公共元素的数量。
例子：

Input :n=2 m=3 k=9 
Output : 1 
First array would be = [ 2, 4, 6, 8 ] 
Second array would be = [ 3, 6, 9 ] 
6 is the only common element
Input :n=1 m=2 k=5 
Output : 2 
 

方法 ：
找到n和m。作为LCM的LCM是n和m的最小公倍数，LCM的所有的倍数将是常见的两个阵列。小于或等于 k 的 LCM 的倍数将等于 k/(LCM(m, n))。
要找到 LCM，首先使用欧几里得算法和 n 的 lcm 计算两个数字的 GCD，m 是 n*m/gcd(n, m)。
下面是上述方法的实现：

C++

// C++ implementation of the above approach
#include 
 
using namespace std;
 
// Recursive function to find
// gcd using euclidean algorithm
int gcd(int a, int b)
{
    if (a == 0)
        return b;
    return gcd(b % a, a);
}
 
// Function to find lcm
// of two numbers using gcd
int lcm(int n, int m)
{
    return (n * m) / gcd(n, m);
}
 
// Driver code
int main()
{
    int n = 2, m = 3, k = 5;
 
    cout << k / lcm(n, m) << endl;
 
    return 0;
}

Java

// Java implementation of the above approach
import java.util.*;
import java.lang.*;
import java.io.*;
 
class GFG
{
 
// Recursive function to find
// gcd using euclidean algorithm
static int gcd(int a, int b)
{
    if (a == 0)
        return b;
    return gcd(b % a, a);
}
 
// Function to find lcm
// of two numbers using gcd
static int lcm(int n, int m)
{
    return (n * m) / gcd(n, m);
}
 
// Driver code
public static void main(String[] args)
{
    int n = 2, m = 3, k = 5;
 
    System.out.print( k / lcm(n, m));
}
}
 
// This code is contributed by mohit kumar 29

Python3

# Python3 implementation of the above approach
 
# Recursive function to find
# gcd using euclidean algorithm
def gcd(a, b) :
 
    if (a == 0) :
        return b;
         
    return gcd(b % a, a);
 
# Function to find lcm
# of two numbers using gcd
def lcm(n, m) :
 
    return (n * m) // gcd(n, m);
 
 
# Driver code
if __name__ == "__main__" :
 
    n = 2; m = 3; k = 5;
 
    print(k // lcm(n, m));
 
# This code is contributed by AnkitRai01

C#

// C# implementation of the above approach
using System;
     
class GFG
{
 
// Recursive function to find
// gcd using euclidean algorithm
static int gcd(int a, int b)
{
    if (a == 0)
        return b;
    return gcd(b % a, a);
}
 
// Function to find lcm
// of two numbers using gcd
static int lcm(int n, int m)
{
    return (n * m) / gcd(n, m);
}
 
// Driver code
public static void Main(String[] args)
{
    int n = 2, m = 3, k = 5;
 
    Console.WriteLine( k / lcm(n, m));
}
}
 
// This code is contributed by Princi Singh

Javascript

 
// javascript implementation of the above approach
// Recursive function to find
// gcd using euclidean algorithm
function gcd(a, b)
{
    if (a == 0)
        return b;
    return gcd(b % a, a);
}
 
// Function to find lcm
// of two numbers using gcd
function lcm(n, m)
{
    return (n * m) / gcd(n, m);
}
 
// Driver code
 
var n = 2, m = 3, k = 5;
 
document.write( parseInt(k / lcm(n, m)));
 
// This code is contributed by Amit Katiyar
 

输出：

0

时间复杂度： O(log(min(n,m)))