找到最小的数 K 使得 K % p = 0 和 q % K = 0

给定两个整数p和q ，任务是找到最小的数K使得K % p = 0和q % K = 0 。如果没有这样的K是可能的，则打印-1 。
例子：

Input: p = 2, q = 8
Output: 2
2 % 2 = 0 and 8 % 2 = 0
Input: p = 5, q = 14
Output: -1

方法：为了使K成为可能， q必须能被p整除。

如果q % p = 0则打印p
否则打印-1 。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function to return the minimum
// value K such that K % p = 0
// and q % k = 0
int getMinVal(int p, int q)
{
 
    // If K is possible
    if (q % p == 0)
        return p;
 
    // No such K is possible
    return -1;
}
 
// Driver code
int main()
{
    int p = 24, q = 48;
    cout << getMinVal(p, q);
 
    return 0;
}

Java

// Java implementation of the approach
import java.io.*;
 
class GFG
{
     
// Function to return the minimum
// value K such that K % p = 0
// and q % k = 0
static int getMinVal(int p, int q)
{
 
    // If K is possible
    if (q % p == 0)
        return p;
 
    // No such K is possible
    return -1;
}
 
// Driver code
public static void main (String[] args)
{
    int p = 24, q = 48;
    System.out.println(getMinVal(p, q));
}
}
 
// This code is contributed by jit_t.

Python3

# Python3 implementation of the approach
 
# Function to return the minimum
# value K such that K % p = 0
# and q % k = 0
def getMinVal(p, q):
 
    # If K is possible
    if q % p == 0:
        return p
 
    # No such K is possible
    return -1
 
# Driver code
p = 24; q = 48
print(getMinVal(p, q))
 
# This code is contributed
# by Shrikant13

C#

// C# implementation of the approach
using System;
 
class GFG
{
     
// Function to return the minimum
// value K such that K % p = 0
// and q % k = 0
static int getMinVal(int p, int q)
{
 
    // If K is possible
    if (q % p == 0)
        return p;
 
    // No such K is possible
    return -1;
}
 
// Driver code
public static void Main ()
{
    int p = 24, q = 48;
    Console.WriteLine(getMinVal(p, q));
}
}
 
// This code is contributed
// by Code_Mech.

PHP

Javascript

输出：

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