打印所有素数集是X素数集的子集的所有数字

给定数字X和N个数字的数组。任务是打印其素数集是X素数集的子集的数组中的所有数字。
例子：

Input: X = 60, a[] = {2, 5, 10, 7, 17} 
Output: 2 5 10 
Set of prime factors of 60: {2, 3, 5} 
Set of prime factors of 2: {2} 
Set of prime factors of 5: {5} 
Set of prime factors of 10: {2, 5} 
Set of prime factors of 7: {7} 
Set of prime factors of 17: {17} 
Hence only 2, 5 and 10’s set of prime factors is a subset of set of prime 
factors of 60. 
Input: X = 15, a[] = {2, 8} 
Output: There are no such numbers 
 

方法：迭代针对阵列中的每个元素，并保持由直到GCD的数量和X的最大公约数除以数变为1的数量和X。如果最后的数字在连续除法后变为1，则打印该数字。
下面是上述方法的实现：

C++

// C++ program to implement
// the above approach
#include 
using namespace std;
 
// Function to print all the numbers
void printNumbers(int a[], int n, int x)
{
 
    bool flag = false;
 
    // Iterate for every element in the array
    for (int i = 0; i < n; i++) {
 
        int num = a[i];
 
        // Find the gcd
        int g = __gcd(num, x);
 
        // Iterate till gcd is 1
        // of number and x
        while (g != 1) {
 
            // Divide the number by gcd
            num /= g;
 
            // Find the new gcdg
            g = __gcd(num, x);
        }
 
        // If the number is 1 at the end
        // then print the number
        if (num == 1) {
            flag = true;
            cout << a[i] << " ";
        }
    }
 
    // If no numbers have been there
    if (!flag)
        cout << "There are no such numbers";
}
 
// Drivers code
int main()
{
    int x = 60;
    int a[] = { 2, 5, 10, 7, 17 };
    int n = sizeof(a) / sizeof(a[0]);
 
    printNumbers(a, n, x);
    return 0;
}

Java

// Java program to implement
// the above approach
class GFG
{
 
// Function to print all the numbers
static void printNumbers(int a[], int n, int x)
{
 
    boolean flag = false;
 
    // Iterate for every element in the array
    for (int i = 0; i < n; i++)
    {
 
        int num = a[i];
 
        // Find the gcd
        int g = __gcd(num, x);
 
        // Iterate till gcd is 1
        // of number and x
        while (g != 1)
        {
 
            // Divide the number by gcd
            num /= g;
 
            // Find the new gcdg
            g = __gcd(num, x);
        }
 
        // If the number is 1 at the end
        // then print the number
        if (num == 1)
        {
            flag = true;
            System.out.print(a[i] + " ");
        }
    }
 
    // If no numbers have been there
    if (!flag)
        System.out.println("There are no such numbers");
}
 
static int __gcd(int a, int b)
{
    if (b == 0)
        return a;
    return __gcd(b, a % b);
     
}
 
// Drivers code
public static void main(String[] args)
{
    int x = 60;
    int a[] = { 2, 5, 10, 7, 17 };
    int n = a.length;
 
    printNumbers(a, n, x);
}
}
 
/* This code contributed by PrinciRaj1992 */

Python3

# Python3 program to implement
# the above approach
from math import gcd
 
# Function to print all the numbers
def printNumbers(a, n, x) :
 
    flag = False
 
    # Iterate for every element in the array
    for i in range(n) :
 
        num = a[i]
 
        # Find the gcd
        g = gcd(num, x)
 
        # Iterate till gcd is 1
        # of number and x
        while (g != 1) :
 
            # Divide the number by gcd
            num //= g
 
            # Find the new gcdg
            g = gcd(num, x)
 
        # If the number is 1 at the end
        # then print the number
        if (num == 1) :
            flag = True;
            print(a[i], end = " ");
 
    # If no numbers have been there
    if (not flag) :
        print("There are no such numbers")
 
# Driver Code
if __name__ == "__main__" :
 
    x = 60
    a = [ 2, 5, 10, 7, 17 ]
    n = len(a)
 
    printNumbers(a, n, x)
     
# This code is contributed by Ryuga

C#

// C# program to implement
// the above approach
using System;
 
class GFG
{
 
// Function to print all the numbers
static void printNumbers(int []a, int n, int x)
{
 
    bool flag = false;
 
    // Iterate for every element in the array
    for (int i = 0; i < n; i++)
    {
 
        int num = a[i];
 
        // Find the gcd
        int g = __gcd(num, x);
 
        // Iterate till gcd is 1
        // of number and x
        while (g != 1)
        {
 
            // Divide the number by gcd
            num /= g;
 
            // Find the new gcdg
            g = __gcd(num, x);
        }
 
        // If the number is 1 at the end
        // then print the number
        if (num == 1)
        {
            flag = true;
            Console.Write(a[i] + " ");
        }
    }
 
    // If no numbers have been there
    if (!flag)
        Console.WriteLine("There are no such numbers");
}
 
static int __gcd(int a, int b)
{
    if (b == 0)
        return a;
    return __gcd(b, a % b);
     
}
 
// Driver code
public static void Main(String[] args)
{
    int x = 60;
    int []a = { 2, 5, 10, 7, 17 };
    int n = a.Length;
 
    printNumbers(a, n, x);
}
}
 
// This code has been contributed by 29AjayKumar

Javascript

 
// Javascript program to implement
// the above approach
 
// Function to print all the numbers
 
// Find the gcd
function __gcd(a, b)
{
    if (b == 0)
        return a;
    return __gcd(b, a % b);
     
}
 
function printNumbers(a, n, x)
{
 
    let flag = false;
 
    // Iterate for every element in the array
    for (let i = 0; i < n; i++) {
 
        let num = a[i];
 
        // Find the gcd
        let g = __gcd(num, x);
 
        // Iterate till gcd is 1
        // of number and x
        while (g != 1) {
 
            // Divide the number by gcd
            num = parseInt(num/g);
 
            // Find the new gcdg
            g = __gcd(num, x);
        }
 
        // If the number is 1 at the end
        // then print the number
        if (num == 1) {
            flag = true;
            document.write(a[i] + " ");
        }
    }
 
    // If no numbers have been there
    if (!flag)
        document.write("There are no such numbers");
}
 
// Drivers code
 
    let x = 60;
    let a = [ 2, 5, 10, 7, 17 ];
    let n = a.length;
 
    printNumbers(a, n, x);
 

输出：

2 5 10