打印所有小于等于N的乘法质数

给定一个整数N，任务是打印所有乘素数≤ñ。

Multiplicative Primes are the primes such that the product of their digits is also a prime. For example; 2, 3, 7, 13, 17, … 
 

例子：

Input: N = 10 
Output: 2 3 5 7
Input: N = 3 
Output: 2 3 
 

方法：使用埃拉托色尼检查所有的素数≤N个筛他们是否是素数相乘，即他们的数字产品也是一个素。如果是，则打印那些乘法质数。
下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function to return the digit product of n
int digitProduct(int n)
{
    int prod = 1;
    while (n) {
        prod = prod * (n % 10);
        n = n / 10;
    }
 
    return prod;
}
 
// Function to print all multiplicative primes <= n
void printMultiplicativePrimes(int n)
{
    // Create a boolean array "prime[0..n+1]". A
    // value in prime[i] will finally be false
    // if i is Not a prime, else true.
    bool prime[n + 1];
    memset(prime, true, sizeof(prime));
 
    prime[0] = prime[1] = false;
    for (int p = 2; p * p <= n; p++) {
 
        // If prime[p] is not changed, then
        // it is a prime
        if (prime[p]) {
 
            // Update all multiples of p
            for (int i = p * 2; i <= n; i += p)
                prime[i] = false;
        }
    }
 
    for (int i = 2; i <= n; i++) {
 
        // If i is prime and its digit sum is also prime
        // i.e. i is a multiplicative prime
        if (prime[i] && prime[digitProduct(i)])
            cout << i << " ";
    }
}
 
// Driver code
int main()
{
    int n = 10;
    printMultiplicativePrimes(n);
}

Java

// Java implementation of the approach
import java.io.*;
 
class GFG
{
 
// Function to return the digit product of n
static int digitProduct(int n)
{
    int prod = 1;
    while (n > 0)
    {
        prod = prod * (n % 10);
        n = n / 10;
    }
    return prod;
}
 
// Function to print all multiplicative primes <= n
static void printMultiplicativePrimes(int n)
{
    // Create a boolean array "prime[0..n+1]". A
    // value in prime[i] will finally be false
    // if i is Not a prime, else true.
    boolean prime[] = new boolean[n + 1 ];
    for(int i = 0; i <= n; i++)
     prime[i] = true;
 
    prime[0] = prime[1] = false;
    for (int p = 2; p * p <= n; p++)
    {
 
        // If prime[p] is not changed, then
        // it is a prime
        if (prime[p])
        {
 
            // Update all multiples of p
            for (int i = p * 2; i <= n; i += p)
                prime[i] = false;
        }
    }
 
    for (int i = 2; i <= n; i++)
    {
 
        // If i is prime and its digit sum is also prime
        // i.e. i is a multiplicative prime
        if (prime[i] && prime[digitProduct(i)])
            System.out.print( i + " ");
    }
}
 
    // Driver code
    public static void main (String[] args)
    {
        int n = 10;
        printMultiplicativePrimes(n);
    }
}
 
// This code is contributed by shs..

Python3

# Python 3 implementation of the approach
from math import sqrt
 
# Function to return the digit product of n
def digitProduct(n):
    prod = 1
    while (n):
        prod = prod * (n % 10)
        n = int(n / 10)
 
    return prod
 
# Function to print all multiplicative
# primes <= n
def printMultiplicativePrimes(n):
     
    # Create a boolean array "prime[0..n+1]".
    # A value in prime[i] will finally be
    # false if i is Not a prime, else true.
    prime = [True for i in range(n + 1)]
 
    prime[0] = prime[1] = False
    for p in range(2, int(sqrt(n)) + 1, 1):
         
        # If prime[p] is not changed,
        # then it is a prime
        if (prime[p]):
             
            # Update all multiples of p
            for i in range(p * 2, n + 1, p):
                prime[i] = False
         
    for i in range(2, n + 1, 1):
         
        # If i is prime and its digit sum
        # is also prime i.e. i is a
        # multiplicative prime
        if (prime[i] and prime[digitProduct(i)]):
            print(i, end = " ")
 
# Driver code
if __name__ == '__main__':
    n = 10
    printMultiplicativePrimes(n)
 
# This code is contributed by
# Surendra_Gangwar

C#

// C# implementation of the approach
class GFG
{
 
// Function to return the digit product of n
static int digitProduct(int n)
{
    int prod = 1;
    while (n > 0)
    {
        prod = prod * (n % 10);
        n = n / 10;
    }
    return prod;
}
 
// Function to print all multiplicative primes <= n
static void printMultiplicativePrimes(int n)
{
    // Create a boolean array "prime[0..n+1]". A
    // value in prime[i] will finally be false
    // if i is Not a prime, else true.
    bool[] prime = new bool[n + 1 ];
     
    for(int i = 0; i <= n; i++)
        prime[i] = true;
 
    prime[0] = prime[1] = false;
    for (int p = 2; p * p <= n; p++)
    {
 
        // If prime[p] is not changed, then
        // it is a prime
        if (prime[p])
        {
 
            // Update all multiples of p
            for (int i = p * 2; i <= n; i += p)
                prime[i] = false;
        }
    }
 
    for (int i = 2; i <= n; i++)
    {
 
        // If i is prime and its digit sum is also prime
        // i.e. i is a multiplicative prime
        if (prime[i] && prime[digitProduct(i)])
            System.Console.Write( i + " ");
    }
}
 
    // Driver code
    static void Main()
    {
        int n = 10;
        printMultiplicativePrimes(n);
    }
}
 
// This code is contributed by chandan_jnu

Javascript

// Javascript implementation of the approach
     
    // Function to return the digit product of n
    function digitProduct(n)
    {
        let prod = 1;
        while (n > 0)
        {
            prod = prod * (n % 10);
            n = Math.floor(n / 10);
        }
        return prod;
    }
     
    // Function to print all
    // multiplicative primes <= n
    function printMultiplicativePrimes(n)
    {
        // Create a boolean array "prime[0..n+1]". A
        // value in prime[i] will finally be false
        // if i is Not a prime, else true.
        let prime = new Array(n + 1);
        for(let i = 0; i <= n; i++)
            prime[i] = true;
       
        prime[0] = prime[1] = false;
        for (let p = 2; p * p <= n; p++)
        {
       
            // If prime[p] is not changed, then
            // it is a prime
            if (prime[p])
            {
       
                // Update all multiples of p
                for (let i = p * 2; i <= n; i += p)
                    prime[i] = false;
            }
        }
       
        for (let i = 2; i <= n; i++)
        {
       
            // If i is prime and its digit sum is also prime
            // i.e. i is a multiplicative prime
            if (prime[i] && prime[digitProduct(i)])
                document.write( i + " ");
        }
    }
     
    // Driver code
    let n = 10;
    printMultiplicativePrimes(n);
     
    // This code is contributed by unknown2108
     

输出：

2 3 5 7