Given an integer N, the task is to print all the multiplicative primes ? N.
Multiplicative Primes are the primes such that the product of their digits is also a prime. For example; 2, 3, 7, 13, 17, …
Examples:
Input: N = 10
Output: 2 3 5 7
Input: N = 3
Output: 2 3
Approach: Using Sieve of Eratosthenes check for all the primes ? N whether they are multiplicative primes i.e. product of their digits is also a prime. If yes then print those multiplicative primes.
Algorithm:
Step 1: Define a function of the static type which will return the int value and take input as an integer number and return the product of its digits.
Step 2: Now Initialize the variable of in type called prod with the initial value 1.
Step 3: Start a while loop which will calculate the product of digits:
a. Multiply prod by the remainder of n divided by 10.
b. Divide n by 10 and update the value of n with the result.
c. Repeat until n becomes 0.
Step 4: Now return prod as the digit product of n
Step 5: Create a function called printMultiplicativePrimes that outputs all multiplicative primes up to n given an integer input of n.
Step 6: Set every value in the prime boolean array, which has a size of n+1, to true during initialization.
Step 7: primes [0] and [1] should be set to false because they are not primes.
Step 8: Get all primes up to the square root of n using a for loop and Update all multiples of p by setting prime[i] to false if prime[p] is true.
Step 9: Print all prime multiplicative numbers up to n using another for loop and Print I if both prime[i] and prime[digitProduct(i)] are true.
Below is the implementation of the above approach:
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;
// Function to return the digit product of nint digitProduct(int n)
{ int prod = 1;
while (n) {
prod = prod * (n % 10);
n = n / 10;
}
return prod;
}// Function to print all multiplicative primes <= nvoid 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 codeint main()
{ int n = 10;
printMultiplicativePrimes(n);
} |
// Java implementation of the approachimport java.io.*;
class GFG
{// Function to return the digit product of nstatic 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 <= nstatic 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.. |
# Python 3 implementation of the approachfrom math import sqrt
# Function to return the digit product of ndef digitProduct(n):
prod = 1
while (n):
prod = prod * (n % 10)
n = int(n / 10)
return prod
# Function to print all multiplicative# primes <= ndef 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 codeif __name__ == '__main__':
n = 10
printMultiplicativePrimes(n)
# This code is contributed by# Surendra_Gangwar |
// C# implementation of the approachclass GFG
{// Function to return the digit product of nstatic 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 <= nstatic 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 |
<?php// PHP implementation of the approach// Function to return the digit product of nfunction digitProduct($n)
{ $prod = 1;
while ($n)
{
$prod = $prod * ($n % 10);
$n = floor($n / 10);
}
return $prod;
}// Function to print all multiplicative// primes <= nfunction printMultiplicativePrimes($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 = array_fill(0, $n + 1, true);
$prime[0] = $prime[1] = false;
for ($p = 2; $p * $p <= $n; $p++)
{
// If prime[p] is not changed, then
// it is a prime
if ($prime[$p])
{
// Update all multiples of p
for ($i = $p * 2; $i <= $n; $i += $p)
$prime[$i] = false;
}
}
for ($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)])
echo $i, " ";
}
}// Driver code$n = 10;
printMultiplicativePrimes($n);
// This code is contributed by Ryuga.?> |
<script>// 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
</script> |
2 3 5 7
Time Complexity: O(n3/2)
Auxiliary Space: O(n)