加性素数

给定数字N ，任务是检查N是否为加数素数。如果N是可加素数，则打印“是”，否则打印“否” 。

An Additive Prime Number is a prime number P such that the sum of digits of P is also a prime number.
For example, 23 is an Additive prime because 2 + 3 = 5 which is a prime number.

例子：

Input: N = 23
Output: Yes
Explanation:
Sum of digits of 23 = 2 + 3 = 5.

Input: N = 10
Output: No

方法：想法是找到数字N的数字总和，并检查其是否为质数。如果sum是质数，则打印“是”，否则打印“否” 。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
  
// Check if N is prime or not
bool isPrime(int n)
{
    // Corner Cases
    if (n <= 1)
        return false;
  
    if (n <= 3)
        return true;
  
    // This is checked to skip
    // middle five numbers
    if (n % 2 == 0 || n % 3 == 0)
        return false;
  
    for (int i = 5; i * i <= n; i = i + 6)
        if (n % i == 0 || n % (i + 2) == 0)
            return false;
  
    return true;
}
  
// Function to get sum of digits
int getSum(int n)
{
    int sum = 0;
    while (n != 0) {
        sum = sum + n % 10;
        n = n / 10;
    }
  
    // Return the sum of digits
    return sum;
}
  
// Function to check whether
// the given number is
// Additive Prime number or not
bool isAdditivePrime(int n)
{
    // If number is not prime
    if (!isPrime(n))
        return false;
  
    // Check if sum of digits
    // is prime or not
    return isPrime(getSum(n));
}
  
// Driver Code
int main()
{
    // Given Number N
    int N = 23;
  
    // Function Call
    if (isAdditivePrime(N))
        cout << "Yes";
    else
        cout << "No";
}

Java

// Java program for the above approach
class GFG{ 
  
// Check if N is prime or not
static boolean isPrime(int n)
{
      
    // Corner Cases
    if (n <= 1)
        return false;
  
    if (n <= 3)
        return true;
  
    // This is checked to skip
    // middle five numbers
    if (n % 2 == 0 || n % 3 == 0)
        return false;
  
    for(int i = 5; i * i <= n; i = i + 6)
       if (n % i == 0 || n % (i + 2) == 0)
           return false;
  
    return true;
}
  
// Function to get sum of digits
static int getSum(int n)
{
    int sum = 0;
    while (n != 0)
    {
        sum = sum + n % 10;
        n = n / 10;
    }
  
    // Return the sum of digits
    return sum;
}
  
// Function to check whether
// the given number is
// Additive Prime number or not
static boolean isAdditivePrime(int n)
{
      
    // If number is not prime
    if (!isPrime(n))
        return false;
  
    // Check if sum of digits
    // is prime or not
    return isPrime(getSum(n));
}
  
// Driver code 
public static void main(String[] args) 
{ 
      
    // Given Number N
    int n = 23;
  
    // Function Call
    if (isAdditivePrime(n))
        System.out.println("Yes");
    else
        System.out.println("No");
} 
} 
  
// This code is contributed by Pratima Pandey

Python3

# Python3 program for the above approach 
  
# Check if N is prime or not 
def isPrime(n):
  
    # Corner Cases 
    if (n <= 1): 
        return False
  
    if (n <= 3):
        return True
  
    # This is checked to skip 
    # middle five numbers 
    if (n % 2 == 0 or n % 3 == 0): 
        return False
          
    i = 5
    while(i * i <= n): 
        if (n % i == 0 or n % (i + 2) == 0): 
            return False
        i = i + 6
  
    return True
  
# Function to get sum of digits 
def getSum(n): 
  
    sum = 0
    while (n != 0):
        sum = sum + n % 10
        n = n / 10
  
    # Return the sum of digits 
    return sum
  
# Function to check whether 
# the given number is 
# Additive Prime number or not 
def isAdditivePrime(n):
  
    # If number is not prime 
    if (not isPrime(n)):
        return False
  
    # Check if sum of digits 
    # is prime or not 
    return isPrime(getSum(n)) 
  
# Driver Code 
  
# Given Number N 
N = 23
  
# Function Call 
if (isAdditivePrime(N)): 
    print ("Yes") 
else:
    print ("No")
  
# This code is contributed by Pratik Basu

C#

// C# program for the above approach
using System;
class GFG{ 
  
// Check if N is prime or not
static bool isPrime(int n)
{
      
    // Corner Cases
    if (n <= 1)
        return false;
  
    if (n <= 3)
        return true;
  
    // This is checked to skip
    // middle five numbers
    if (n % 2 == 0 || n % 3 == 0)
        return false;
  
    for(int i = 5; i * i <= n; i = i + 6)
       if (n % i == 0 || n % (i + 2) == 0)
           return false;
  
    return true;
}
  
// Function to get sum of digits
static int getSum(int n)
{
    int sum = 0;
    while (n != 0)
    {
        sum = sum + n % 10;
        n = n / 10;
    }
  
    // Return the sum of digits
    return sum;
}
  
// Function to check whether
// the given number is
// Additive Prime number or not
static bool isAdditivePrime(int n)
{
      
    // If number is not prime
    if (!isPrime(n))
        return false;
  
    // Check if sum of digits
    // is prime or not
    return isPrime(getSum(n));
}
  
// Driver code 
public static void Main() 
{ 
      
    // Given Number N
    int n = 23;
  
    // Function Call
    if (isAdditivePrime(n))
        Console.Write("Yes");
    else
        Console.Write("No");
} 
} 
  
// This code is contributed by Code_Mech

输出：

Yes