数字的递归总和是否为质数

给定一个数字n，我们需要找到该数字每个数字的总和，直到该数字成为一个数字为止。如果总和是素数，我们需要打印“是”，如果不是素数，则需要打印“否”。
例子：

Input : 5602
Output: No
Explanation:
Step 1- 5+6+0+2 = 13
Step 2- 1+3 = 4 
4 is not prime

Input : 56
Output : Yes
Explanation:
Step 1- 5+6 = 11
Step 2- 1+1 = 2 hence 2 is prime 

这个想法很简单，我们可以快速找到数字的递归和。

int recDigSum(int n)
{
    if (n == 0) 
       return 0;
    return (n % 9 == 0) ? 9 : (n % 9);
}

一旦计算了递归总和，只需简单地检查它是否为2、3、5或7(这些只是个位数的素数)就可以检查它是否为素数。

C++

// CPP code to check if
// recursive sum of
// digits is prime or not.
#include
using namespace std;
  
// Function for recursive 
// digit sum 
int recDigSum(int n)
{
    if (n == 0)
        return 0;
    else
    {
        if (n % 9 == 0)
            return 9;
        else
            return n % 9;
    }
}
      
// function to check if prime
// or not the single digit 
void check(int n)
{
    // calls function which
    // returns sum till
    // single digit 
    n = recDigSum(n);
      
    // checking prime
    if (n == 2 or n == 3 or n == 5 or n == 7)
        cout << "Yes";
    else
        cout << "No";
}
  
// Driver code
int main()
{
   int n = 5602;
    check(n);
}
  
// This code is contributed by Shreyanshi.

Java

// java code to check if
// recursive sum of
// digits is prime or not.
import java.io.*;
  
class GFG 
{
  
    // Function for recursive 
    // digit sum 
    static int recDigSum(int n)
    {
        if (n == 0)
            return 0;
        else
        {
            if (n % 9 == 0)
                return 9;
            else
                return n % 9;
        }
    }
          
    // function to check if prime
    // or not the single digit 
    static void check(int n)
    {
        // calls function which
        // returns sum till
        // single digit 
        n = recDigSum(n);
          
        // checking prime
        if (n == 2 || n == 3 || n == 5 || n == 7)
            System.out.println ( "Yes");
        else
            System.out.println("No");
    }
      
    // Driver code
    public static void main (String[] args) 
    {
        int n = 5602;
        check(n);
      
    }
}
  
// This code is contributed by vt_m

Python

# Python code to check if recursive sum
# of digits is prime or not.
def recDigSum(n):
      
    if n == 0:
        return 0
    else:
        if n % 9 == 0:
            return 9 
        else:
            return n % 9
      
# function to check if prime or not
# the single digit 
def check(n):
      
    # calls function which returns sum 
    # till single digit 
    n = recDigSum(n)
      
    # checking prime 
    if n == 2 or n == 3 or n == 5 or n == 7:
        print "Yes"
    else:
        print "No"
  
      
# driver code 
n = 5602
check(n)

C#

// C# code to check if
// recursive sum of
// digits is prime or not.
using System;
  
class GFG 
{
  
    // Function for recursive 
    // digit sum 
    static int recDigSum(int n)
    {
        if (n == 0)
            return 0;
        else
        {
            if (n % 9 == 0)
                return 9;
            else
                return n % 9;
        }
    }
          
    // function to check if prime
    // or not the single digit 
    static void check(int n)
    {
        // calls function which
        // returns sum till
        // single digit 
        n = recDigSum(n);
          
        // checking prime
        if (n == 2 || n == 3 
            || n == 5 || n == 7)
            Console.WriteLine ( "Yes");
        else
            Console.WriteLine("No");
    }
      
    // Driver code
    public static void Main () 
    {
        int n = 5602;
        check(n);
      
    }
}
  
// This code is contributed by anuj_67.

输出：

No