Dado un número N , la tarea es verificar si N es un número primo aditivo o no. Si N es un número primo aditivo , escriba «Sí» , de lo contrario, escriba «No» .
Un número primo aditivo es un número primo P tal que la suma de los dígitos de P también es un número primo .
Por ejemplo, 23 es un número primo aditivo porque 2 + 3 = 5, que es un número primo.
Ejemplos:
Entrada: N = 23
Salida: Sí
Explicación:
Suma de dígitos de 23 = 2 + 3 = 5.Entrada: N = 10
Salida: No
Planteamiento: La idea es encontrar la suma de los dígitos del número N y comprobar si es primo o no. Si la suma es un número primo, imprima «Sí» , de lo contrario, imprima «No» .
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
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
Javascript
<script>
// JavaScript program for the above approach
// Check if N is prime or not
function 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 || n % 3 == 0)
return false;
for(let 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
function getSum(n)
{
let 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
function isAdditivePrime(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
// Given Number N
let n = 23;
// Function Call
if (isAdditivePrime(n))
document.write("Yes");
else
document.write("No");
// This code is contributed by susmitakundugoaldanga
</script>
Producción:
Yes