Dados dos enteros positivos n1 y n2, la tarea es verificar si ambos son primos primos o no. Imprime ‘SÍ’ si ambos números son primos primos; de lo contrario, imprime ‘NO’.
Primos primos: en matemáticas, los primos primos son números primos que difieren en 4. Supongamos que ‘p’ es un número primo y si (p + 4) también es un número primo, ambos números primos se llamarán primos primos.
Los primos primos por debajo de 100 son:
(3, 7), (7, 11), (13, 17), (19, 23), (37, 41), (43, 47), (67, 71), (79, 83), (97 , 101)
Ejemplos:
Input: n1 = 7, n2 = 11
Output: YES
Both are prime numbers and they differ by 4.
Input: n1 = 5, n2 = 11
Output: NO
Both are prime numbers but they differ by 6.
Enfoque: una solución simple es verificar si ambos números son primos y difieren en 4 o no. Si son primos y difieren en 4 entonces serán primos primos, de lo contrario no.
A continuación se muestra la implementación del enfoque anterior:
CPP
// CPP program to check Cousin prime
#include <bits/stdc++.h>
using namespace std;
// Function to check if a number is prime or not
bool isPrime(int n)
{
// Corner cases
if (n <= 1)
return false;
if (n <= 3)
return true;
// This is checked so that we can skip
// middle five numbers in below loop
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;
}
// Returns true if n1 and n2 are Cousin primes
bool isCousinPrime(int n1, int n2)
{
// Check if they differ by 4 or not
if (abs(n1 - n2) != 4)
return false;
// Check if both are prime number or not
else
return (isPrime(n1) && isPrime(n2));
}
// Driver code
int main()
{
// Get the 2 numbers
int n1 = 7, n2 = 11;
// Check the numbers for cousin prime
if (isCousinPrime(n1, n2))
cout << "YES" << endl;
else
cout << "NO" << endl;
return 0;
}
JAVA
// JAVA program to check Cousin prime
import java.util.*;
class GFG {
// Function to check if a number is prime or not
static boolean isPrime(int n)
{
// Corner cases
if (n <= 1)
return false;
if (n <= 3)
return true;
// This is checked so that we can skip
// middle five numbers in below loop
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;
}
// Returns true if n1 and n2 are Cousin primes
static boolean isCousinPrime(int n1, int n2)
{
// Check if they differ by 4 or not
if (Math.abs(n1 - n2) != 4)
return false;
// Check if both are prime number or not
else
return (isPrime(n1) && isPrime(n2));
}
// Driver code
public static void main(String[] args)
{
// Get the 2 numbers
int n1 = 7, n2 = 11;
// Check the numbers for cousin prime
if (isCousinPrime(n1, n2))
System.out.println("YES");
else
System.out.println("NO");
}
}
Python
# Python program to check Cousin prime
import math
# Function to check whether a
# number is prime or not
def isPrime( n ):
# Corner cases
if n <= 1:
return False
if n <= 3:
return True
# This is checked so that we
# can skip middle five numbers
# in below loop
if n % 2 == 0 or n % 3 == 0:
return False
for i in range(5, int(math.sqrt(n)+1), 6):
if n % i == 0 or n %(i + 2) == 0:
return False
return True
# Returns true if n1 and n2 are Cousin primes
def isCousinPrime( n1, n2) :
# Check if they differ by 4 or not
if(not (abs(n1-n2)== 4)):
return False
# Check if both are prime number or not
else:
return (isPrime(n1) and isPrime(n2))
# Driver code
# Get the 2 numbers
n1 = 7
n2 = 11
# Check the numbers for cousin prime
if (isCousinPrime(n1, n2)):
print("YES")
else:
print("NO")
C#
// C# Code for Cousin Prime Numbers
using System;
class GFG {
// Function to check if the given
// number is prime or not
static bool isPrime(int n)
{
// Corner cases
if (n <= 1)
return false;
if (n <= 3)
return true;
// This is checked so that we can skip
// middle five numbers in below loop
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;
}
// Returns true if n1 and n2 are Cousin primes
static bool isCousinPrime(int n1, int n2)
{
// Check if the numbers differ by 4 or not
if (Math.Abs(n1 - n2) != 4) {
return false;
}
else {
return (isPrime(n1) && isPrime(n2));
}
}
// Driver program
public static void Main()
{
// Get the 2 numbers
int n1 = 7, n2 = 11;
// Check the numbers for cousin prime
if (isCousinPrime(n1, n2))
Console.WriteLine("YES");
else
Console.WriteLine("NO");
}
}
PHP
<?php
// PhP program to check Cousin prime
// Function to check if the given
// Number is prime or not
function isPrime($n)
{
// Corner cases
if ($n <= 1) return false;
if ($n <= 3) return true;
// This is checked so that we can skip
// middle five numbers in below loop
if ($n % 2 == 0 || $n % 3 == 0)
return false;
for ($i = 5; $i * $i <= $n; $i = $i + 6)
if ($n % $i == 0 || $n % ($i + 2) == 0)
return false;
return true;
}
// Returns true if n1 and n2 are Cousin primes
function isCousinPrime($n1, $n2)
{
// Check if the given number differ by 4 or not
if(abs($n1-$n2)!=4)
return false;
// Check if both numbers are prime or not
else
return (isPrime($n1) && isPrime($n2));
}
// Driver code
// Get the 2 numbers
$n1 = 7;
$n2 = 11;
// Check the numbers for cousin prime
if (isCousinPrime($n1, $n2))
echo "YES";
else
echo "NO";
?>
Javascript
<script>
// Javascript program to check Cousin prime
// Function to check if a number is prime or not
function isPrime(n)
{
// Corner cases
if (n <= 1)
return false;
if (n <= 3)
return true;
// This is checked so that we can skip
// middle five numbers in below loop
if (n % 2 == 0 || n % 3 == 0)
return false;
for (var i = 5; i * i <= n; i = i + 6) {
if (n % i == 0 || n % (i + 2) == 0) {
return false;
}
}
return true;
}
// Returns true if n1 and n2 are Cousin primes
function isCousinPrime(n1, n2)
{
// Check if they differ by 4 or not
if(Math.abs(n1 - n2) != 4)
return false;
// Check if both are prime number or not
else
return (isPrime(n1) && isPrime(n2));
}
// Driver code
// Get the 2 numbers
var n1 = 7, n2 = 11;
// Check the numbers for cousin prime
if (isCousinPrime(n1, n2))
document.write( "YES");
else
document.write( "NO" );
</script>
Producción:
YES
Complejidad de tiempo: O(n1 1/2 )
Espacio Auxiliar: O(1)