Dado un número n, verifica si es un número primo o no usando la recursividad.
Ejemplos:
Input : n = 11 Output : Yes Input : n = 15 Output : No
La idea se basa en el método de la escuela para buscar números primos .
C++
// CPP Program to find whether a Number
// is Prime or Not using Recursion
#include <bits/stdc++.h>
using namespace std;
// Returns true if n is prime, else
// return false.
// i is current divisor to check.
bool isPrime(int n, int i = 2)
{
// Base cases
if (n <= 2)
return (n == 2) ? true : false;
if (n % i == 0)
return false;
if (i * i > n)
return true;
// Check for next divisor
return isPrime(n, i + 1);
}
// Driver Program
int main()
{
int n = 15;
if (isPrime(n))
cout << "Yes";
else
cout << "No";
return 0;
}
Java
// java Program to find whether a Number
// is Prime or Not using Recursion
import java.util.*;
class GFG {
// Returns true if n is prime, else
// return false.
// i is current divisor to check.
static boolean isPrime(int n, int i)
{
// Base cases
if (n <= 2)
return (n == 2) ? true : false;
if (n % i == 0)
return false;
if (i * i > n)
return true;
// Check for next divisor
return isPrime(n, i + 1);
}
// Driver program to test above function
public static void main(String[] args)
{
int n = 15;
if (isPrime(n, 2))
System.out.println("Yes");
else
System.out.println("No");
}
}
// This code is contributed by Sam007.
Python3
# Python 3 Program to find whether
# a Number is Prime or Not using
# Recursion
# Returns true if n is prime, else
# return false.
# i is current divisor to check.
def isPrime(n, i = 2):
# Base cases
if (n <= 2):
return True if(n == 2) else False
if (n % i == 0):
return False
if (i * i > n):
return True
# Check for next divisor
return isPrime(n, i + 1)
# Driver Program
n = 15
if (isPrime(n)):
print("Yes")
else:
print("No")
# This code is contributed by
# Smitha Dinesh Semwal
C#
// C# Program to find whether a Number
// is Prime or Not using Recursion
using System;
class GFG
{
// Returns true if n is prime, else
// return false.
// i is current divisor to check.
static bool isPrime(int n, int i)
{
// Base cases
if (n <= 2)
return (n == 2) ? true : false;
if (n % i == 0)
return false;
if (i * i > n)
return true;
// Check for next divisor
return isPrime(n, i + 1);
}
// Driver code
static void Main()
{
int n = 15;
if (isPrime(n, 2))
Console.Write("Yes");
else
Console.Write("No");
}
}
// This code is contributed by Sam007
PHP
<?php
// PHP Program to find whether a Number
// is Prime or Not using Recursion
// Returns true if n is prime, else
// return false.
// i is current divisor to check.
function isPrime($n, $i = 2)
{
// Base cases
if ($n <= 2)
return ($n == 2) ? true : false;
if ($n % $i == 0)
return false;
if ($i * $i > $n)
return true;
// Check for next divisor
return isPrime($n, $i + 1);
}
// Driver Code
$n = 15;
if (isPrime($n))
echo("Yes");
else
echo("No");
// This code is contributed by Ajit.
?>
Javascript
<script>
// JavaScript program to find whether a Number
// is Prime or Not using Recursion
// Returns true if n is prime, else
// return false.
// i is current divisor to check.
function isPrime(n, i)
{
// Base cases
if (n <= 2)
return (n == 2) ? true : false;
if (n % i == 0)
return false;
if (i * i > n)
return true;
// Check for next divisor
return isPrime(n, i + 1);
}
// Driver code
let n = 15;
if (isPrime(n, 2))
document.write("Yes");
else
document.write("No");
</script>
Producción:
No
Publicación traducida automáticamente
Artículo escrito por Shahnawaz_Ali y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA