Comprobar si un número dado es un número feo o no

Dado un número entero N , la tarea es averiguar si el número dado es un número feo o no. 

Los números feos son números cuyos únicos factores primos son 2, 3 o 5.

Ejemplos: 

Entrada: N = 14 
Salida: No 
Explicación: 
14 no es feo ya que incluye otro factor primo 7.

Entrada: N = 6 
Salida: Sí 
Explicación: 
6 es feo ya que incluye 2 y 3. 
 

Enfoque: La idea es usar la recursividad para resolver este problema y verificar si un número es divisible por 2, 3 o 5. Si es así, entonces divida el número por eso y verifique recursivamente si un número es un número feo o no. Si en algún momento no existe dicho divisor, devuelve falso, de lo contrario, verdadero.

A continuación se muestra la implementación del enfoque anterior:
 

// C++ implementation to check if a number is an ugly number
// or not
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to check if a number is an ugly number or not
int isUgly(int n)
{
    // Base Cases
    if (n == 1)
        return 1;
    if (n <= 0)
        return 0;
 
    // Condition to check if the number is divided by 2, 3,
    // or 5
    if (n % 2 == 0)
        return isUgly(n / 2);
    if (n % 3 == 0)
        return isUgly(n / 3);
    if (n % 5 == 0)
        return isUgly(n / 5);
 
    // Otherwise return false
    return 0;
}
// Driver Code
int main()
{
    int no = isUgly(14);
    if (no == 1)
        cout << "Yes" << endl;
    else
        cout << "No" << endl;
    return 0;
}
 
// This code is contributed by Sania Kumari Gupta (kriSania804)

// C implementation to check if a number is an ugly number
// or not
 
#include <stdio.h>
 
// Function to check if a number is an ugly number or not
int isUgly(int n)
{
    // Base Cases
    if (n == 1)
        return 1;
    if (n <= 0)
        return 0;
 
    // Condition to check if the number is divided by 2, 3,
    // or 5
    if (n % 2 == 0)
        return isUgly(n / 2);
    if (n % 3 == 0)
        return isUgly(n / 3);
    if (n % 5 == 0)
        return isUgly(n / 5);
 
    // Otherwise return false
    return 0;
}
// Driver Code
int main()
{
    int no = isUgly(14);
    if (no == 1)
        printf("Yes");
    else
        printf("No");
    return 0;
}
 
// This code is contributed by Sania Kumari Gupta (kriSania804)

// Java implementation to
// check if a number is ugly number
class GFG {
 
    // Function to check the ugly
    // number
    static int isUgly(int n)
    {
        // Base Cases
        if (n == 1)
            return 1;
        if (n <= 0)
            return 0;
 
        // Condition to check if
        // a number is divide by
        // 2, 3, or 5
        if (n % 2 == 0) {
            return (isUgly(n / 2));
        }
        if (n % 3 == 0) {
            return (isUgly(n / 3));
        }
        if (n % 5 == 0) {
            return (isUgly(n / 5));
        }
        return 0;
    }
 
    // Driver Code
    public static void main(String args[])
    {
        int no = isUgly(14);
        if (no == 1)
            System.out.println("Yes");
        else
            System.out.println("No");
    }
}

# Python3 implementation to check
# if a number is an ugly number
# or not
 
# Function to check if a number
# is an ugly number or not
def isUgly(n):
 
    # Base Cases
    if (n == 1):
        return 1
    if (n <= 0):
        return 0
 
    # Condition to check if the
    # number is divided by 2, 3, or 5
    if (n % 2 == 0):
        return (isUgly(n // 2))
         
    if (n % 3 == 0):
        return (isUgly(n // 3))
     
    if (n % 5 == 0):
        return (isUgly(n // 5))
 
    # Otherwise return false
    return 0
 
# Driver Code
if __name__ == "__main__":
 
    no = isUgly(14)
     
    if (no == 1):
        print("Yes")
    else:
        print("No")
 
# This code is contributed by chitranayal

// C# implementation to check
// if a number is ugly number
using System;
 
class GFG{
 
// Function to check the ugly
// number
static int isUgly(int n)
{
     
    // Base Cases
    if (n == 1)
        return 1;
    if (n <= 0)
        return 0;
 
    // Condition to check if
    // a number is divide by
    // 2, 3, or 5
    if (n % 2 == 0)
    {
        return (isUgly(n / 2));
    }
    if (n % 3 == 0)
    {
        return (isUgly(n / 3));
    }
    if (n % 5 == 0)
    {
        return (isUgly(n / 5));
    }
    return 0;
}
 
// Driver Code
public static void Main(String []args)
{
    int no = isUgly(14);
     
    if (no == 1)
        Console.WriteLine("Yes");
    else
        Console.WriteLine("No");
}
}
 
// This code is contributed by amal kumar choubey

<script>
 
// Javascript implementation to check
// if a number is an ugly
// number or not
 
// Function to check if a number
// is an ugly number or not
function isUgly(n)
{
    // Base Cases
    if (n == 1)
        return 1;
    if (n <= 0)
        return 0;
 
    // Condition to check if the
    // number is divided by 2, 3, or 5
    if (n % 2 == 0) {
        return (isUgly(n / 2));
    }
    if (n % 3 == 0) {
        return (isUgly(n / 3));
    }
    if (n % 5 == 0) {
        return (isUgly(n / 5));
    }
 
    // Otherwise return false
    return 0;
}
// Driver Code
  
    let no = isUgly(14);
    if (no == 1)
        document.write("Yes");
    else
        document.write("No");
     
 
// This code is contributed by Mayank Tyagi
 
</script>

No