Multiplica un número por 15 sin usar los operadores * y /

Dado un número entero N , la tarea es multiplicar el número por 15 sin utilizar los operadores de multiplicación * y división / .

Ejemplos:

Entrada: N = 10 
Salida: 150

Entrada: N = 7 
Salida: 105 

Método 1: podemos multiplicar el número entero N por 15 utilizando operadores bit a bit. Primero, desplace el número a la izquierda en 4 bits, lo que equivale a (16 * N) , luego reste el número original N del número desplazado, es decir , ((16 * N) – N), que es igual a 15 * N.

A continuación se muestra la implementación del enfoque anterior: 

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return (15 * N) without
// using '*' or '/' operator
long multiplyByFifteen(long n)
{
    // prod = 16 * n
    long prod = (n << 4);
 
    // ((16 * n) - n) = 15 * n
    prod = prod - n;
 
    return prod;
}
 
// Driver code
int main()
{
    long n = 7;
 
    cout << multiplyByFifteen(n);
 
    return 0;
}

// Java implementation of the approach
class GFG {
 
    // Function to return (15 * N) without
    // using '*' or '/' operator
    static long multiplyByFifteen(long n)
    {
        // prod = 16 * n
        long prod = (n << 4);
 
        // ((16 * n) - n) = 15 * n
        prod = prod - n;
 
        return prod;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        long n = 7;
        System.out.print(multiplyByFifteen(n));
    }
}

# Python3 implementation of the approach
 
# Function to return (15 * N) without
# using '*' or '/' operator
def multiplyByFifteen(n):
     
    # prod = 16 * n
    prod = (n << 4)
     
    # ((16 * n) - n) = 15 * n
    prod = prod - n
     
    return prod
     
# Driver code
n = 7
print(multiplyByFifteen(n))

// C# implementation of the approach
using System;
class GFG {
 
    // Function to return (15 * N) without
    // using '*' or '/' operator
    static long multiplyByFifteen(long n)
    {
        // prod = 16 * n
        long prod = (n << 4);
 
        // ((16 * n) - n) = 15 * n
        prod = prod - n;
 
        return prod;
    }
 
    // Driver code
    public static void Main()
    {
        long n = 7;
        Console.Write(multiplyByFifteen(n));
    }
}

<?php
// PHP implementation of the approach
 
// Function to return (15 * N) without
// using '*' or '/' operator
function multiplyByFifteen($n)
{
    // prod = 16 * n
    $prod = ($n << 4);
 
    // ((16 * n) - n) = 15 * n
    $prod = $prod - $n;
 
    return $prod;
}
 
// Driver code
 
    $n = 7;
 
    echo multiplyByFifteen($n);
 
// This code is contributed by anuj_67..
?>

<script>
 
// JavaScript implementation of the approach
 
// Function to return (15 * N) without
// using '*' or '/' operator
function multiplyByFifteen(n)
{
    // prod = 16 * n
    let prod = (n << 4);
 
    // ((16 * n) - n) = 15 * n
    prod = prod - n;
 
    return prod;
}
 
// Driver code
 
    let n = 7;
 
    document.write(multiplyByFifteen(n));
 
// This code is contributed by Surbhi Tyagi.
 
</script>

105

Método 2: También podemos calcular la multiplicación (15 * N) como la suma de (8 * N) + (4 * N) + (2 * N) + N que se puede obtener realizando operaciones de desplazamiento a la izquierda como (8 * N ) = (N << 3) , (4 * N) = (n << 2) y (2 * N) = (n << 1) .

A continuación se muestra la implementación del enfoque anterior:

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return (15 * N) without
// using '*' or '/' operator
long multiplyByFifteen(long n)
{
    // prod = 8 * n
    long prod = (n << 3);
 
    // Add (4 * n)
    prod += (n << 2);
 
    // Add (2 * n)
    prod += (n << 1);
 
    // Add n
    prod += n;
 
    // (8 * n) + (4 * n) + (2 * n) + n = (15 * n)
    return prod;
}
 
// Driver code
int main()
{
    long n = 7;
 
    cout << multiplyByFifteen(n);
 
    return 0;
}

// Java implementation of the approach
class GFG {
 
    // Function to return (15 * N) without
    // using '*' or '/' operator
    static long multiplyByFifteen(long n)
    {
        // prod = 8 * n
        long prod = (n << 3);
 
        // Add (4 * n)
        prod += (n << 2);
 
        // Add (2 * n)
        prod += (n << 1);
 
        // Add n
        prod += n;
 
        // (8 * n) + (4 * n) + (2 * n) + n = (15 * n)
        return prod;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        long n = 7;
        System.out.print(multiplyByFifteen(n));
    }
}

# Python3 implementation of the approach
 
# Function to perform Multiplication
def multiplyByFifteen(n):
     
    # prod = 8 * n
    prod = (n << 3)
     
    # Add (4 * n)
    prod += (n << 2)
     
    # Add (2 * n)
    prod += (n << 1)
     
    # Add n
    prod += n
     
    # (8 * n) + (4 * n) + (2 * n) + n = (15 * n)
    return prod
     
# Driver code
n = 7
print(multiplyByFifteen(n))

// C# implementation of the approach
using System;
class GFG {
 
    // Function to return (15 * N) without
    // using '*' or '/' operator
    static long multiplyByFifteen(long n)
    {
        // prod = 8 * n
        long prod = (n << 3);
 
        // Add (4 * n)
        prod += (n << 2);
 
        // Add (2 * n)
        prod += (n << 1);
 
        // Add n
        prod += n;
 
        // (8 * n) + (4 * n) + (2 * n) + n = (15 * n)
        return prod;
    }
 
    // Driver code
    public static void Main()
    {
        long n = 7;
        Console.Write(multiplyByFifteen(n));
    }
}

<script>
 
// Javascript implementation of the approach  
 
// Function to return (15 * N) without
// using '*' or '/' operator
function multiplyByFifteen(n)
{
     
    // prod = 8 * n
    var prod = (n << 3);
 
    // Add (4 * n)
    prod += (n << 2);
 
    // Add (2 * n)
    prod += (n << 1);
 
    // Add n
    prod += n;
 
    // (8 * n) + (4 * n) + (2 * n) + n = (15 * n)
    return prod;
}
 
// Driver code
var n = 7;
 
document.write(multiplyByFifteen(n));
 
// This code is contributed by shikhasingrajput
 
</script>

105

Tiempo Complejidad: O(1)
Espacio Auxiliar: O(1)