Encuentra el múltiplo de x más cercano a a^b

Dados tres números enteros a , b y x , la tarea es obtener el múltiplo de x más cercano a a ^b .
Ejemplos: 
 

Entrada: a = 5, b = 4, x = 3 
Salida: 624 
5 ⁴ = 625 y 624 es el múltiplo de 3 más cercano a 625
Entrada: a = 349, b = 1, x = 4 
Salida: 348 
 

Acercarse: 

• Calcula a ^b y guárdalo en una variable, digamos num .
• Luego, calcula ⌊num / x⌋ y guárdalo en un piso variable .
• Ahora el elemento más cercano a la izquierda será el más cercano a la izquierda = x * piso .
• Y el elemento más cercano a la derecha será la derecha más cercana = x * (piso + 1) .
• Finalmente, el número más cercano entre ellos será min(num – más cercano a la izquierda, más cercano a la derecha – num) .

A continuación se muestra la implementación del enfoque anterior:
 

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
#define ll long long int
 
// Function to return the multiple of x
// which is closest to a^b
ll getClosest(int a, int b, int x)
{
    ll num = pow(a, b);
 
    int floor = num / x;
 
    // Closest element on the left
    ll numOnLeft = x * floor;
 
    // Closest element on the right
    ll numOnRight = x * (floor + 1);
 
    // If numOnLeft is closer than numOnRight
    if ((num - numOnLeft) < (numOnRight - num))
        return numOnLeft;
 
    // If numOnRight is the closest
    else
        return numOnRight;
}
 
// Driver code
int main()
{
    int a = 349, b = 1, x = 4;
    cout << getClosest(a, b, x) << endl;
    return 0;
}

//Java implementation of the approach
 
public class GFG {
 
// Function to return the multiple of x
// which is closest to a^b
    static long getClosest(int a, int b, int x) {
        long num = (long) Math.pow(a, b);
 
        int floor = (int) (num / x);
 
        // Closest element on the left
        long numOnLeft = x * floor;
 
        // Closest element on the right
        long numOnRight = x * (floor + 1);
 
        // If numOnLeft is closer than numOnRight
        if ((num - numOnLeft) < (numOnRight - num)) {
            return numOnLeft;
        } // If numOnRight is the closest
        else {
            return numOnRight;
        }
    }
 
    public static void main(String[] args) {
        int a = 349, b = 1, x = 4;
        System.out.println(getClosest(a, b, x));
 
    }
}

# Python3 implementation of the approach
 
# Function to return the multiple of x
# which is closest to a^b
def getClosest(a, b, x) :
    num = pow(a, b)
 
    floor = num // x
 
    # Closest element on the left
    numOnLeft = x * floor
 
    # Closest element on the right
    numOnRight = x * (floor + 1)
 
    # If numOnLeft is closer than numOnRight
    if ((num - numOnLeft) <
        (numOnRight - num)):
        return numOnLeft
 
    # If numOnRight is the closest
    else :
        return numOnRight
 
# Driver code
if __name__ == "__main__" :
     
    a, b, x = 349, 1, 4
    print(getClosest(a, b, x))
 
# This code is contributed by Ryuga

// C# implementation of the approach
using System;
 
// #define ll long long int
 
class GFG
{
// Function to return the multiple of x
// which is closest to a^b
static long getClosest(int a, int b, int x)
{
    int num = (int)Math.Pow(a, b);
 
    int floor = (int)(num / x);
 
    // Closest element on the left
    int numOnLeft = (int)(x * floor);
 
    // Closest element on the right
    int numOnRight = (int)(x * (floor + 1));
 
    // If numOnLeft is closer than numOnRight
    if ((num - numOnLeft) < (numOnRight - num))
        return numOnLeft;
 
    // If numOnRight is the closest
    else
        return numOnRight;
}
 
// Driver code
public static void Main()
{
    int a = 349, b = 1, x = 4;
    Console.WriteLine(getClosest(a, b, x));
}
}
 
// This code is contributed
// by Akanksha Rai

<?php
// PHP implementation of the above approach
 
// Function to return the multiple of x
// which is closest to a^b
function getClosest($a, $b, $x)
{
    $num = pow($a, $b);
 
    $floor = (int)($num / $x);
 
    // Closest element on the left
    $numOnLeft = $x * $floor;
 
    // Closest element on the right
    $numOnRight = $x * ($floor + 1);
 
    // If numOnLeft is closer than numOnRight
    if (($num - $numOnLeft) <
        ($numOnRight - $num))
        return $numOnLeft;
 
    // If numOnRight is the closest
    else
        return ceil($numOnRight);
}
 
// Driver code
$a = 349;
$b = 1;
$x = 4;
echo getClosest($a, $b, $x);
 
// This code is contributed by jit_t
?>

<script>
 
// Javascript implementation of the approach
 
 
// Function to return the multiple of x
// which is closest to a^b
function getClosest( a,  b,  x)
{
    let num = Math.pow(a, b);
 
    let floor = Math.floor(num / x);
 
    // Closest element on the left
    let numOnLeft = x * floor;
 
    // Closest element on the right
    let numOnRight = x * (floor + 1);
 
    // If numOnLeft is closer than numOnRight
    if ((num - numOnLeft) < (numOnRight - num))
        return numOnLeft;
 
    // If numOnRight is the closest
    else
        return numOnRight;
}
 
 
    // Driver Code
     
    let a = 349, b = 1, x = 4;
    document.write(getClosest(a, b, x) + "</br>");
     
</script>

348

Complejidad de tiempo: O(1)

Espacio Auxiliar: O(1)