Menor potencia de 4 mayor o igual a N

Dado un número entero N , la tarea es encontrar la potencia de cuatro más pequeña mayor o igual que N .

Ejemplos: 

Entrada: N = 12 
Salida: 16 
2 ⁴ = 16 que es el siguiente 
número mayor requerido después de 12.

Entrada: N = 81 
Salida: 81 
 

Acercarse:  

1. Encuentre la raíz cuarta de la n dada.
2. Calcule su valor mínimo usando la función floor() .
3. Si n es en sí mismo una potencia de cuatro, entonces devuelve n.
4. De lo contrario, agregue 1 al valor mínimo.
5. Devuelve la cuarta potencia de ese número.

A continuación se muestra la implementación del enfoque anterior:  

// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the smallest power
// of 4 greater than or equal to n
int nextPowerOfFour(int n)
{
    int x = floor(sqrt(sqrt(n)));
    // If n is itself is a power of 4 then return n
    if (pow(x, 4) == n)
        return n;
    else {
        x = x + 1;
        return pow(x, 4);
    }
}
 
// Driver code
int main()
{
    int n = 122;
    printf("%d", nextPowerOfFour(n));
    return 0;
}
 
// This code is contributed by Sania Kumari Gupta

// C implementation of above approach
#include <math.h>
#include <stdio.h>
 
// Function to return the smallest power
// of 4 greater than or equal to n
int nextPowerOfFour(int n)
{
    int x = floor(sqrt(sqrt(n)));
    // If n is itself is a power of 4 then return n
    if (pow(x, 4) == n)
        return n;
    else {
        x = x + 1;
        return pow(x, 4);
    }
}
 
// Driver code
int main()
{
    int n = 122;
    printf("%d", nextPowerOfFour(n));
    return 0;
}
 
// This code is contributed by Sania Kumari Gupta

// Java implementation of above approach
import java.util.*;
import java.lang.Math;
import java.io.*;
 
class GFG
{
     
// Function to return the smallest power
// of 4 greater than or equal to n
static int nextPowerOfFour(int n)
{
    int x = (int)Math.floor(Math.sqrt(Math.sqrt(n)));
 
    // If n is itself is a power of 4
    // then return n
    if (Math.pow(x, 4) == n)
        return n;
    else {
        x = x + 1;
        return (int)Math.pow(x, 4);
    }
}
 
// Driver code
public static void main (String[] args)
              throws java.lang.Exception
{
    int n = 122;
    System.out.println(nextPowerOfFour(n));
}
}
 
// This code is contributed by nidhiva

# Python3 implementation of above approach
import math
 
# Function to return the smallest power
# of 4 greater than or equal to n
def nextPowerOfFour(n):
    x = math.floor((n**(1/2))**(1/2));
 
    # If n is itself is a power of 4
    # then return n
    if ((x**4) == n):
        return n;
    else:
        x = x + 1;
        return (x**4);
 
# Driver code
 
n = 122;
print(nextPowerOfFour(n));
 
# This code is contributed by Rajput-Ji

// C# implementation of above approach
using System;
class GFG
{
     
// Function to return the smallest power
// of 4 greater than or equal to n
static int nextPowerOfFour(int n)
{
    int x = (int)Math.Floor(Math.Sqrt(Math.Sqrt(n)));
 
    // If n is itself is a power of 4
    // then return n
    if (Math.Pow(x, 4) == n)
        return n;
    else
    {
        x = x + 1;
        return (int)Math.Pow(x, 4);
    }
}
 
// Driver code
public static void Main ()
{
    int n = 122;
    Console.WriteLine(nextPowerOfFour(n));
}
}
 
// This code is contributed by anuj_67..

<script>
 
// Javascript implementation of above approach
 
// Function to return the smallest power
// of 4 greater than or equal to n
function nextPowerOfFour(n)
{
    let x = Math.floor(Math.sqrt(
            Math.sqrt(n)));
 
    // If n is itself is a power of 4
    // then return n
    if (Math.pow(x, 4) == n)
        return n;
    else
    {
        x = x + 1;
        return Math.pow(x, 4);
    }
}
 
// Driver code
let n = 122;
document.write(nextPowerOfFour(n));
 
// This code is contributed by mohit kumar 29
 
</script>

256

Complejidad de tiempo: O(logx) donde x es sqrt(sqrt(n))

Espacio Auxiliar: O(1)