大于n的2的最小最小幂(不使用算术运算运算符)

给定一个非负整数n 。问题是在不使用算术运算运算符的情况下找到大于n的最小完美幂2。
例子 ：

Input : n = 10
Output : 16

Input : n = 128
Output : 256

算法 ：

C++

// C++ implementation of smallest perfect power
// of 2 greater than n
#include 
 
using namespace std;
 
// Function to find smallest perfect power
// of 2 greater than n
unsigned int perfectPowerOf2(unsigned int n)
{
    // To store perfect power of 2
    unsigned int per_pow = 1;
 
    while (n > 0)
    {
        // bitwise left shift by 1
        per_pow = per_pow << 1;
 
        // bitwise right shift by 1
        n = n >> 1;
    }
 
    // Required perfect power of 2
    return per_pow;
}
 
// Driver program to test above
int main()
{
    unsigned int n = 128;
    cout << "Perfect power of 2 greater than "
        << n << ": " << perfectPowerOf2(n);
    return 0;
}

Java

// JAVA Code for Smallest perfect
// power of 2 greater than n
import java.util.*;
 
class GFG {
     
    // Function to find smallest perfect
    // power of 2 greater than n
    static int perfectPowerOf2( int n)
    {
        // To store perfect power of 2
         int per_pow = 1;
      
        while (n > 0)
        {
            // bitwise left shift by 1
            per_pow = per_pow << 1;
             
            n = n >> 1;
        }
      
        // Required perfect power of 2
        return per_pow;
    }
     
    // Driver program
    public static void main(String[] args)
    {
         int n = 12;
         System.out.println("Perfect power of 2 greater than "
                            + n + ": " + perfectPowerOf2(n));
        }
    }
         
    //This code is contributed by Arnav Kr. Mandal.

Python3

# Python3 implementation of smallest
# perfect power of 2 greater than n
 
# Function to find smallest perfect
# power of 2 greater than n
def perfectPowerOf2( n ):
     
    # To store perfect power of 2
    per_pow = 1
     
    while n > 0:
     
        # bitwise left shift by 1
        per_pow = per_pow << 1
         
        # bitwise right shift by 1
        n = n >> 1
         
    # Required perfect power of 2
    return per_pow
 
# Driver program to test above
n = 128
print("Perfect power of 2 greater than",
            n, ":",perfectPowerOf2(n))
 
# This code is contributed by "Sharad_Bhardwaj".

C#

// C# Code for Smallest perfect
// power of 2 greater than n
using System;
 
class GFG {
 
    // Function to find smallest perfect
    // power of 2 greater than n
    static int perfectPowerOf2(int n)
    {
        // To store perfect power of 2
        int per_pow = 1;
 
        while (n > 0)
        {
            // bitwise left shift by 1
            per_pow = per_pow << 1;
 
            n = n >> 1;
        }
 
        // Required perfect power of 2
        return per_pow;
    }
 
    // Driver program
    public static void Main()
    {
        int n = 128;
        Console.WriteLine("Perfect power of 2 greater than " +
                           n + ": " + perfectPowerOf2(n));
    }
}
 
// This code is contributed by Sam007

PHP

 0)
    {
        // bitwise left shift by 1
        $per_pow = $per_pow << 1;
 
        // bitwise right shift by 1
        $n = $n >> 1;
    }
 
    // Required perfect power of 2
    return $per_pow;
}
 
    // Driver code
    $n = 128;
    echo "Perfect power of 2 greater than ".
          $n . ": ".perfectPowerOf2($n);
 
// This code is contributed by mits
?>

Javascript

 
// JavaScript implementation of smallest perfect power
// of 2 greater than n
 
// Function to find smallest perfect power
// of 2 greater than n
function perfectPowerOf2(n)
{
    // To store perfect power of 2
    let per_pow = 1;
 
    while (n > 0)
    {
        // bitwise left shift by 1
        per_pow = per_pow << 1;
 
        // bitwise right shift by 1
        n = n >> 1;
    }
 
    // Required perfect power of 2
    return per_pow;
}
 
// Driver program to test above
 
    let n = 128;
    document.write("Perfect power of 2 greater than "
        + n + ": " + perfectPowerOf2(n));
 
 
// This code is contributed by Surbhi Tyagi.
 

输出：

Perfect power of 2 greater than 128: 256

时间复杂度： O(num) ，其中num是n的二进制表示形式中的位数。