K的最大和最小幂分别小于和大于等于N

给定正整数N和K ，任务是找出K的最大和最小幂分别大于或小于N。
例子：

Input: N = 3, K = 2 
Output: 2 4 
Highest power of 2 less than 3 = 2 
Smallest power of 2 greater than 3 = 4
Input: N = 6, K = 3 
Output: 3 9 
Highest power of 3 less than 6 = 2 
Smallest power of 3 greater than 6 = 9 
 

方法：

1. 计算以底数K为单位的N的对数( log _K N )以获得指数幂，以使升至该指数的K为K的最高幂，小于N。
2. 对于K的最小乘方小于N，找到从上一步计算出的K的下一个乘方

下面是上述方法的实现：

C++

// C++ implementation of the approach
 
#include 
using namespace std;
 
// Function to return the highest power
// of k less than or equal to n
int prevPowerofK(int n, int k)
{
    int p = (int)(log(n) / log(k));
    return (int)pow(k, p);
}
 
// Function to return the smallest power
// of k greater than or equal to n
int nextPowerOfK(int n, int k)
{
    return prevPowerofK(n, k) * k;
}
 
// Function to print the result
void printResult(int n, int k)
{
    cout << prevPowerofK(n, k)
         << " " << nextPowerOfK(n, k)
         << endl;
}
 
// Driver code
int main()
{
    int n = 25, k = 3;
 
    printResult(n, k);
 
    return 0;
}

Java

// Java implementation of the approach
import java.io.*;
 
class GFG{
 
// Function to return the highest power
// of k less than or equal to n
static int prevPowerofK(int n, int k)
{
    int p = (int)(Math.log(n) / Math.log(k));
    return (int) Math.pow(k, p);
}
 
// Function to return the smallest power
// of k greater than or equal to n
static int nextPowerOfK(int n, int k)
{
    return prevPowerofK(n, k) * k;
}
 
// Function to print the result
static void printResult(int n, int k)
{
    System.out.println(prevPowerofK(n, k) + " " +
                       nextPowerOfK(n, k));
}
 
// Driver Code
public static void main (String args[])
{
    int n = 25, k = 3;
    printResult(n, k);
}
}
 
// This code is contributed by shivanisinghss2110

Python3

# Python3 implementation of the approach
import math
 
# Function to return the highest power
# of k less than or equal to n
def prevPowerofK(n, k):
 
    p = int(math.log(n) / math.log(k))
    return int(math.pow(k, p))
 
# Function to return the smallest power
# of k greater than or equal to n
def nextPowerOfK(n, k):
 
    return prevPowerofK(n, k) * k
 
# Function to print the result
def printResult(n, k):
 
    print(prevPowerofK(n, k), nextPowerOfK(n, k))
 
# Driver code
n = 6
k = 3
 
printResult(n, k)
 
# This code is contributed by divyamohan123

C#

// C# implementation of the approach
using System;
class GFG{
 
// Function to return the highest power
// of k less than or equal to n
static int prevPowerofK(int n, int k)
{
    int p = (int)(Math.Log(n) / Math.Log(k));
    return (int) Math.Pow(k, p);
}
 
// Function to return the smallest power
// of k greater than or equal to n
static int nextPowerOfK(int n, int k)
{
    return prevPowerofK(n, k) * k;
}
 
// Function to print the result
static void printResult(int n, int k)
{
    Console.WriteLine(prevPowerofK(n, k) + " " +
                      nextPowerOfK(n, k));
}
 
// Driver Code
public static void Main(String []args)
{
    int n = 25, k = 3;
    printResult(n, k);
}
}
 
// This code is contributed by gauravrajput1

Javascript

 
 
// Javasript implementation of the approach
 
// Function to return the highest power
// of k less than or equal to n
function prevPowerofK(n, k)
{
    var p = parseInt(Math.log(n) / Math.log(k));
    return parseInt(Math.pow(k, p));
}
 
// Function to return the smallest power
// of k greater than or equal to n
function nextPowerOfK(n, k)
{
    return prevPowerofK(n, k) * k;
}
 
// Function to print the result
function printResult(n, k)
{
    document.write(prevPowerofK(n, k)
         + " " + nextPowerOfK(n, k) + "<br>");
          
}
 
// Driver code
var n = 25, k = 3;
printResult(n, k);
 
 

输出：

9 27