找到最接近a ^ b的x的倍数

给定三个整数a ， b和x ，任务是获得x的最接近a ^b的倍数。

例子：

Input: a = 5, b = 4, x = 3
Output: 624
5⁴ = 625 and 624 is the multiple of 3 which is closest to 625

Input: a = 349, b = 1, x = 4
Output: 348

方法：

• 计算^b并将其存储在变量num中。
• 然后，计算⌊num/x⌋并将其存储在变量floor中。
• 现在，左侧最接近的元素将是closetLeft = x * floor 。
• 右边最接近的元素将是closetRight = x *(floor + 1) 。
• 最后，其中最接近的数字将是min(num-最近的左，最近的正确-num) 。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
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

//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

# 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#

// 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

输出：

348

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