Find the multiple of x which is closest to a^b

Given three integers a, b and x, the task is to get the multiple of x which is closest to ab.

Examples:

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

Approach:

  • Calculate ab and store it in a variable say num.
  • Then, calculate ⌊num / x⌋ and store it in a variable floor.
  • Now the closest element at the left will be closestLeft = x * floor.
  • And the closest element on the right will be closestRight = x * (floor + 1).
  • Finally, the closest number among them will be min(num – closestLeft, closestRight – num).

Below is the implementation of the above approach:

C++

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// 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;
}

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Java

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//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));
  
    }
}

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Python3

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# 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

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C#

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// 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

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PHP

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<?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
?>

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Output:

348


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