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Find multiple of x closest to or a ^ b (a raised to power b)
  • Difficulty Level : Expert
  • Last Updated : 13 Apr, 2021

We are given 3 numbers a, b and x. We need to output the multiple of x which is closest to a^b. 
Note : b can be a negative number
Examples : 
 

Input : x = 2, a = 4, b = -2
Output : 0
Explanation : a^b = 1/16. 
Closest multiple of 2 to 1/16 is 0.

Input : x = 4, a = 349, b = 1
Output : 348
Explanation :a^b = 349
The closest multiple of 4 to 349 is 348.

 

Observations : 
 

1. When b is negative and a is 1, then a ^ b turns out 
to be 1 and hence, closest multiple of x becomes either
 x * 0 or x * 1.

2. When b is negative and a is more than 1, then a ^ b
turns out to be less than 1 and hence closest multiple
of x becomes 0.

3. When b is positive, calculate a ^ b, then let 
mul = Integer (a^b / x), then closest multiple of x is 
mul * x or (mul + 1) * x.

 

C++




// C++ Program to find closest
// multiple of x to a^b
#include <bits/stdc++.h>
using namespace std;
 
// function to find closest multiple 
// of x to a^b
void multiple(int a, int b, int x)
{   
    if (b < 0) {
        if (a == 1 && x == 1)
            cout << "1";
 
        else
            cout << "0";
    }
 
    // calculate a ^ b / x
    int mul = pow(a, b);
     
    int ans = mul / x;
     
    // Answer is either (ans * x) or
    // (ans + 1) * x
    int ans1 = x * ans;
    int ans2 = x * (ans + 1);
 
    // Printing nearest answer
    cout << (((mul - ans1) <= (ans2 - mul)) ?
                                ans1 : ans2);
}
 
// Driver Program
int main()
{
    int a = 349, b = 1, x = 4;
 
    multiple(a, b, x);
    return 0;
}

Java




// java Program to find closest
// multiple of x to a^b
import java.io.*;
 
public class GFG {
 
    // function to find closest
    // multiple of x to a^b
    static void multiple(int a, int b, int x)
    {
        if (b < 0)
        {
            if (a == 1 && x == 1)
                System.out.println("1");
            else
                System.out.println("0");
        }
     
        // calculate a ^ b / x
        int mul = (int)Math.pow(a, b);
         
        int ans = mul / x;
         
        // Answer is either (ans * x) or
        // (ans + 1) * x
        int ans1 = x * ans;
        int ans2 = x * (ans + 1);
     
        // Printing nearest answer
        System.out.println(((mul - ans1)
                        <= (ans2 - mul))
                         ? ans1 : ans2);
    }
     
    // Driver Program
    static public void main (String[] args)
    {
        int a = 349, b = 1, x = 4;
 
        multiple(a, b, x);
    }
}
 
// This code is contributed by vt_m.

C#




// C# Program to find closest
// multiple of x to a^b
using System;
 
public class GFG {
         
    // function to find closest multiple
    // of x to a^b
    static void multiple(int a, int b, int x)
    {
        if (b < 0) {
            if (a == 1 && x == 1)
                Console.WriteLine("1");
     
            else
                Console.WriteLine("0");
        }
     
        // calculate a ^ b / x
        int mul = (int)Math.Pow(a, b);
         
        int ans = mul / x;
         
        // Answer is either (ans * x) or
        // (ans + 1) * x
        int ans1 = x * ans;
        int ans2 = x * (ans + 1);
     
        // Printing nearest answer
        Console.WriteLine(((mul - ans1)
                       <= (ans2 - mul))
                         ? ans1 : ans2);
    }
     
    // Driver Program
    static public void Main ()
    {
        int a = 349, b = 1, x = 4;
 
        multiple(a, b, x);
    }
}
 
// This code is contributed by vt_m.

PHP




<?php
// PHP Program to find closest
// multiple of x to a^b
 
// function to find
// closest multiple
// of x to a^b
function multiple($a, $b, $x)
{
    if ($b < 0)
    {
        if ($a == 1 && $x == 1)
            echo "1";
 
        else
            echo "0";
    }
 
    // calculate a ^ b / x
    $mul = pow($a, $b);
     
    $ans = $mul / $x;
     
    // Answer is either
    // (ans * x) or
    // (ans + 1) * x
    $ans1 = $x * $ans;
    $ans2 = $x * ($ans + 1);
 
        $k = ((($mul - $ans1) <= ($ans2 - $mul))
                               ? $ans1 : $ans2);
        echo($k);
}
 
    // Driver Code
    $a = 348;
    $b = 1;
    $x = 4;
 
    multiple($a, $b, $x);
 
// This code is contributed by ajit
?>

Python3




# Python3 Program to
# find closest multiple
# of x to a^b
import math
 
# function to find closest
# multiple of x to a^b
def multiple(a, b, x):
    if (b < 0):
        if (a == 1 and x == 1):
            print("1");
        else:
            print("0");
             
    # calculate a ^ b / x
    mul = int(pow(a, b));
     
    ans = int(mul / x);
     
    # Answer is either (ans * x)
    # or (ans + 1) * x
    ans1 = x * ans;
    ans2 = x * (ans + 1);
     
    # Printing nearest answer
    if ((mul - ans1) <= (ans2 - mul)):
        print(ans1);
    else:
        print(ans2);
 
# Driver Code
a = 349;
b = 1;
x = 4;
multiple(a, b, x);
 
# This code is contributed
# by mits

Javascript




<script>
// javascript Program to find closest
// multiple of x to a^bpublic
 
    // function to find closest
    // multiple of x to a^b
    function multiple(a , b , x) {
        if (b < 0) {
            if (a == 1 && x == 1)
                document.write("1");
            else
                document.write("0");
        }
 
        // calculate a ^ b / x
        var mul = parseInt( Math.pow(a, b));
 
        var ans = mul / x;
 
        // Answer is either (ans * x) or
        // (ans + 1) * x
        var ans1 = x * ans;
        var ans2 = x * (ans + 1);
 
        // Printing nearest answer
        document.write(((mul - ans1) <= (ans2 - mul)) ? ans1 : ans2);
    }
 
    // Driver Program
        var a = 349, b = 1, x = 4;
 
        multiple(a, b, x);
 
// This code is contributed by gauravrajput1
</script>

Output: 
 

348

This article is contributed by Rohit Thapliyal. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
 

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