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 Programint 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^bimport 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^busing 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^bimport 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 Codea = 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
Time Complexity: O(log b), to find power
Auxiliary Space: O(1),as no extra space is required
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