Check if n is divisible by power of 2 without using arithmetic operators
Given two positive integers n and m. The problem is to check whether n is divisible by 2m or not without using arithmetic operators.
Examples:
Input : n = 8, m = 2 Output : Yes Input : n = 14, m = 3 Output : No
Approach: If a number is divisible by 2 then it has its least significant bit (LSB) set to 0, if divisible by 4 then two LSB’s set to 0, if by 8 then three LSB’s set to 0 and so on. Keeping this in mind, a number n is divisible by 2m if (n & ((1 << m) – 1)) is equal to 0 else not.
C++
// C++ implementation to chech whether n// is divisible by pow(2, m)#include <bits/stdc++.h>using namespace std;// function to chech whether n// is divisible by pow(2, m)bool isDivBy2PowerM(unsigned int n, unsigned int m){ // if expression results to 0, then // n is divisible by pow(2, m) if ((n & ((1 << m) - 1)) == 0) return true; // n is not divisible return false;}// Driver program to test aboveint main(){ unsigned int n = 8, m = 2; if (isDivBy2PowerM(n, m)) cout << "Yes"; else cout << "No"; return 0;} |
Java
// JAVA Code for Check if n is divisible // by power of 2 without using arithmetic// operatorsimport java.util.*;class GFG { // function to chech whether n // is divisible by pow(2, m) static boolean isDivBy2PowerM(int n, int m) { // if expression results to 0, then // n is divisible by pow(2, m) if ((n & ((1 << m) - 1)) == 0) return true; // n is not divisible return false; } /* Driver program to test above function */ public static void main(String[] args) { int n = 8, m = 2; if (isDivBy2PowerM(n, m)) System.out.println("Yes"); else System.out.println("No"); }}// This code is contributed by Arnav Kr. Mandal. |
Python3
# Python3 implementation to chech# whether n is divisible by pow(2, m)# function to chech whether n# is divisible by pow(2, m)def isDivBy2PowerM (n, m): # if expression results to 0, then # n is divisible by pow(2, m) if (n & ((1 << m) - 1)) == 0: return True # n is not divisible return False# Driver program to test aboven = 8m = 2if isDivBy2PowerM(n, m): print("Yes")else: print( "No") # This code is contributed by "Sharad_Bhardwaj". |
C#
// C# Code for Check if n is divisible// by power of 2 without using arithmetic// operatorsusing System;class GFG { // function to chech whether n // is divisible by pow(2, m) static bool isDivBy2PowerM(int n, int m) { // if expression results to 0, then // n is divisible by pow(2, m) if ((n & ((1 << m) - 1)) == 0) return true; // n is not divisible return false; } /* Driver program to test above function */ public static void Main() { int n = 8, m = 2; if (isDivBy2PowerM(n, m)) Console.Write("Yes"); else Console.Write("No"); }}// This code is contributed by Sam007 |
PHP
<?php// PHP implementation to chech whether// n is divisible by pow(2, m)// function to chech whether n// is divisible by pow(2, m)function isDivBy2PowerM($n, $m){ // if expression results to 0, then // n is divisible by pow(2, m) if (($n & ((1 << $m) - 1)) == 0) return true; // n is not divisible return false;} // Driver Code $n = 8; $m = 2; if (isDivBy2PowerM($n, $m)) echo "Yes"; else echo "No";// This code is contributed by ajit?> |
Javascript
<script>// JavaScript implementation to chech whether n// is divisible by pow(2, m)// function to chech whether n// is divisible by pow(2, m)function isDivBy2PowerM(n, m){ // if expression results to 0, then // n is divisible by pow(2, m) if ((n & ((1 << m) - 1)) == 0) return true; // n is not divisible return false;}// Driver program to test above let n = 8, m = 2; if (isDivBy2PowerM(n, m)) document.write("Yes"); else document.write("No");// This code is contributed by Surbhi Tyagi.</script> |
Output:
Yes
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