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 above int main() { unsigned int n = 8, m = 2; if (isDivBy2PowerM(n, m)) cout << "Yes"; else cout << "No"; return 0; } |
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Java
// JAVA Code for Check if n is divisible // by power of 2 without using arithmetic // operators import 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. |
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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 above n = 8m = 2if isDivBy2PowerM(n, m): print("Yes") else: print( "No") # This code is contributed by "Sharad_Bhardwaj". |
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C#
// C# Code for Check if n is divisible // by power of 2 without using arithmetic // operators using 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 |
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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 ?> |
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Output:
Yes
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Improved By : jit_t
