在不使用算术运算运算符的情况下检查n是否可以被2的整数除

给定两个正整数n和m 。问题是不使用算术运算运算符检查n是否可被2 ^m整除。
例子：

Input : n = 8, m = 2
Output : Yes

Input : n = 14, m = 3
Output : No

方法：如果一个数字可被2整除，则其最低有效位(LSB)设置为0，如果被4整除，则两个LSB设置为0，如果被8则将三个LSB设置为0，依此类推。请记住，如果(n＆(((1 << m)– 1))等于0，则数字n可以被2 ^m整除。

C++

// C++ implementation to chech whether n
// is divisible by pow(2, m)
#include 
 
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;
}

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.

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 = 8
m = 2
if 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
// 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

Javascript

 
// 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.
 

输出：

Yes