有效地检查n是否为4的倍数

给定数字n 。问题是不使用算术运算运算符就可以有效地检查n是否为4的倍数。
例子：

Input : 16
Output : Yes

Input : 14
Output : No

方法： 4的倍数在二进制表示中始终以00作为其后两位。我们必须检查n的最后两位是否未设置。
如何检查最后两位是否未设置。
如果n＆3 == 0，则后两位未置位，否则要么置位要么置位。

C++

// C++ implementation to efficiently check whether n
// is a multiple of 4 or not
#include 
using namespace std;
 
// function to check whether 'n' is
// a multiple of 4 or not
string isAMultipleOf4(int n)
{
   
    // if true, then 'n' is a multiple of 4
    if ((n & 3) == 0)
        return "Yes";
 
    // else 'n' is not a multiple of 4
    return "No";
}
 
// Driver program to test above
int main()
{
    int n = 16;
    cout << isAMultipleOf4(n);
    return 0;
}

Java

// Java implementation to efficiently check
// whether n is a multiple of 4 or not
 
class GFG {
    // method to check whether 'n' is
    // a multiple of 4 or not
    static boolean isAMultipleOf4(int n)
    {
        // if true, then 'n' is a multiple of 4
        if ((n & 3) == 0)
            return true;
 
        // else 'n' is not a multiple of 4
        return false;
    }
 
    // Driver method
    public static void main(String[] args)
    {
        int n = 16;
        System.out.println(isAMultipleOf4(n) ? "Yes" : "No");
    }
}

Python 3

# Python 3 implementation to
# efficiently check whether n
# is a multiple of 4 or not
 
# function to check whether 'n'
# is a multiple of 4 or not
def isAMultipleOf4(n):
 
    # if true, then 'n' is
    # a multiple of 4
    if ((n & 3) == 0):
        return "Yes"
 
    # else 'n' is not a
    # multiple of 4
    return "No"
 
# Driver Code
if __name__ == "__main__":
 
    n = 16
    print (isAMultipleOf4(n))
 
# This code is contributed
# by ChitraNayal

C#

// C# implementation to efficiently check
// whether n is a multiple of 4 or not
using System;
 
class GFG {
     
    // method to check whether 'n' is
    // a multiple of 4 or not
    static bool isAMultipleOf4(int n)
    {
        // if true, then 'n' is a multiple of 4
        if ((n & 3) == 0)
            return true;
 
        // else 'n' is not a multiple of 4
        return false;
    }
 
    // Driver method
    public static void Main()
    {
        int n = 16;
        Console.WriteLine(isAMultipleOf4(n) ? "Yes" : "No");
    }
}
 
// This code is contributed by vt_m.

Javascript

 
// Javascript implementation to efficiently check
// whether n is a multiple of 4 or not
     
    // method to check whether 'n' is
    // a multiple of 4 or not
    function isAMultipleOf4(n)
    {
        // if true, then 'n' is a multiple of 4
        if ((n & 3) == 0)
            return true;
   
        // else 'n' is not a multiple of 4
        return false;
    }
     
    // Driver method
    let n = 16;
    document.write(isAMultipleOf4(n) ? "Yes" : "No");
     
    //This code is contributed by rag2127
     

输出：

Yes

我们可以概括以上解决方案吗?
同样，我们可以检查2的其他幂。例如，如果n＆7为0，则数字n将是8的倍数。通常可以说。

// x must be a power of 2 for below
// logic to work
if (n & (x -1) == 0)
   n is a multiple of x
Else
   n is NOT a multiple of x

https://youtu.be/ke