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，则后两位未设置，否则将两者或其中之一置位。

C/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.

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