C++程序不使用*，/，+，-，％运算符将数字除以3

对于给定的正数，我们需要将数字除以3，而不使用任何这些*，/，+，-％运算符

例子：

Input : 48 
Output : 16

Input : 16  
Output : 5 

算法

1. 取一个数字num，sum = 0
2. while(num> 3)，将数字左移两位，并且sum = add(num >> 2，sum)。创建一个函数add(x，y)以按位运算符将两个数相加
   • 取x，y
   • 运行do while循环，直到a不为零
   • a = x＆y，b = x ^ y ;, y = b;
   • 在终止循环返回值b之后，得出x和y的和
3. 取num与3的按位与
   对于新的num = add(num >> 2，num＆3)
4. 在终止while循环后，打印得出最终结果的和值。

For example if your number is 10 then convert to binary 10 => 00001010
If 10 > 3 then shift the binary number 2 bits, Now num will be 00000010 i.e, 2
Now sum will be 2
num = (shift the number by 2 bits) + (number BITWISE AND 3)
num = 2+2
Now the number will be 4 then, 4 > 3 => true so loop will be repeated
4 => 00000100 then shift the binary number 2 bits
Now sum = 2 + 1 ie, sum = 3 (this value is returned)
num = (shift the number(00000100) by 2 bits) + (number BITWISE AND 3)
num = 1 + 0
ie remainder = 1

// C++ program for divided a number by
// 3 without using operator
#include 
using namespace std;
  
// A function to add 2 numbers without using +
int add(int x, int y)
{
    int a, b;
    do {
        a = x & y;
        b = x ^ y;
        x = a << 1;
        y = b;
    } while (a);
  
    return b;
}
  
// Function to divide by 3
int divideby3(int num)
{
    int sum = 0;
  
    while (num > 3) {
        sum = add(num >> 2, sum);
        num = add(num >> 2, num & 3);
    }
  
    if (num == 3)
        sum = add(sum, 1);
  
    return sum;
}
  
// Driver program
int main(void)
{
    int num = 48;
    cout << "The number divided by 3 is ";
    cout << divideby3(num);
  
    return 0;
}

输出：

The number divided by 3 is 16