algorithm - 在不使用除法、乘法或模数的情况下将数字除以 3

Question

Without using /, % and * operators, write a function to divide a number by 3. itoa() is available.

以上是我在面试中被问到的，我真的想不出答案。我想将数字转换为字符串并添加所有数字，但这只会告诉我数字是否可整除。或者，通过重复减法，它也可以告诉我余数。但是，如何获得除法商？

Answer 1

下面的代码接受 2 个整数，并将第一个除以第二个。它支持负数。

int divide (int a, int b) {
    if (b == 0)
        //throw division by zero error

    //isPos is used to check whether the answer is positive or negative
    int isPos = 1;
    //if the signs are different, the answer will be negative
    if ((a < 0 && b > 0) || (a > 0 && b < 0))
        int isPos = 0;


    a = Math.abs(a);
    b = Math.abs(b);
    int ans = 0;
    while (a >= b) {
        a = a-b;
        ans++;
    }
    if (isPos)
        return 0-ans;
    return ans;
}