在不使用 * ，/ ，+ ,-,% 操作符的情况下，将一个数除以3

Question

How would you divide a number by 3 without using *, /, +, -, %, operators?

The number may be signed or unsigned.

转载于:https://stackoverflow.com/questions/11694546/divide-a-number-by-3-without-using-operators

Answer 1

This is a simple function which performs the desired operation. But it requires the + operator, so all you have left to do is to add the values with bit-operators:

// replaces the + operator
int add(int x, int y)
{
    while (x) {
        int t = (x & y) << 1;
        y ^= x;
        x = t;
    }
    return y;
}

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; 
}

As Jim commented this works, because:

• n = 4 * a + b
• n / 3 = a + (a + b) / 3

• So sum += a, n = a + b, and iterate

• When a == 0 (n < 4), sum += floor(n / 3); i.e. 1, if n == 3, else 0