How would you divide a number by 3 without using *
, /
, +
, -
, %
, operators?
The number may be signed or unsigned.
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