Is it possible to compute $(a+b)\times(c+d)$ with a calculator without memory? I'm implementing a simple calculator which can only do operations sequentially, one at a time.  Expressions of the form:
value1 <operator1> value2 <operator2> value3 <operator3> value4

are interpreted as
((value1 <operator1> value2) <operator2> value3) <operator3> value4

This is just like an old-school calculator, where all you have is the previous answer or value to work with.
This would work fine for something like
$$a \times b + c + d$$
Since I'm not implementing the concept of parentheses, nor operator precedence, nor memory, could an expression such as
$$(a + b) \times (c + d)$$
be expressed with this calculator?
I've tried rewriting this formula into a parenthesis-less form, but I can't seem to find a solution. However, that doesn't mean that a solution doesn't exist.
Furthermore, if the previous expression can be rewritten in the sequential form, then the next question is: can all expressions be rewritten that way?
 A: Assuming none of the numbers are zero, here is one general way to do it. First, expand out all parentheses so that what you have is a sum of products of numbers (possibly with some subtraction and division thrown in the mix). Then repeat the following:

Let $A$ be the sum of all the terms so far, and let $abc$ be the next term. Then you can add that by typing $$A\div b\div c+a\times b\times c$$Hopefully you see how to add a term which is the product of more than three numbers, and also how to deal with subtraction and division.

With this in mind, and with some small optimisation, we calculate $(a+b)(c+d)$ by typing
$$
a+b\times c\div d +a+b\times d
$$
Edit: More detailed general algorithm.
First expand all brackets so that the entire expression is a sum of terms on the following form:
$$
\pm\frac{a_1\cdot a_2\cdots a_{i}}{a_{i+1}+\cdots+a_n}
$$
A single expression of this form is trivial to calculate. If there are more than one, we can add them one by one by the following recursive process:
Say we have already added a number of these terms and gotten the result $A$, and the next term is the one I've written above. Then we can incorporate the new term into our sum by typing
$$
\div a_2\div\cdots\div a_i\times a_{i+1}\times\cdots \times a_n\\\pm a_1\times\cdots \times a_i\div a_{i+1}\div \cdots \div a_n
$$where the $\pm$ is the same sign as in the term above.
Small side note: This method cannot handle denominators with sums or differences in them. I don't know if that's even possible. If someone comes up with a way of calculating, say $\frac1{a+b}$ under these constrictions, please let me know.
