Editorial

For $x\le y$ notice that $||x-y|-y|=|y-x-y|=x$, which means that after two operations $x$ hasn't changed. In such case, the answer is $\{\begin{array}{ll}x& \text{if }k\text{ is even},\\ y-x& \text{if }k\text{ is odd}.\end{array}$

For $x>y$ we may reduce the problem by subtracting $y$ from $x$ a fixed number of times.