Every time increase $$$d$$$ by $$$k$$$ and use a second operation, it will be end in at most $$$O(\sqrt{\dfrac zk})$$$ times.

If you choose to increase $$$d$$$ by $$$i$$$ and always choose the second operation from then on, the additional cost is $$$ix+\lceil\dfrac z{d+i}\rceil y$$$ . You can find that there are at most $$$O(\sqrt z)$$$ different $$$\lceil\dfrac z{d+i}\rceil$$$ and always use the smallest $$$i$$$ with the same $$$\lceil\dfrac z{d+i}\rceil$$$ will be better. So now is $$$O(\sqrt{\dfrac zk}\min(\sqrt z,k))$$$ .

If $$$\sqrt z\le k$$$ , then $$$O(\sqrt{\dfrac zk}\cdot\sqrt z)=O(\dfrac z{\sqrt k})\le O(z^{\frac 34})$$$ ;

If $$$\sqrt z \gt k$$$ ,then $$$O(\sqrt{\dfrac zk}\cdot k)=O(\sqrt{zk}) \lt O(z^{\frac 34})$$$ .

However, E can be solve in $$$O(z^{3/4})$$$ , only by using some optimizations in brute force. My Code