Clarifications

No! C++ is the best!

Nobody has mentioned that before(?) This round is just an ordinary one because I can exactly solve problem G as usual.

Yeah same, I tend to quit if the problem is about expected value of something

How delighted I was when I solved D, how frustrated I was when I saw E

$$$F$$$'s expected value was just sum of probabilities for each node. But the main work (in my solution) was how to create a DSU variant to build a tree which can be used the propagate (add) the final answer for each node.

Yes, they should be banned from AtCoder

The problem is solved by considering global operations separately and checking the last update when modifying individual characters.

Regarding global operations, it is sufficient to record the time and type of the last operation.

Finally, if the individual update is made after the global update, it is printed directly; otherwise, the last global update is applied to the individual update before printing.

among the operations of type 2 and 3 only the last operation matters

no lol the problem is simply convex, that is why

I'm not sure if it would help but I'm writing my way of thought anyways. Keep in mind that it is not really formal, I suggest your refer to the editorial for more formality.

Let $$$E_i$$$ be the expected number of money needed to get $$$i$$$ points when Takahashi adopts the optimal strategy.

At first Takahashi has $$$0$$$ points. So there are $$$M$$$ points he needs (at least), so the expected number of money needed is $$$E_M$$$. What's left is to compute $$$E$$$.

Say Takahashi has $$$0$$$ points (i.e needs $$$M$$$ points). At that point and based on his strategy he will choose some $$$k\in\{1, \dots , N\}$$$ and spin the $$$k$$$-th wheel. Doing that, he will pay $$$C_k$$$ (for sure) and will gain $$$S_{k_j}$$$ points ($$$1\leq j\leq P_k$$$) with probability $$$\frac{1}{P_k}$$$. Then he would have $$$S_{k_j}$$$ points, so he will need $$$M - S_{k_j}$$$ more points. That implies that he will then need an average of $$$E_{M - S_{k_j}}$$$ money. So the expected number of money when having $$$0$$$ points ($$$M$$$ points remaining) is

(keep in mind that $$$E_i = 0$$$ for $$$i \leq 0$$$, but you can just have $$$E_0=0$$$ and always take the index $$$i$$$ as $$$\max\{0, i\}$$$)

Similarly to $$$E_M$$$ we can compute all of $$$E$$$. This yields a $$$O(MNP)$$$ solution ($$$P=\max_i P_i$$$), which is basically the editorial's solution looking at it slightly differently. I think understanding one solution will make one easily understand the other.