When I first read A I thought I only needed to check that the number of plays, clears were non-decreasing. Fortunately during the implementation I realized I also needed plays-clears to be non-decreasing because each clear means one more time played. Took exactly 10 minutes because I double-checked after realizing I was missing a condition at first.

Then in B we can quickly see that if the average of the largest $$$m$$$ elements is $$$x$$$ we can do the desired. Say $$$k$$$ is the largest $$$k$$$ with this property, that is $$$\dfrac{a_1+\dots + a_k}{k} \ge x > \dfrac{a_1+\dots+a_k+a_{k+1}}{k+1}$$$ where the numbers $$$a_i$$$ are sorted in non-increasing order.

We can prove that the answer is $$$k$$$ since by showing any reform preserves that the sum of any $$$k+1$$$ elements is smaller than $$$x(k+1)$$$. The idea is thinking about the largest $$$k+1$$$ elements and checking how they change with a reform, all the numbers that change are replaced by this new average or numbers that weren't previously among the $$$k+1$$$ largest so we can check they have a smaller sum than before. I'll edit in the formal proof tomorrow.

The problem took around $$$10$$$ minutes as well. Fortunately there were no implementation mistakes in either.

As for C, I misread the problem as monsters exploding from both sides which I couldn't solve in the remaining time. The actual problem was monster $$$i$$$ just affecting $$$i+1$$$ so only one explosion damage $$$\min(b[i-1],a[i])$$$ is going to be wasted and we just minimize it.

This one went slightly better. The first two problems of A were very easy, we just needed to create a prefix/suffix and check that we didn't get any contradictions. The third part is not much harder, we ignore the asterisks other than the first and last and glue the insides together but I thought this wouldn't always fit in $$$10^4$$$ and thought I needed to combine the strings.

After a some time I decided to temporarily skip to B. Since the sums of rows of the Pascal triangle are powers of $$$2$$$ it was natural to consider the binary representation of the desired number, however since we need to advance through rows that we aren't interested in we may overcount by a little bit. For a while I was trying to fix this in the beginning of the number, until I eventually realized I could just fix it at the end by traveling through a bunch of $$$1$$$'s since we can overcount by at most 30. Formally I represented $$$N-31$$$ in binary, traveled through $$$1$$$'s until I found a row with sum $$$2^i$$$ appearing in the representation which I completely visited. At the end, if I went through $$$r$$$ rows that I didn't care about I just advance through $$$31-r$$$ $$$1$$$'s.

The implementation took me a lot of time, and I checked my math a lot of times to make sure I was using $$$\le 500$$$ steps. Fortunately everything passed the first time.

I went back to implement the easy parts of A, and since I was still thinking the last part was hard I decided to implement the brute force approach for $$$C$$$, unfortunately I finished a couple minutes after the contest ended.

Although I think I could've solved A on a better day, I ended up in position 2552 which is much better than I expected!