Yes

Why am I getting TLE for problem A for an O(1) solution.

Why is it so hard ?

Who else solved problem A (Shuffled Anagrams) using max flow?

Dude what the heck was that difficulty curve?? For me, B was the hardest problem.

Solution for the problem D "Increasing Sequence Card Game" test set 3 (N > 10^6): https://www.quora.com/What-is-the-sum-of-the-series-1+-1-2-+-1-3-+-1-4-+-1-5-up-to-infinity-How-can-it-be-calculated

It was by far the easiest problem of them all. Which awarded the largest amount of points.

I tried doing problem C(Palindromic Crossword) using some dfs kind algorithm(I stored nearest blockages for each row and column) and kept getting runtime error, can someone point out the mistake in my logic? Thanks link: https://pastebin.com/Xx3KHbGU

I have a much different solution to $$$D$$$ that allows for a more general statement that makes the problem slightly more interesting (and definitely harder).

Using one-indexing, if we try to find the number of times card $$$i$$$ is at position $$$j$$$ and ends up in the final hand, we find that this number is $$${i-1 \choose j-1}(j-1)! (N-j)!$$$. In short, this is choosing $$$j$$$ cards lower than $$$i$$$ and placing them in front of position $$$j$$$, then permuting everything but card $$$i$$$ and position $$$j$$$. To find the total number of times a single card $$$i$$$ is in the final hand, sum over all positions $$$j$$$ and expand. This nets you $$$\sum_{j \le i} \frac{(i-1)!}{(i-j)!(j-1)!} (j-1)! (N-j)!$$$. Simplify to get $$$(i-1)!\sum_{j \le i} \frac{(N-j)!}{(i-j)!}$$$. Notice that the subtraction of numerator and denominator removes $$$j$$$ from the summation, so we can try turning the fraction into a binomial coefficient. $$$(i-1)! (N-i)! \sum_{j \le i} \frac{(N-j)!}{(i-j)! (N-i)!} \rightarrow (i-1)! (N-i)! \sum_{j \le i} {N-j \choose i-j} \rightarrow (i-1)! (N-i)! {N \choose i-1}$$$ where the last transformation is by a binomial identity. Simplifying this expression trivially we, we find that the contribution is exactly $$$\frac{N!}{N-i+1}$$$, but the entire expression will be divided by $$$N!$$$ so the contribution to the expected value of card $$$i$$$ is exactly $$$\frac{1}{N-i+1}$$$

The problem could have given a set $$$S$$$ and asked to find the expected value of the intersection between the final hand and $$$S$$$ or, if you want to keep the $$$H_n$$$ approximation, you can ask for the intersection between the final hand and all cards not in $$$S$$$. Even so, it's easily bruteforceable if one knows about the linearity of expectation, but the statement no longer screams to be brute forced.

for last problem Why I was getting WA for 3rd test case when I was trying to solve this recurrence $$$f(n)=f(n-1)+1/n$$$ by matrix exponentiation.

for me A was harder than D.

For D problem, I found a solution easier than analysis solution. Suppose that we found answer for n-1 (base case is n=1 which is 1). Think all of the permutations from n-1, increase every number with 1. Now, we should just insert 1 into somewhere into them, there are two cases: 1. Insert 1 at the beginning, it gives us (n-1)!, because there are (n-1)! permutations. 2. Insert 1 at anywhere expect the beginning, it won't be included in any increasing subsequence. So it gives us for every answer from n-1, (n-1) times of it + from the first case there is 1 more f(n-1) too. So f(n)/n! = (f(n-1) / (n-1)!) * n + (n-1)! / n! = f(n-1) + 1/n. Then for n <= 1e6, brute force, otherwise use formula for harmonic sums, which is f(n) = ln(n + 1) + (Euler–Mascheroni constant =~ 0.5772). (edit: fixed plagiarism)

It was a nice round! I secured rank 1085 in it

Is C some form of a well-known problem ? I noticed a lot of people do C but fail at D.

For me is a wonderful contest experience, although terrible at first.

At first I failed both problem A and B from time to time, and cannot pass a single test set in the first 1 hour 10 minutes. You can imagine how desperate I was when contest pass almost half and my score was zero!

Then I pass the small test set for problem B, and then problem C (A typical DFS problem), my ranking rise from 2000+ to 600+, then I pass problem D due to my clever observation, and my current ranking rise to 250+.

Finally I look back on problem A, first I use brute force to pass small test set to guarantee 4 points, then I pass large test set 8 minutes till the end of the contest.

My final score is 87/100, with the ranking 205, which is the highest ranking of all times. So guys, never give up until last minute in competitive programming.

For anyone interested, here's a link to my screencast. I've also recorded solution explanations at this link.