The problems are algorithmic problems with some financial terms(which will be well explained) and a bit of maths. Don't know about the t-shirts.

This reminds me a story about Feynman (the physicist), he was taking a psychological examination. This is an excerpt from it:

Then at some point near the end he says, “How much do you value life?”

“Sixty‑four.”

“Why did you say ‘sixty‑four’?”

“How are you supposed to measure the value of life?”

“No! I mean, why did you say ‘sixty‑four,’ and not ‘seventy‑three,’ for instance?”

“If I had said ‘seventy‑three,’ you would have asked me the same question!”

It is a company challenge, hence not rated, I think.

No

Yeah, I decided to not solve that problem and try to learn with the other problems.

That's the evilest thing I've ever seen in my life T.T

I actually think it was a good exercise in problem statement decryption. You don't always get a perfectly explained problem or a simple statement — it was good practice in my opinion :)

It seems it can change even during contest

Yes, I agree and it is not fair at all. They didn't even announce anywhere that they changed the scoring system. It seriously hampered the rank list, mostly the top 10. HackerRank should revert the change and explain what is going on.

I totally agree. Doing such things moves an online judge/community from the "serious" category to "not serious", I think.

What are you even talking about?

Regardless of the scoring strategy that you had in mind before the contest (and I think it's far from clear that all the long contests use the last submit time), the rule that was in place at the time the would-be winners were determining their strategy was sum-of-times. This is a sponsored contest with large monetary prizes and it was over in 3-4 hours (as to who wins these prizes). Then, after the contest (effectively), you went and changed the rules to what they had been supposed to be in your heads? Nobody cares about what that may have been!

Now the only thing you can do to partially salvage your reputation is to change it back and issue an apologetic clarification, IMHO.

And now other 3 people are affected by bug in scoring system. I think the most fair decision will be to split prizes between 6 people, or ignore time at all and split it between everyone who solved all problems, or even don't give any prizes in this contest.

If you enjoy solving problems and can't solve all of them within 2/3 hours... I think it is better than solving problems of contests which already ended?

How did you solve Trade analysis?

Generate all possible valid subsets consisting of a's and b's of length n/2. There can be at most n/2 choose n/4 distinct valid subsets. Now for each subset count how many ways they occur as two distinct subsequence in the string with DP.

Don't know if this is the intended solution or not but that's the way I solved it.

source code

code

I solved it with FFT too. http://pastebin.com/Spk2vZRb

But It's definitely an overkill.

Another solution using a polynomial:

Let f(x) = (1+x*a1)...(1+x*an). It's well known that the sum of a polynomial's coefficients is f(1). However, we have this weighted by |T| thing. So, let's consider f'(x) instead. Notice that the coefficient of x^(t-1) is equal to t*(sum of products of t things from a). So, f'(1) is our answer. Also, note that f'(x) can be written as sum of a_i * f(x) / (1+x*a_i). So, f'(1) can be found pretty easily from this sum. For example, see this code here: http://pastebin.com/JnN4yWCf

Sort the elements by P*(100-C). Now if we pick M elements, then it's optimal to make sure that the first K elements can be sold. Use heap to build two arrays prefix[] and suffix[], where prefix[i] = sum of K largest Price*100 in the first i elements, and suffix[i] = sum of (M-K) largest Price*Prob in the last (N-i+1) elements. Finally answer = max(prefix[i] + suffix[i+1]).

Code: http://pastebin.com/uFwdA9Q3

I only use backtracking with meet in the middle principle. Backtrack for the first 2^(N/2) possible configurations, save the difference of them in an array count[]. By difference I mean the remaining characters if we trim the similar prefix. Each configuration can be represented by a binary mask smaller than 2^(N/2), which is small. Do the same brute force backward, now use count[] to re-calculate the answer.

Code: http://pastebin.com/akXMX9Dc