Facebook Hacker Cup 2017 was just announced!

Quoting:

We're excited to announce the dates for the online rounds of the 2017 Facebook Hacker Cup!

- Qualification Round (72 hours): January 6, 2017 4pm PST - January 9, 2017 4pm PST
- Round 1 (24 hours): January 14, 2017 10am PST - January 15, 2017 10am PST
- Round 2 (3 hours): January 21, 2017 10am PST - 1pm PST
- Round 3 (3 hours): January 28, 2017 10am PST - 1pm PST

The exact date and location for the onsite finals are still being worked out, but we expect the finals to take place in April or May. Happy hacking!

Just noticed a bug in virtual participation in gym's SWERC 2016.

For some reason, their submissions to problems B and E are double counted with different times.

Hi all,

SWERC 2016's problems, solutions and I/O are now available on the contest website.

Any feedback is welcome.

The ACM ICPC world finals programming environment has been updated with Python specifications. Both Python 2 and 3 will be available with modules provided in that page.

There's no information about grading in Python. In a previous Codeforces blog post, it is mentioned that it's not guaranteed the problems are solvable in Python.

So, does anyone intend to use Python?

Also, do you think it would help bringing more contestants to ICPC regionals?

The round 1 of Facebook Hacker Cup 2016 starts in a little over 4 hours at 10am PST.

Everyone who solved at least one problem in the qualification round (CF post) is eligible to compete. Click here to enter Round 1.

Note that advancement criteria to round 2 was changed. The criteria now is to achieve at least 30 points in Round 1, instead of same points as 500th. These changes were published in the competition FAQ.

Hi everyone,

I saw this question on Quora about Problem E — Emoticons from Dhaka Preliminary Round 2015 (problemset). The statement is as follows:

Given a string S (|S| <= 10⁵) of characters '^' or '_', find the number of disjoint '^_^' subsequences. Disjoint means that we can't use the same character in 2 subsequences. For example, if S is ^^__^^ then the answer is 2. However, for S equal to _^^ the answer is 0.

My approach was to process the string left to right and keep count of the number of sequences "^" and "^_" seen so far. When we see a "^", then we can form "^_^" if there is any "^_" subsequence before or keep it as an unmatched "^" to form a "^_" subsequence later.

I think the intended solution is a greedy algorithm, but I couldn't find a greedy approach that works in all cases. Any thoughts?

Sample input (number of tests <= 5000 but number of characters in a input file <= 2.1 * 10⁶)

7
_^^_^^_
^__^__^
______
^^__^^
^_^^__^
^_^_^^
^_^_^_^^^^_^_^

Sample output

1
1
0
2
2
2
4

Hi everyone,

The ACM ICPC Southwestern European Regional Contest will be held this weekend in Porto, Portugal. Who's competing this year?

I'll be one of the judges, can't wait :)

Cheers