Codeforces Round 292 took place late on Tuesday (problems, results, top 5 on the left). Haghani has pulled off an impressive victory given that he was not the rating favorite - but of course he did climb much higher in the ratings as a result, now he's ranked 28th. Congratulations! rng_58, on the other hand, could've probably been the rating favorite if not for his recent strategy of solving the hardest problem first. Once again he became the only person to solve the said hardest problem, but couldn't solve two other problems in time and only placed 10th.

TopCoder SRM 650 started next morning less than 8 hours after the Codeforces round ended (problems, results, top 5 on the left), so big congratulations to Endagorion on winning it in addition to placing 3rd earlier, and for being the only one to solve the hard problem!

There were also four Open Cup rounds in January and February. The problems from those rounds are not published yet, but it's been a month now and it's boring to wait longer, so I will try to explain the problem statements in full below.

Open Cup 2014-15 Grand Prix of Japan took place three weeks ago (results, top 5 on the left). Problem I was, on the surface, a repeat of a problem from Petr Mitrichev Contest 9 back from 2011: it simply asked to count the number of pairs of strings (s, t) such that the length of s is N, the length of t is M, each character is from an alphabet of size A, and t is a substring of s. When this problem was given on my contest, N, M and A were up to 100, and you needed to output a floating-point number - the probability that a random pair has this property. No team could solve this problem during the contest - take a look at problem H in standings. The reference solution made heavy use of the fact that we need to output a fraction and are allowed 10

TopCoder SRM 650 started next morning less than 8 hours after the Codeforces round ended (problems, results, top 5 on the left), so big congratulations to Endagorion on winning it in addition to placing 3rd earlier, and for being the only one to solve the hard problem!

There were also four Open Cup rounds in January and February. The problems from those rounds are not published yet, but it's been a month now and it's boring to wait longer, so I will try to explain the problem statements in full below.

Open Cup 2014-15 Grand Prix of Japan took place three weeks ago (results, top 5 on the left). Problem I was, on the surface, a repeat of a problem from Petr Mitrichev Contest 9 back from 2011: it simply asked to count the number of pairs of strings (s, t) such that the length of s is N, the length of t is M, each character is from an alphabet of size A, and t is a substring of s. When this problem was given on my contest, N, M and A were up to 100, and you needed to output a floating-point number - the probability that a random pair has this property. No team could solve this problem during the contest - take a look at problem H in standings. The reference solution made heavy use of the fact that we need to output a fraction and are allowed 10

^{-9}error, since for longer substrings the probability is almost zero, and for shorter substrings we can iterate over all possible prefix functions of the substring, and when the prefix function is fixed counting superstrings is a relatively straightforward dynamic programming on the Knuth-Morris-Pratt automaton.Fast forward 3.5 years, and now the harder version is given on a contest: now N is up to 200, M is up to 50, A is up to 1000, and you need to find the answer modulo 10

^{9}+7, so you can't squeeze by with approximation arguments anymore. And one team does solve this problem - congratulations to Alex, Slava and Artem!

The key to solving the harder version is to notice that we don't, in fact, need the entire prefix function of the substring to count the number of superstrings containing it: it turns out that we can just look at the set of prefixes of the substring that are also its suffixes, in other words at the set of its borders. Can you see why the borders are enough?

Thanks for reading, and check back next week for the full explanation of this problem's solution, and for a new hard problem!