Good evening Codeforces, let me briefly describe solutions for all problems of today's morning yandex algorithm round. Thanks everyone who participated, I apologize for making problems a bit too tough and viscous for a 100 minutes contest. Anyway, I hope everyone found something interesting.

I would like to thank lperovskaya for organising this competition and managing Yandex.contest, snarknews and SergeiFedorov for their help with the problemset, Endagorion, PavelKunyavskiy, AleX, glebushka98, gustokashin, map and boris for testing. Special thanks to Gassa and my girlfriend Marina Kruglikova for fixing mistakes and disambiguations in English and Russian statements.

Let's get it started.

Problem A. Odysseus Sails Home.

There is no tricky idea behind this problem: one just needs to check if the vector (x_f - x_s, y_f - y_s) can be represented as a convex combination of vectors (x_i, y_i). One of the easiest approaches for the general case is to try all pairs of wind vectors and check if the target vector lies inside the cone they form. However, the devil is in the details. One shouldn't forget to:

• Check if it's possible to get to Ithaca using only one wind direction;

• Special if for case (x_f, y_f) = (x_s, y_s);

• Ignore wind vectors (x_i, y_i) = (0, 0);

• Avoid usage of doubles — everything fits in long long.

Time complexity: O(n²). O(n) solution also exists.

Problem B. Chariot Racing.

For the fixed value t = const we can easily calculate the intersection of all segments (chariots) as max(0, min(b_i + v_i * t) - max(a_i + v_i * t)). The problem was to find maximum for t ≥ 0.

Both min(b_i + v_i * t) and max(a_i + v_i * t) are convex functions. min(b_i + v_i * t) is concave upward, because it's derivative only decreases, as faster chariots overtake slower one. Similar max(a_i + v_i * t) is convex down. This means function min(b_i + v_i * t) - max(a_i + v_i * t) is concave upward, and this, in turn is sufficient condition for applying ternary search.

Ternary search is enough to solve the problem, but the solution which does binary search on derivative is faster and more stable. We need to find the maximum t such that the chariot i with minimum b_i + v_i * t goes faster then the chariot j with maximum a_j + v_j * t.

The only special case (for some solutions only) was n = 1.

Time complexity: O(n·logMaxC).

Problem C. Equality and Roads.

First check if the graph is connected. If no, print <<\t{NO}>> for all queries.

For connected graph count the number of connected components if only 0-edges are allowed (denote it as a) and the number of connected components if only 1-edges are allowed (denote it as b). Then, for the given x it's possible to construct the desired span if and only if condition b - 1 ≤ x ≤ n - a holds.

Lets proof the above statement. It's pretty obvious that this condition is necessary, but the sufficiensy isn't that clear. In my opinion, the easiest way is to present the algorithm. It consists of five steps:

1. Create DSU, add all 1-edges.

2. Add all 0-edges, remember which of them caused joins. There will be b - 1 such edges.

3. Clear the DSU, add all 0-edges remembered on step 2.

4. Add 0-edges until there are exactly x of them in the current span.

5. Add 1-edges until the graphs becomes connected. That will always happen because of the step 1 and 2.

Time complexity: O(n).

Problem D. Sequences of Triangles.

Thanks to snarknews — the author and developer of this problem.

Let f(x) be the longest sequence that ends on the triangle with the hypotenuse of length x. If we generate all Pithagorean triples with a, b, c ≤ L the dynamic programming approach will be the one that is easy to implement here.

Exhaustive description of the generation process and Euclid's formula could be found here, I would like to skip copying it to analysis.

Problem E. Strong Squad.

Thanks to SergeiFedorov — the author of this problem.

Perimeter p is equal to 2·(h + w), where h is the number of rows and w is the number of columns in the resulting rectangle. Build a bipartite graph on rows and columns where there is an edge connecting row x to column y if and only if soldier(x, y) = 0. What we should find now is the maximum independent set, such that in both set of rows and set of columns there is a least one vertex chosen (we are not allowed to choose rectangles 0 × m or n × 0).

The [well-known fact](https://en.wikipedia.org/wiki/K%C3%B6nig%27s_theorem_(graph_theory)) (yes, just say that the fact is well-known if you don't want to proove it) is that the size of the maximum independent subset in bipartite graph is equal to |V| + |U| - ν(G), where ν(G) stands for the maximum matching. To meet the condition that there should be at least one vertex chosen in both parts we should:

1. Try all the available pairs;

2. Exclude vertices already connected to chosen pair;

3. Find maximum matching.

Time complexity: O((nm)²·min(n, m)) or O(n⁵) for the worst case n = m. Is reduced to O(n^4.5) by using Dinic's algorithm.

About Time Limit: though for n = m = 100 the number of operations will be about C·10¹⁰, C seems to be pretty small. We can show that comes from the fact that the worst case is then only half of the cells is filled with zeroes. Also at least comes from the Kuhn's algorithm itself. The actual timing for the most straightforward Kuhn was about 2.5 from 5 seconds TL. Any optimisations sped it up to less than 1 second.

Problem F. Lexicographically Smallest String.

We want to pick up two indices and revert corresponding substring to make the resulting string as small as possible. To start with let's reduce a degree of freedom from 2 to 1.

Statement 1. If we represent string S as S = c + S' and character c is not greater than any character in S', than Answer(S) = c + Answer(S'). Otherwise, the answer always reverts some prefix.

Proof. Assume Answer(S) < c + Answer(S'), that means we should revert some prefix of S, i.e. some substring (1, i). If we revert substring (2, i), the length of the prefix that consists of only c will increase, and the result will become lexicographically smaller. Assumption is incorrect. If c is greater than some character os S' it's obvious we should revert some prefix that ends at position with the smallest possible character.

Now we need to solve the following task: revert some prefix of the string S to make it as small as possible. First, we need to take a look at string S^r and it's suffixes. We will call two suffixes strongly comparable if none of them is a prefix of another. Similary, we will say that one suffix is strongly lesser or strongly greater than another, if they are strongly comparable and the corresponding inequality holds for classiс comparison.

Statement 2. If S = A + B = C + D and A^r is strictly lesser than C^r than A^r + B is stricly lesser than C^r + D. Proof: obvious.

From statement 2 it follows that we should only consider suffixes of S^r that are greater than any other suffix they are strictly comparable to. From definition of strict comparison it comes out all the remaining suffixes are prefixes of the longest one remaining. We can try to revert them all and choose the best, but this may be too slow.

From the string theory and prefix-function algorithm we know, that if the longest suffix of the string S' that is equal to it's prefix has length l than S' = w + w + ... + w + w', where w' is prefix of w and w has length n - l.

Statement 3. If we present the longest suffix S' that is not majorated by any other suffix in the above form, than the set of other not majorated suffixes will be equal to the set of suffixes of S' that have the form w + w + ... + w + w'.

Proof. If the string w is not prime, than there exists some suffix of S' of length more than n - |w| that is not striclty lesser than S', but this contradicts to the way we found w. That means the string w is prime and only suffixes of form w + w + ... + w + w' are not strictly comparable to S'.

Statement 4. If |A| = |B| then A + A + C < A + C + B < C + B + B or A + A + C > A + C + B > C + B + B or A + A + C = A + C + B = C + B + B for any string C.

Proof. Compare strings A + C and C + B. The result of this comparison will determine the case, as we could always change A + C to C + B or vice versa.

Applying the statement 4 to the case where C = w', A = w, B = w^r we conclude that we should only try to revert the longest and the shorest prefix of S such that corresponding suffixes of S^r are not strongly greater than any other suffix of S^r.

To find those suffixes of S^r one should perform Duval's algorithm to find Lindon decomposition of the string S^r. We are interested in the last prime string in the decomposition and the last pre-prime string in algorithm's workflow (or the last prime for the string S + #, where # is some character striclty smaller than any other character in S as was mentioned in comments by Al.Cash.

Time complexity: O(|S|).

Доброго времени суток, уважаемые пользователи Codeforces!

Хочу ещё раз обратить внимание всех потенциально заинтересованных участников на такое мероприятие как Открытая всероссийская олимпиада школьников. Напоминаю я не просто так, а по отличному поводу — сегодня там появились три новые задачи, а в ближайшие дни ожидаются ещё две. Задачи подготовлены жюри московских олимпиад, мы честно старались сделать их интересными и разнообразными.

Если кто-то пропустил начало олимпиады, то в этом нет ничего страшного, до конца ещё целый месяц, вполне достаточно, чтобы сдать все предложенные задачи. Время сдачи никак не влияет на положение в таблице. Лучшие X участников будут приглашены поучаствовать в очном финале, для которого задачи отбираются с ещё большей любовью и заботой, так что не упустите свой шанс.

Для тех, кто руководствуется более приземлёнными соображениями, чем качество задач, дополнительно сообщаю, что заключительный этап традиционно имеет первый уровень, а значит призеры и победители могут получить льготы при поступлении в вуз в соответствии с порядком приема в вузы.

Призеры отборочного этапа олимпиады никаких льгот не имеют — поступить не выходя из дома всё-таки не получится.

Более подробную информацию об олимпиаде вы можете найти по приведённой выше ссылке. Любые оставшиеся вопросы смело задавайте в комментариях.

P.S. Тех, кто будет в комментариях обсуждать решения задач, ждёт справедливое возмездие и отрицательная карма!

UPD: добавлены две новые задачи, больше задачи добавляться не будут.

UPD2: Почти наверное традиционного продления олимпиады не будет. Планируется пригласить порядка 400 участников. Обратите внимание, сейчас для попадания в топ-400 требуется набрать совсем немного баллов. Жюри рекомендует прочитать все задачи — почти в каждой имеются простые группы с немалым количеством баллов.

In the clarification during just finished 222 round it has been announced that this round is going to be rated for both divisions, however, div1 coders are able only to register for "out of competition" participation. Is it a mistake or the concept has changed?

It'll be a little unfair to make the last round of the old year div2-only :)

Hello everybody!

Today's round will be held by SergeiFedorov and me. We have done our best to make it diverse (different task complexity and themes) and rated (we know it's important). Your questions, wishes and constructive criticism (destructive we already can do :-)) leave in comments.

The contestants will plunge into the cold February of Nvodske and will be to help one's friends to survive the cold. Just imagine all responsibility!)

On this occasion I would like to thank all Codeforces team for the great job, they are doing, Artem Rakhov (RAD) for his help in task preparations, Maria Belova (Delinur) for problems translation, my mom, my dad, our soundman and Tuscany winemakers, for us didn't fall ill while preparing this round.

Distribution will be something like:

div. 1: 500-500-1000-1500-3000 (yeah, it really costs 3000)

div. 2: 500-1000-1500-1500-3000

Good luck & have fun!