Hi there!

Here is editorial for the round #78.

Russian editorial will be published a bit later - I supposed that there are more people who can understand English (or who can do English -> Russian web page translation) than vice versa.

Again, I'm very sorry for the situation with problem B (div. 1) / D (div. 2). Also, I'm sorry that I underestimated difficulty of the problems. At least, I hope the problems appeared interesting for many contestants.

Here the problem was to round a number up according to the usual mathematical rules with the exception that if the last digit of integer part is equal to 9, you should output "GOTO Vasilisa.". One may notice that to check whether number's fractional part is not less than 0.5 only one digit just after the decimal point should be analysed. If it is '5' or greater - add one to the last digit of the integer part, and the problem is solved. Probably, the simplest way to deal with the input data was using of the string variables.

The problem was to accurately check what is required in the problem statement. First of all, check whether all volumes in the input are equal. In this case output "Exemplary pages.". Otherwise find two cups with largest and smallest volumes. Suppose their numbers are a and b, and their volumes are v[a] and v[b]. Now suppose that before pouring their volumes were equal to V. Then they contained 2V units of juice before (and after) pouring. So, you need to check whether (v[a] + v[b]) is divisible by 2. If this is not so - output "Unrecoverable confihuration.". Otherwise assign to the cups presumable old volume v[a] = v[b] = (v[a] + v[b])/2. Now if only one pouring have been made, volumes of juice in all cups should be equal, and you print corresponding message "... ml. from ... to ...". If volumes are not equal, print "Unrecoverable configuration" instead.

In this problem you were required to find the number of sufficiently different colorings of a cube faces with predefined six colors. The most trivial solution is to introduce some ordering of the cube faces (say, 0 - front, 1 - back, 2 - up, 3 - down, 4 - left, 5 - right), then consider 720 = 6! arrangements of colors over these 6 faces. Each arrangement is some permutation of characters from the input. For each arrangement we find all its 24 rotations - and get 24 strings. Lexicographically smallest string will be representative of this coloring. The answer is the number of different representatives.

For this problem I assumed numerical solution. But there are several cases to consider. Below without loss of generality we assume a <= b.

1. l <= a <= b. In this case the answer is restricted by the length of the coffin, so the answer is l and it is clear that the coffin l x l can be brought through the corridor (a, b) - let's denote corridor's sizes in this way.

2. a < l <= b. In this case the answer is a, and it is clear that no larger number can be an answer. Indeed, otherwise the coffin (w > a) x (l > a) is impossible to drag through the corridor (a, b).

3. a <= b < l. This is the most general case, where we should rotate the coffin inside the corridor where it has a kink. To maximise the width of the coffin, we want to move it in such a way that one corner of the coffin touches one outer wall of the corridor (suppose bottommost on the picture), and another corner adjacent to the same long side of the coffin touches another outer wall of the corridor (leftmost on the picture). Let's introduce coordinate system in such a way that bottommost wall be OX axis, and leftmost wall - OY axis. Suppose that during the "rotation" process one corner of the coffin is at the point (x,0) (0 <= x <= l), then another corner should be at the point (0,sqrt(l*l-x*x)). And the answer we search for is min {distance from the segment (x,0) - (0,sqrt(l*l-x*x)) to the point (a,b) }, where you take min{} over all 0 <= x <= l. Let this distance at point x be f(x). Since f(x*) is minimal in some point x* and increases everywere to the left and to the right from x*, one may use ternary search to find its minimum.

Exact solution for this problem is also possible: you can reduce the problem to minimizing the dot product of the vectors (a-x,b) and (-x,sqrt(l*l-x*x)) over x. But this leads to the neccessity to find the roots of the fourth-degree polynomial, which is not the best idea during the contest.

This problem was about famous puzzle "Hanoi towers", but diameters of some discs might be equal. How to solve that? A good thing to do is to write BFS solution to check optimality of your ideas for small inputs (by the way, BSF works quickly for almost all towers that have up to 10 discs) and then try to create an algo which solves the puzzle in an optimal way.

Let C (x1, x2, ..., xn) be a solution (under "solution" here we mean optimal number of moves - the moves itself is easy to get with one recursive procedure; also "solution" is the number of moves to move group of discs from one peg to any other (and not some particular) ) to the puzzle when we have a puzzle with x1 equal largest discs, x2 equal second largest discs and so on. And let U (x1, x2, ..., xn) be a solution to the puzzle when you are allowed not to save the order of the discs (you should still follow the restriction of the initial puzzle not to put larger discs onto the smaller ones, but at the end discs of the same diameter may be in any order).

Then one of the optimal solutions to the problem is the following:

C (x1, x2, ..., xn) = U (x1, x2, ..., xn) if x1 = 1 (*)

C (x1, x2, ..., xn) = 2*x1 - 1 if n = 1 (**)

C (x1, x2, ..., xn) = U (x2, ..., xn) + x1 + U (x2, ..., xn) + x1 + C (x2, ..., xn). (***)

U (x1, x2, ..., xn) = U (x2, ..., xn) + x1 + U (x2, ..., xn) (****)

Why so? One can notice that U() is "almost" solution for our problem: it "flips" order of the bottommost group of equal discs, the order of the rest of the discs remains the same! (try to understand why)

That's why (*) is correct.

The (**) is quite obvious.

The (***) does the following: move (x2, ..., xn) from peg 1 to peg 2 without saving the order. Then move x1 equal discs from peg 1 to peg 3, then move (x2, ..., xn) from peg 2 to peg 1 without saving the order (but it occurs that after we apply U() to the same group of discs twice, the order restored!), then move x1 equal discs from peg 3 to peg 2, and then use C() to move (x2, ..., xn) from peg 1 to peg 2 (here we use C() since we should preserve the order). So, (***) is correct.

And (****) is quite straightforward expression for U(): move all discs but the largest group with the same algo, then move largest discs (that's why if x1 > 1, the group of discs "flips"), and then move all discs but the largest group onto the same peg with x1.

This problem was about optimally playing this simple-at-first-glance game. The key thing to recognize in the statement was that it is not always optimal to name card which you don't have. Sometimes it is optimal to confuse the opponent by naming card which you have on hand. In this case... yes, he may think that the card you named is card on the table and lose during the next turn. Now the problem is to understand when to use the strategy of reduction of opponent's cards, when to bluff in the abovementioned sense and when to try to determine which card is on the table. But instead of "when" the right question is "how frequently" since we have nothing else but usual constant-sum matrix game, and optimal strategy is the mixture of these three. Let's construct a matrix first. Player 1 has three pure strategies: "playing" (when he plays the game and really tries to determine opponent's cards and card on the table), "guessing" (when he guesses which card is lying on the table) and "bluffing" (when he tries to confuse his opponent to force him to lose by naming card in his own hand). In turn, if the first player used "bluffing" strategy, or during the "playing" strategy named card on the table, his opponent has two strategies: "check" (i.e. to believe the first player that he doesn't own the card he named and guess it as the card on the table) and "move on" (i.e. to decide that it was a "bluffing" strategy and the game should be continued, but with notice that the first player has named card on hands). Let's denote P(m,n) probability to win the game when the first player has m cards and the second player has n cards. Then P(m,n) is the value of the matrix game with the following matrix (rows - strategies of the first player, two numbers in the rows - probabilities to win when the second player uses strategies "check" and "move on" correspondingly:

                                "check"                                    "move on"

"playing"        n/(n+1)*(1-P(n-1,m))        1/(n+1) + n/(n+1)*(1-P(n-1,m))

"guessing"                 1/(n+1)                                     1/(n+1)

"bluffing"                       1                                       1-P(n,m-1)

How to get these numbers in the matrix? Consider the first row: "playing" strategy of the first player, "check" strategy of the second. First just names one of the n+1 cards. With probability 1/(n+1) he names card on the table, seconds checks it and wins (so, probability to with for the first is 0), with probability n/(n+1) the first names one of the cards on hands of the second player, so the game continues, second wins with prob. P(n-1,m) in this case. Then the overall probability for the first to win with such combination of pure strategies is n/(n+1)*(1-P(n-1,m)). In the same manner we fill other cells of the matrix. Finally we solve the game (this can be done straightforwardly, or with one formula if one notices that the "guessing" strategy is suboptimal everywhere when m>=1 and n>=1 and that the game doesn't have saddle points) and get answer to the problem - P(m,n).

And the last thing to note: when m==0 it is clear that during his move the second wins, so the first should guess, and P(0,n) = 1/(n+1). When n==0 P(m,0)==1 sinse we just do one rightguessing.