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By Edvard, history, 3 years ago, translation, , 678A - Johny Likes Numbers

The problem was suggested by Abdrakhman Ismail Ismail_A.

We should find minimal x, so x·k > n. Easy to see that . To learn more about floor/ceil functions I reccomend the book of authors Graham, Knuth, Patashnik "Concrete Mathematics". There is a chapter there about that functions and their properties.

С++ solution

Complexity: O(1).

678B - The Same Calendar

The problem was suggested by Arthur Jaworski KingArthur.

Two calendars are same if and only if they have the same number of days and starts with the same day of a week. So we should simply iterate over years and maintain the day of a week of January, 1st (for example). Easy to see that the day of a week increases by one each year except of the leap years, when it increases by two.

C++ solution

Complexity: O(1) — easy to see that we will not iterate more than some small fixed constant times.

678C - Joty and Chocolate

The problem was suggested by Sheikh Monir skmonir.

Easy to see that we can paint with both colours only tiles with the numbers multiple of lcm(a, b). Obviously that tiles should be painted with more expensive colour. So the answer equals to .

C++ solution

Complexity: O(log(max(a, b))).

678D - Iterated Linear Function

The problem was suggested by Zi Song Yeoh zscoder.

The problem can be solved using closed formula: it's need to calculate the sum of geometric progression. The formula can be calculated using binary exponentiation.

I'll describe more complicated solution, but it's more general. If we have a set of variables and at each step all variables are recalculating from each other using linear function, we can use binary matrix exponentiation. There is only one variable x in our problem. The new variable x' is calculating using formula A·x + B. Consider the matrix z = [[A, B], [0, 1]] and the vector v = [0, 1]. Let's multiply z and v. Easy to see that we will get the vector v' = [x', 1]. So to make n iterations we should multiply z and v n times. We can do that using binary matrix exponentiation, because matrix multiplication is associative.

As an exercise try to write down the matrix for the Fibonacci numbers and calculate the n-th Fibonacci number in O(logn) time. The matrix and the vector is under the spoiler.

The matrix and the vector for the Fibonacci numbers
C++ solution

Complexity: O(logn).

678E - Another Sith Tournament

The problem was suggested and prepared by Alexey Dergunov dalex.

Let's solve the problem using dynamic programming. zmask, i — the maximal probability of Ivans victory if the siths from the mask already fought and the i-th sith left alive. To calculate that DP we should iterate over the next sith (he will fight against the i-th sith): .

C++ solution

Time complexity: O(2nn2).

Memory complexity: O(2nn).

678F - Lena and Queries

The problem was suggested by AmirMohammad Dehghan amd.

Let's interpret the problem geometrically: the pairs from the set are the lines and the problem to find to topmost intersection of the vertical line with the lines from the set.

Let's split the queries to blocks. Consider the lines added before the current block and that will not deleted in the current block. Let's build the lower envelope by that lines. Now to calculate the answer to the query we should get maximum over the lines from the envelope and the lines from the block before the current query that is not deleted yet. There are no more than lines from the block, so we can iterate over them. Let's find the answers from the envelope for all queries of the third type from the block at once: we should sort them and iterate over envelope using two pointers technique.

C++ solution

Complexity: . Tutorial of Educational Codeforces Round 13  Comments (32)
 » Please write the editorial for E
 » About problem E: After some interesting discussion, I was wondering what exactly can we say about the optimal solution?Let's assume that optimal sequence can be represented by a sequence a1, a2, ..., an of fighters in which a1 fights a2, then the winner fights a3, etc. Intuitively, the probability someone wins is higher the more to the right of the sequence it is: this implies an should be Ivan in an optimal sequence.Now who do we want at the second to last position? That person will be given a greater chance to fight Ivan, so we want it to be the one that has the smallest chance to win against him. Generalizing to all n players, we have the following O(n2) greedy solution: construct the sequence of fighters from the last to the first. For each pick, pick the available fighter that makes probability of Ivan winning in current sequence highest possible.Code: 18489412Of course, there's no such thing as "proof by AC", so I'm still curious if this solution is actually correct. If it is, I think it is quite interesting :)
•  » » Some history: the setting was taken from Timus 1218. In 1218, we discovered a NlogN solution and included that problem into our contest at Petrozavodsk camp, where 7 teams out of 50 solved it.As my teammates just said, it became a tradition: all "Jedi Tournament" problems can be solved better than author initially thought. You could have done the same with your solution.
•  » » Actually I've discussed with dalex that. And I thought about the same thing. But both of us agreed that greedy solution is incorrect here.
•  » » » If you discussed this before the contest, shouldn't you have included a test that fails the greedy solution?I'm also curious to know how you reached this conclusion.
•  » » » » I thought about different greedy solution. This one passes random tests and must be correct.
•  » » » » Oh you misunderstood. We have not a countertest, we just thought that greedy solution is incorrect.
 » Another solution for problem F in O(N*log2N).
•  » » Could u please go for more detail about it?? I have difficulty in reading and understanding ur code :P
•  » » »
•  » » Do you use the technique with storing the blocks of the sizes 2k?
•  » » » Problem to find the topmost intersection of the vertical line with the lines from the set can be solved with Convex Hull Trick.I builded a segment tree of CHT over time interval. For ith insertion query update the segment tree with this line in range [ i, time till ith line exists ]. For every query of 3rd type it is equivalent to point query at that point of time.
•  » » » » Nice. But I think it's more complicated.
•  » » » » » During contest I tried with the same approach as mentioned in editorial (Submission). But it timed out because my complexity was O(N1.5 * logN). So I had to go this way but it seems that logN factor can be removed.
•  » » 3 years ago, # ^ | ← Rev. 2 →   I solved the problem using your idea but with complexity O(N*logN). Submission here: 18842789
•  » » » Insertion in logN nodes is log2N, right ?P.S: Your N*logN is slower than mine N*log2N.
•  » » » » 3 years ago, # ^ | ← Rev. 4 →   Actually if you insert the lines in sorted order the overall complexity for insertion is because you insert in nodes in total linear time. This was marX and I did (using your idea, btw) 18845711. You probably didn't look deeper into marX code.
 » Can anyone explain a little more about the recurrence relation in E?
•  » »
 » Can someone explain the editorials for Problem E in a somewhat detail manner, it would be a great help.
•  » » 6 weeks ago, # ^ | ← Rev. 2 →   $z(mask, i)$ is not the probability that the $i$-th sith wins. Remember that we are calculating the probability of the $0$-th sith being the ultimate winner. It simply says that all siths in the mask have fought some match and the only sith alive out of those currently is the $i$-th sith. Ivan corresponds to the $0$th bit. There are $2$ cases:$1$. $0$-th bit is $0$, which means Ivan hasn't fought yet.$2$. $0$-th bit is $1$. In this case, if $i \ne 0$, this means that Ivans fought but was killed. So, the answer for this state is $0$. Otherwise, Ivan is the survivor from all matches conducted so far and we proceed with the tournament.