How to ask a question in the new GCJ system? Is is possible?

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How to ask a question in the new GCJ system? Is is possible?

Hi!

Last times there was a post about quality of WF problemset. I also agree that they could be better, there were many other contests which consisted of definitely more interesting/better prepared problems. So here comes my question: what is your favorite ACM-style contest?

I'm asking mostly about quality of tasks, maybe there were very interesting? Or maybe something different caught your attention? Of course there is a lot of interesting tasks, but I'm definitely not asking about this kind of contests where there are 9 very easy tasks and then 2 very hard (but so nice) ones.

For example my favorite problemset was on CERC 2017, I think that there were many very interesting problems and they really needed no knowledge. I also liked problems from WF 2017.

Of course you don't have to choose some contest from the ICPC family. Maybe some snackdown contest was so cool for you? Or maybe you liked some contest from training camp in your high school? Write about it in comments.

Hi guys!

On IOI I've learned many new things, so now I want to give you a challenge. You probably remember my and Errichto's eliminations to VK Cup 2016. Let's focus on this problem: 674C - Levels and Regions

Main solution was solving it in O(n*k), because k wasn't greater than 50. What if k is just lower or equal to n? :D

There are two key observations.

- Obviously, higher
*k*means smaller result. - For every
*i*(1 <*i*<*n*) there is*res*(*i*+ 1) -*res*(*i*) ≤*res*(*i*) -*res*(*i*- 1).

The above implies that for every *i* (that 1 ≤ *i* ≤ *n*) there exists such real value *X* that *res*(*i*) + *i*·*X* ≤ *res*(*j*) + *j*·*X* for every . Also, for fixed *i* good *X*'s are between *res*(*i* + 1) - *res*(*i*) and *res*(*i*) - *res*(*i* - 1), inclusive (except for trivial cases *k* = 1 and *k* = *n*).

So, when we have fixed *X*, we can calculate the minimum value of dividing sequence into intervals (we can use any number of them), where cost for an interval is the same as in the basic version of task plus *X*. Using the convex hull trick, we can do it in *O*(*n*). Then we check how many intervals we've used to do it. If we've used too many, we know that *X* should be greater (to force the algorithm to use smaller number of intervals). If we've used at most *k* intervals, we should make *X* lower. So now we see that we should use binary search to find good *X*!

After finding good *X*, we just calculate the optimum value for full sequence and subtract *k*·*X*. This gives us *O*(*n*·*log*) solution.

Editorial was created by Errichto, but he said that he has enough contribution, so I'm posting it for you. ;)

(invented by GlebsHP — thanks!)

You are supposed to implement what is described in the statement. When you read numbers *t*_{i}, check if two consecutive numbers differ by more than 15 (i.e. *t*_{i} - *t*_{i - 1} > 15). If yes then you should print *t*_{i - 1} + 15. You can assume that *t*_{0} = 0 and then you don't have to care about some corner case at the beginning. Also, you can assume that *t*_{n + 1} = 91 or *t*_{n + 1} = 90 (both should work — do you see why?). If your program haven't found two consecutive numbers different by more than 15 then print 90. If you still have problems to solve this problem then check codes of other participants.

(invented by Errichto)

Some prefix of problems must belong to one division, and the remaining suffix must belong to the other division. Thus, we can say that we should choose the place (between two numbers) where we split problems. Each pair *a*_{i}, *b*_{i} (let's say that *a*_{i} < *b*_{i}) means that the splitting place must be between *a*_{i} and *b*_{i}. In other words, it must be on the right from *a*_{i} and on the left from *b*_{i}.

For each pair if *a*_{i} > *b*_{i} then we swap these two numbers. Now, the splitting place must be on the right from *a*_{1}, *a*_{2}, ..., *a*_{m}, so it must be on the right from *A* = *max*(*a*_{1}, *a*_{2}, ..., *a*_{m}). In linear time you can calculate *A*, and similarly calculate *B* = *min*(*b*_{1}, ..., *b*_{m}). Then, the answer is *B* - *A*. It may turn out that *A* > *B* though but we don't want to print a negative answer. So, we should print *max*(0, *B* - *A*).

(invented by Errichto)

We are going to iterate over all intervals. Let's first fix the left end of the interval and denote it by *i*. Now, we iterate over the right end *j*. When we go from *j* to *j* + 1 then we get one extra ball with color *c*_{j + 1}. In one global array *cnt*[*n*] we can keep the number of occurrences of each color (we can clear the array for each new *i*). We should increase by one *cnt*[*c*_{j + 1}] and then check whether *c*_{j + 1} becomes a new dominant color. But how to do it?

Additionally, let's keep one variable *best* with the current dominant color. When we go to *j* + 1 then we should whether *cnt*[*c*_{j + 1}] > *cnt*[*best*] or (*cnt*[*c*_{j + 1}] = = *cnt*[*best*] and *c*_{j + 1} < *best*). The second condition checks which color has smaller index (in case of a tie). And we must increase *answer*[*best*] by one then because we know that *best* is dominant for the current interval. At the end, print values *answer*[1], *answer*[2], ..., *answer*[*n*].

(invented by Errichto)

There is no solution if *n* = 4 or *k* ≤ *n*. But for *n* ≥ 5 and *k* ≥ *n* + 1 you can construct the following graph:

Here, cities (*x*1, *x*2, ..., *x*_{n - 4}) denote other cities in any order you choose (cities different than *a*, *b*, *c*, *d*). You should print (*a*, *c*, *x*1, *x*2, ..., *x*_{n - 4}, *d*, *b*) in the first line, and (*c*, *a*, *x*1, *x*2, ..., *x*_{n - 4}, *b*, *d*) in the second line.

Two not very hard challenges for you. Are you able to prove that the answer doesn't exist for *k* = *n*? Can you solve the problem if the four given cities don't have to be distinct but it's guaranteed that *a* ≠ *b* and *c* ≠ *d*?

(invented by Stonefeang)

When we repeat something and each time we have probability *p* to succeed then the expected number or tries is , till we succeed. How to calculate the expected time for one region [*low*, *high*]? For each *i* in some moment we will try to beat this level and then there will be *S* = *t*_{low} + *t*_{low + 1} + ... + *t*_{i} tokens in the bag, including *t*_{i} tokens allowing us to beat this new level. The probability to succeed is , so the expected time is . So, in total we should sum up values for *i* < *j*. Ok, we managed to understand the actual problem. You can now stop and try to find a slow solution in *O*(*n*^{2}·*k*). Hint: use the dynamic programming.

- Let
*dp*[*i*][*j*] denote the optimal result for prefix of*i*levels, if we divide them into*j*regions. - Let
*pre*[*i*] denote the result for region containing levels 1, 2, ...,*i*(think how to calculate it easily with one loop). - Let
*sum*[*i*] denote the sum of*t*_{j}for all 1 ≤*j*≤*i*. - Let
*rev*[*i*] denote the sum of for all 1 ≤*j*≤*i*.

Now let's write formula for *dp*[*i*][*j*], as the minimum over *l* denoting the end of the previous region:

So we can use convex hull trick to calculate it in *O*(*n*·*k*). You should also get AC with a bit slower divide&conquer trick, if it's implemented carefully.

(invented by Stonefeang)

Let's say that every company has one parent (a company it follows). Also, every copmany has some (maybe empty) set of children. It's crucial that sets of children are disjoint.

For each company let's keep (and always update) one value, equal to the sum of:

- the income from its own fanpage
- the income from its children's fanpages

It turns out that after each query only the above sum changes only for a few values. If *a* starts to follows *b* then you should care about *a*, *b*, *par*[*a*], *par*[*b*], *par*[*par*[*a*]]. And maybe *par*[*par*[*b*]] and *par*[*par*[*par*[*a*]]] if you want to be sure. You can stop reading now for a moment and analyze that indeed other companies will keep the same sum, described above.

Ok, but so far we don't count the income coming from parent's fanpage. But, for each company we can store all its children in one set. All children have the same "income from parent's fanpage" because they have the same parent. So, in set you can keep children sorted by the sum described above. Then, we should always puts the extreme elements from sets in one global set. In the global set you care about the total income, equal to the sum described above and this new "income from parent". Check codes for details. The complexity should be , with big constant factor.

(invented by Errichto)

Let *dp*[*v*][*h*] denote the probability that subtree *v* (if attacked now) would have height at most *h*. The first observation is that we don't care about big *h* because it's very unlikely that a path with e.g. 100 edges will survive. Let's later talk about choosing *h* and now let's say that it's enough to consider *h* up to 60.

When we should answer a query for subtree *v* then we should sum up *h*·(*dp*[*v*][*h*] - *dp*[*v*][*h* - 1]) to get the answer. The other query is harder.

Let's say that a new vertex is attached to vertex *v*. Then, among *dp*[*v*][0], *dp*[*v*][1], *dp*[*v*][2], ... only *dp*[*v*][0] changes (other values stay the same). Also, one value *dp*[*par*[*v*]][1] changes, and so does *dp*[*par*[*par*[*v*]]][2] and so on. You should iterate over *MAX*_*H* vertices (each time going to parent) and update the corresponding value. TODO — puts here come formula for updating value.

The complexity is *O*(*q*·*MAX*_*H*). You may think that *MAX*_*H* = 30 is enough because is small enough. Unfortunately, there exist malicious tests. Consider a tree with paths from root, each with length 31. Now, we talk about the probability of magnitude:

which is more than 10^{ - 6} for *d* = 30.

http://www.wolframalpha.com/input/?i=1+-+(1-(1%2F2)%5Ed)%5E(N%2Fd)+for+N+%3D+500000+and+d+%3D+30

(invented by Stonefeang)

Let's start with *O*(*q*·*p*^{2}) approach, with the dynamic programming. Let *dp*[*days*][*beds*] denote the maximum number of barrels to win if there are *days* days left and *beds* places to sleep left. Then:

Here, *i* represents the number of bears who will go to sleep. If the same *i* bears drink from the same *X* barrels and this exact set of bears go to sleep then on the next day we only have *X* barrels to consider (wine is in one of them). And for *X* = *dp*[*days* - 1][*beds* - *i*] we will manage to find the wine then.

And how to compute the dp faster? Organizers have ugly solution with something similar to meet in the middle. We calculate dp for first *q*^{2 / 3} days and later we use multiply vectors by matrix, to get further answers faster. The complexity is equivalent to *O*(*p*·*q*) but only because roughly *q* = *p*^{3}. We saw shortest codes though. How to do it guys?

You may wonder why there was 2^{32} instead of 10^{9} + 7. It was to fight with making the brute force faster. For 10^{9} + 7 you could add *sum* + = *dp*[*a*][*b*]·*dp*[*c*][*d*] about 15 times (using unsigned long long's) and only then compute modulo. You would then get very fast solution.

(invented by qwerty787788)

Let's first consider a solution processing query in O(n) time, but using O(1) extra memory. If p = 51%, it's a well known problem. We should store one element and some balance. When processing next element, if it's equal to our, we increase balance. If it's not equal, and balance is positive, we decrease it. If it is zero, we getting new element as stored, and setting balance to 1.

To generalize to case of elements, which are at least 100/k%, we will do next. Let's store k elements with balance for each. When getting a new element, if it's in set of our 5, we will add 1 to it's balance. If we have less, than 5 elements, just add new element with balance 1. Else, if there is element with balance 0, replace it by new element with balance one. Else, subtract 1 from each balance. The meaning of such balance becomes more mysterious, but it's not hard to check, that value is at least 100/k% of all elements, it's balance will be positive.

To generalize even more, we can join two of such balanced set. To do that, we sum balances of elements of all sets, than join sets to one, and then removing elements with smallest balance one, by one, untill there is k elements in set. To remove element, we should subtract it's balance from all other balances.

And now, we can merge this sets on segment, using segment tree. This solution will have complexity like *n* * *log*(*n*) * *MERGE*, where MERGE is time of merging two structures. Probably, when k is 5, *k*^{2} / 2 is fastest way. But the profit is we don't need complex structures to check which elements are really in top, so solution works much faster.

Tutorial of VK Cup 2016 - Round 3

Hello everybody!

I'm glad to announce that Round 1 of VK Cup 2016 will take place this Monday, and me (Stonefeang) and Kamil Dębowski (Errichto) are the problemsetters!

There will be an official round for teams from VK cup, but if you are not eligible to participate in it, then you can compete (alone, not in team) in one of two additional editions (one for div.1 and one for div.2), so everybody is invited to take part in the competition! Just register in your category here. All three rounds will be **rated**. Div.1 and Div.2 editions will look like normal CF round, but will have common problems with official edition.

If you can't register before the round, then you will be able to do it during the contest (but not for the entire duration, you can cheсk it here). Let's thank Mike for this great feature!

We want to thank GlebsHP for help in preparing the problems and MikeMirzayanov because without him we wouldn't have such a great platform as Codeforces, where we all can train and develop our passion.

You will again help Limak, your favorite bear. This time it may be harder, because evil Radewoosh will try to disturb him.

We wish you good luck and great fun! Can't wait to see you during the contest! :D

**UPD** Scoring will be:

For VK: **500 — 750 — 1000 — 1500 — 2000 — 3000**

For Div.2: **500 — 1000 — 1500 — 2000 — 2500**

For Div.1: **500 — 1000 — 1500 — 2000 — 3000**

**UPD** Editorial is ready.

**UPD** Congratulations to the winners!

In official VK:

1.Never Lucky: subscriber and tourist

2.SobolevTeam: Seyaua and sdya

3.LNU Penguins: witua and RomaWhite

4.Dandelion: Um_nik and Umqra

5.uıɟɟnɯ ɐuɐuɐq ǝɥʇ ɟo uɹnʇǝɹ╰(º o º╰): enot.1.10 and romanandreev

In Div.1

1.dotorya

2.kcm1700

3.izrak

4.KrK

5.Swistakk

And in Div.2

1.osmanorhan

2.nhho

3.fudail225

5.agaga

4.alanM

Also let's thank qwerty787788 and AlexFetisov for testing problems, without them it would be much harder to prepare contest, so give them an applause!

Announcement of VK Cup 2016 - Round 1

Announcement of VK Cup 2016 - Round 1 (Div.1 Edition)

Announcement of VK Cup 2016 - Round 1 (Div. 2 Edition)

Guys, what memes about waiting for system tests do you have? :D

Can you post them in comments?

These are not mine. I've found them somewhere in blogs.

We can easily calculate the sum of money that we need to buy all the bananas that we want, let's name it x.

If n > = x the answer is 0, because we don't need to borrow anything.

Otherwise the answer is x - n.

Let's count the number of badges with coolness factor 1, 2 and so on. Then, let's look at the number of badges with value equal to 1. If it's greater than 1, we have to increase a value of every of them except for one. Then, we look at number of badges with value 2, 3 and so on up to 2n - 2 (because maximum value of badge which we can achieve is 2n - 1). It is easy to see that this is the correct solution. We can implement it in O(n), but solutions that work in complexity O(n^2) also passed.

It's easy to count who wins and after how many "fights", but it's harder to say, that game won't end. How to do it?

Firstly let's count a number of different states that we can have in the game. Cards can be arranged in any one of n! ways. In every of this combination, we must separate first soldier's cards from the second one's. We can separate it in n + 1 places (because we can count the before and after deck case too).

So war has (n + 1)! states. If we'd do (n + 1)! "fights" and we have not finished the game yes, then we'll be sure that there is a state, that we passed at least twice. That means that we have a cycle, and game won't end.

After checking this game more accurately I can say that the longest path in the state-graph for n = 10 has length 106, so it is enough to do 106 fights, but solutions that did about 40 millions also passed.

Alternative solution is to map states that we already passed. If we know, that we longest time needed to return to state is about 100, then we know that this solution is correct and fast.

Firstly we have to note, that second soldier should choose only prime numbers. If he choose a composite number x that is equal to p * q, he can choose first p, then q and get better score. So our task is to find a number of prime factors in factorization of n.

Now we have to note that factorization of number a! / b! is this same as factorization of numbers (b + 1)*(b + 2)*...*(a - 1)*a.

Let's count number of prime factor in factorization of every number from 2 to 5000000.

First, we use Sieve of Eratosthenes to find a prime diviser of each of these numbers. Then we can calculate a number of prime factors in factorization of a using the formula:

primefactors[a] = primefactors[a / primediviser[a]] + 1

When we know all these numbers, we can use a prefix sums, and then answer for sum on interval.

There are few ways to solve this task, but I'll describe the simplest (in my opinion) one.

Let's build a flow network in following way:

Make a source.

Make a first group of vertices consisting of n vertices, each of them for one city.

Connect a source with ith vertex in first group with edge that has capacity ai.

Make a sink and second group of vertices in the same way, but use bi except for ai.

If there is a road between cities i and j or i = j. Make two edges, first should be connecting ith vertex from first group, and jth vertex from second group, and has infinity capacity. Second should be similar, but connect jth from first group and ith from second group.

Then find a maxflow, in any complexity.

If maxflow is equal to sum of ai and is equal to sum of bi, then there exists an answer. How can we get it? We just have to check how many units are we pushing through edge connecting two vertices from different groups.

I told about many solutions, because every solution, which doesn't use greedy strategy, can undo it's previous pushes, and does it in reasonable complexity should pass.

Tutorial of Codeforces Round #304 (Div. 2)

Hi everyone!

I am pleased to announce that Codeforces Round #304 (Div.2), of which I am the author, will take place today. This will be my first round, so I hope that it will be cool and interesting. Traditionally Div.1 participants can take part out of the competition.

I want to thank znirzejskwarka, Dakurels and Zlobober for help with preparing the problems, thank Delinur for translating the problems, and thank to MikeMirzayanov and all who created polygon for this great system.

I wish you all good luck!

**UPD** Scoring will be 500-1000-1250-1500-2250.

**UPD** editorial

**UPD** Congratulations for winners in div.2:

And in div.1:

Announcement of Codeforces Round #304 (Div. 2)

Codeforces (c) Copyright 2010-2018 Mike Mirzayanov

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