Hi!

Tomorrow, at Feb/07/2019 16:35 (Moscow time) we will host Codeforces Global Round 1.

It is the first round of a new series of Codeforces Global Rounds. The rounds are open for everybody, the rating will be updated for everybody.

The prizes for this round:

- 30 best participants get a t-shirt.
- 20 t-shirts are randomly distributed among those with ranks between 31 and 500, inclusive.

The prizes for the 6-round series in 2019:

- In each round top-100 participants get points according to the table.
- The final result for each participant is equal to the sum of points he gets in the four rounds he placed the highest.
- The best 20 participants over all series get sweatshirts and place certificates.

The problems of this round were developed by a team of authors: aitch, simonlindholm, grphil, V--gLaSsH0ldEr593--V, GreenGrape, Nebuchadnezzar, _kun_ and me. Thanks arsijo and _kun_ for their help in the round's coordination, and also majk, pashka, Jeel_Vaishnav, Ashishgup and UnstoppableSolveMachine for testing the round!

Good luck!

Congratulations to the winners!

Among reds and yellows, there is GreenGrape too. :D

Upvotes? Great community!

They did surgery on GreenGrape

I hope the difficulty gap between problems will be proper.

Just love the unstable contests. GreenGrape is here so we can hope for hell of an unstable contest!I mean contest with weak pretests O_o

maybe the purpose is to encourage us to hack others...

I don't understand why people give down vote in almost everything. Everyone can tell their opinion. Now a days, if someone is asking for help they also get down voted.

There is no disadvantages(at least I think) for downvoted comments unless you are posting something spam, illegal, NSFW, or whatever. So don't keep those downvotes in your mind :)

how many problems will be there?

eight

What is the number of problems? And round duration???

in the previous blog they said every global round will be 7-9 problem

Exactly!

Me in this round.

How I spent two hours on problem D...

Also, me using DSU for B.

Same

It's not like I am going to get any GP score, but it seems a bit unreasonable to rank 4 out of 6 contests, especially with the fact that rounds are not announced weeks ahead of the schedule for participants to allocate time slot for the contest.

Ranking 4 out of 6 is a good idea because of the reason you wrote.

They clearly meant that 4 out of 6 is still unreasonable.

Not if they are announced only half a week beforehand IMO, at this point I can't even tell if the time slot is fixed. I'd imagine I can't participate 1 or 2 of them later into the year since they are midnight rounds for me, and another 1 or 2 of them because it clashes with other activities that I could have possibly adjusted my schedule for it had it announced earlier.

But "4 out of 6" only helps you, compared to "6 out of 6".

I don't mean it's dragging me back, but it's just not good enough to be viable for everyone.

What I agree with is that they should announce rounds earlier if we must compete for some total score.

Nice contest after chinese new year :D Good luck all !

wishing **

Good luck** to everyone who is going to participate. - !NOOB :|Thanks, everyone who did devote me :)

Why am i getting negative contribution without doing anything.

This is life.

That's how codeforces comments works.

dhur bedi

I'm not so sure if this is just me or the case for others too, but I feel like weekends would be more convenient for such rounds, people might be more busy during weekdays. Just an opinion I might be wrong :) Best of luck to those participating in this one!

On saturday there will be atcoder.

On sunday will be opencup.

So, it is quite reasonable why it is not on weekends.

Oh true! Seems quite logical now. It's impossible to please everyone anyways so that's alright. Thanks!

Counterpoint: The Atcoder round isn't for reds and Opencup isn't public either.

There used to be a time when questions at the level of C were thought-provoking. Now the contests are like type first 3 fast and sometimes the transition in difficulty is so drastic that as you move from 1 level to another, the difficulty increases manifold. I believe A should be solved by 80%, B should be solved by 60%, C by 20% and D by like 7-8%. And C level questions need to level up.

I thought today's C was quite challenging and B was not so straightforward to implement.

If you are able to solve C's easier than before, it's because you are getting better.

None of the first three problems were typing questions, because they all required some thought. (Easy or difficult, depending on your level).

A typing question is like this is the scenario — implement it.

All the first 3 questions required some deductions.

This comment was made before the contest.

Ah I see.

None of the first three problems were typing questionsLol. C was a typing challenge and you wrote so much code. No wonder you took so long to solve it.

Is that what you call "competitive" programming? LMAO.

See my submission here

Your solution is essentially the same.

I prefer writing code which is clean and makes the logic clear.

Get a life, man :)

Surely stalking what time I make code submissions and tracking down every single comment I make on here isn't the highlight of your life.

I prefer writing code which is clean and makes the logic clear.Rewriting functionality that the STL already provides for you doesn't make your code cleaner. It makes it worse.

LMAO. Shame on you to call yourself a programmer.

Shame on you to call yourself a human being.

What is the reason for randomly giving T-shirts to 20 people, instead of top 50? Just wondering...

And what is the random generator?

Well shit. I can't participate again because the time is too early.

How many problems and duration ?

7-9 problems for 2-3 hours

Getting One

T-shirtcan make your motivation level rise to themoons! Best of luck everyone.Will it be in ACM-rules? Or in CF rules?

There

was(thanks for the correction!) a 'и' in the English post. Now that I learned Russian a little, I do know it means "and" in Russian, but for those who don't know Russian, it's a bit confusing (especially when it's in the problem statement -- which I've encountered long ago, though it was clear from the context that it means "and" ;)Codeforces has its own variant of the language. Its people can gradually learn extra words like «и» и «Разбор».

Great, I've finally learned something other than сука блять.

Should we expect an interactive problem?

seem difficult for practicer like me to compete t-shirt...

its impossible

Feb 7, 2019: Day of emancipation div2 and div3....

Hi everyone!

I got my old PC fixed just in time to do the contest. Excited to do this contest and this is such a nostalgic feeling. Let us enjoy it together, everyone.

Also, 1 like for getting my main PC fixed. or my PC dies. Your upvotes are really needed. Who wants a poor PC that went through such a harsh life of having 50 chrome tabs open every moment to die? Your likes are needed people.

Oh, at this rate you guys are gonna even kill my old computer and I won't be able to do the contest. Keep it up everyone, 100 downvotes and my battery will blow up.

Too late guys, I already finished the contest. There's no point in blowing up my battery anymore.

Protip: you can put a whole persistent OS on a USB drive, and boot from it anywhere (except where BIOS is locked). You don't need your own PC at all.

Oh that's actually really cool. If I had something like that, I could be using an updated Chrome with an updated IOS which I could really use for a VPN. I could also use my main vimrc. But I still needed a repaired PC to use at my home. If I didn't have one, I would either give up the contest, or try to use a smartphone which is kinda ridiculous (has anybody done that?), or do the contest at a place with a PC which would work actually, but I don't know of a place like that off the top of my mind.

tl dr; thank you for the protip.

It really is Chinese-friendly. I hope that I'll be not so green after this round.

Good luck everyone :D

Happy Rose day to all...

Chup re, Majnu Kahin Ke!

Will the round be sorted by penalties or will points be awarded?

10,000+ registration is now become common for codeforces ):

Are the problems sorted by ther difficulty ?

Wish you good pretests and high rating!

Traditional randomization scripts!

The second script is used to feed random places into the first script. Seed will be the score of the top1 contestant of today's contest, length is the number of people with ranks 31..500 (likely 470) and nwinners is 20.

get_ranklist.pyrandgen.cppWell, congratulations for the tourist for being the first one again ;) and deciding the seed today: 11806

./randgen 11806 470 209 41 71 77 113 138 151 160 180 184 199 232 235 257 275 325 368 383 440 453

So the random winners are:

And the non-random winners are:

Congratulations!

I think wery0 liked this round wery much. Thanks, problemsetters (especially Mr GreenGrape) for good tasks and fast editorial. Also, I want to say thanks to MikeMirzayanov for great(grape) platforms: Codeforces and Polygon!!!!

grapeforces?

How to solve E

Look at what happens to

a_{ j}-a_{ j - 1}(i.e., the difference array) after an operation.Can you elaborate more, I did similar approach but obviously wrong since I got WA :D

I think sum of squares of difference array remains constant

when you make an operation, you actually make a swap in the difference array so you just have to check if the two arrays are the same

Oh yeah, so that would be wrong. The correct approach would be to sort the difference arrays and check if both are same.

by "the two arrays are the same" i meant the two difference arrays. You obviously have to sort them because if not any good test would have two identical vectors ;)

Sorry, I was not saying that your approach is wrong. I was pointing out the mistake in my approach where I said that the sum of squares of difference array remains constant

Seems to me that you can shift the difference array arbitrarily around, since you can swap any two positions. Also, the first and last position of c and t must be the same anyhow. Hence, you just need to check if t can be made.

Did we just have to check if there exists

c_{ j}-c_{ j - 1}=t_{ i}-c_{ i - 1}such thatj≥i?Actually we just need to sort the difference arrays and check if they are same.

Wow! That's a great observation.

How did you notice that? :)

Many people used pow function in A which will overflow but those ran fine. Can someone provide a test case on which these overflow solutions will be wrong if any?

[deleted]

99 ^ 100.000 it's 200.000 digits long, no way it will fit

100

^{99999}is not equal to 10^{10}.That's wrong. 1e2^1e5 = 1e200000.

Its not multiplication, hence do-not add powers

it's 1e(2*1e5) not 1e10. ((a^x)^y) = (a^(x*y))

I guess everybody is doing their part...

sir, (10

^{2})^{105}is not equal to 10^{10}.my part: 100

^{105}isn't equal to 10^{10}sir91 11 1 1 1 1 1 1 1 1 1 1 1

I hacked two solutions with

`pow`

using`99 99 1 0 0 0 ... 0`

.Is overflow treated as error? Values mod2 will be fine even after overflow, right?

`pow`

doesn't operate on integer. You will get a floating point number, that is very unprecise and converting it to an integer will return something bad.Feelsbad then, bcs i saw 3 people using this :(

Is this for just the precision problems with the function

`pow`

?If yes, so what about the result being not fitting in long long? how to hack it, please?

When

`pow(x,y)`

is very large then it will become`inf`

which will make result`inf`

.In C++`inf%2`

is`0`

which means they have defined it as even.So using any big number for which answer is`odd`

will break the normal solution.Luckily overflow results in values modulo 2^31 or 2^63 so it doesn't change the parity.

Overflow solution will not be wrong because overflow will preserve parity.

(Because it happens in cycles of 2^{32} or 2^{64})

A was more interesting than usual. :)

How to solve D?

Did not solve it, but if im not mistaken you can decrease each value to ={1,2,3}. And then apply DP.

Nope. You should set that value to {1,2,3,4}. Consider the following testcase: 12 4 1 2 3 1 2 3 2 3 4 2 3 4.

Just note that you don't need to remove the same consecutive triplet more than 2 times, because otherwise you could have just gotten the same result by removing each number separately.

~~This in turn allows for a DP solution. Just go through the numbers in sorted order and keep track of how many consecutive triplets the previous number and the previous previous number have been a part of.~~EDIT: Some clarifications. The dp state is how many you have left of the previous number and how many you have left of the previous previous number. And because you only want to remove the same consecutive triplet at most twice you can assume that you have at most 2 of the previous previous number.

Very nice problem C!

how to solve it?

Hint: Brute force f(a) for small values of a and look for a pattern.

i don't know how to deal 2^even-1

just brute force it for all the 2^even-1

I guess you can prove that the largest divisor is always the answer

After I brute forced them, I hard coded them into my solution.

where i can see the all solutions

For

a= 2^{ k}- 1,the binary form is 11111....Consider a 1 ≤b<a,we will find that ,so the gcd is .Its greatest value isa's largest divisor(not includingaitself).Iterating divisor can be done in .I am not submitted yet but what i get is basically what XOR is — odd number of ones gives you set bit.example 0001 1110 gives you 1111 so take the first case -all the bits is not set

so first you need to check whether the given number having all the bits is set or not if they are not then just add those bits which are not set. example — a given number is a= 5 — its binary representation is (101).its given that you need to take a number which is less than current number(a) so you just need to take b as sum of those whose bit is not set in current number (a) a=5 binary = 101, b=2 its binary=10 when you xor it you will get all bits set means when you xor (a^b) (101 ^10)=you will get 111 and the and of those ( and the and of (a&b) is zero so when you compute gcd of(aXORb ,a and b) you will get a^b itself as (a and b) is zero so here is the first case

2 nd case -

suppose all the bits is set example 15 its bits representation is 1111 so you just need to check what highest number is divisible by a example

just run a loop max int out=1; for(int i=2;i*i<=in;i++){ if(in%i==0){ if(out<in/i)out=in/i; if(out<i)out=i; }

} cout<<out<<endl; and thats all .You will get your maximum gcd ..

how to solve B?

Note that if we are allowed n pieces of tape (when k = n), we can end up using a minimum total length of n cm. (We would use 1 cm for each broken segment).

Consider the case where k = n-1. Then we need one piece of tape to cover (at least) 2 broken segments. We want to minimize the total length, so we want this piece of tape to cover the two adjacent segments that are closest.

In general, if we have k pieces of tape, then we will have to merge (n-k) adjacent segments.

Let's define the adjacent difference of segments i and (i+1) as (b[i+1] — b[i] — 1).

So for b[i] = 3 and b[i+1] = 4, the adjacent difference would be 0 and we can tape over both with 1 piece of tape of length 2.

We can generalize this by sorting all adjacent differences then taking the sum of the (n-k) smallest ones and adding n to that sum to get our final answer.

I'm very bad at ad-hoc problems but this particular one seems to offer two general observations that are worth for me to learn: (1) applying the pigeonhole principle to constrain the solution: "if you can use only n-1 tapes than at least two of the broken segments must be joined together by one tape"and more importantly (2) adding up the results in a special order: in this case, adding the in-between lengths separately from the n target segments. I was of course trying to add up all contiguous things together.

Great problem. If anyone can think of ad-hoc problems with a similar flavor, perhaps where you have to apply idea #2, I would love to know.

Best problem C I've ever seen~

Guess, who is the writer!

GreenGrape

lol :

what about d ,e ,f , g, h . they are not good ?

Well I meant among all the contests I've ever seen this is the best C problem.All the problems are good!

deleted

Can someone explain why binary searching for the answer (minimum length) with a greedy function wrong in case of B problem? I kept on getting WA on pretest 7. A hint would suffice.

What is your greedy function?

For a fixed length, i did a sliding window, starting from a broken point and covering as many points as I can until the total used length is <= maximum length in current call to the greedy function. Then I would start from the next index and do the same until either all points are covered or can't be covered.

How do you decide whether to start a new piece of tape at the next broken point vs extending the current piece of tape to cover the next broken segment?

I added the description of my solution to B here.

Understood the mistake. Thanks.

Can you please give some test case where the binary search approach will fail... I also wrote solution using Binary Search but WA on pretest 7

10 500000 4

1 30003 61255 101250 141246 171244 202492 242483 282475 305289

Correct Ans: 185309 with these indices forming groups [1,2,3][4][5,6,7,8][9,10]

B was harder than C for me. LOL

i found c to be easier than a and b !

how to solve B?no idea :(

you need to find k minimum difference pairs from neighbourhoods

the problem is almost similar to atcoder beginner contest 117 problem C which i solved just today .find the maxx sum = arr[last] — arr[first] you need to find the gap between each array elemets , sort them , and then start sweeping them k times . subtract difference[i] each time from maxx sum .

Well...I think the pretest of problem A is too weak because I passed the pretest while writing

`x&1`

wrong as`x^1`

...E is one of the best problem E ever on Codeforces, until I realized that one can fail system test because they forget to check if

c[1] ≠t[1] orc[n] ≠t[n].GreenGrape influence is always real.Hmm, you don't really have to check

c[n] ≠t[n]Does this count as cheating? I saw a submission for C where the user created a map of all the possible input a's to the answer. Probably these answers were brute forced earlier and logged?

Nope, in some problems that's the actual intended solution

If you say this then tourist is cheating.

haha , why ?

As what a considerable amount of people have done, he also hardcoded the answer for input in form 2

^{ n}- 1.yes .

Almost everyone did this

Precomputation is a very legitimate strategy. ICPC World Finals 2018 Problem J solution used precomputation.

Spoiled >:(

What is test case 35 in D? Many people including me are failing on that

Problem E was on AtCoder: link.

good that's why i like mokoto , his mind is awesome , problem b was also similar to abc 117 c

thanks to mokoto i solved it !

That's how I solve pE within 5 minutes lol

Any idea how to solve D .

even the submission that i didnt have time to submit is WA .. now I can rest in peace

I love this contest, where the difficulty is in ideas not just complex code.

When can we konw the final result ?

https://codeforces.com/contest/1110/standings

Just for curiosity, why are there no hacking session for this contes_- _t? The contests I participated since were rated for Div2 so... just curious.

Hacking sessions are only available for Educational Rounds and Div.3 contests. ;)

thanks :) glad to check ACs right after!

So I was using ideone for this contest since I'm not home and I didn't want to install codeblocks on this pc and I got a message that bruce_knight copied my code for B. The message said I could be getting banned for this ? What are my options ?

Even tourist couldn't solve

H?There must be something wrong with tourist or that question...

There must be something wrong with tourist...

You should know that the author might prepare it for more than 2 hours. But tourist just have less than 1.5 hours

I think there are some problems that have appeared in another place. For example: E: BZOJ5071 I'm sorry that the language for this problem is Chinese, but this problem is similar to Problem E （My English is not good）

that is the strategy of tourist

Can someone explain the logic behind E?

Consider the difference array

d, i.e,d_{ i}=a_{ i + 1}-a_{ i}, then any operation is swapping two elements in the difference array.In api expected rating change shows +34 but in rating changes it shows -4

I had almost solved problem G but there was not enough time.... It's an interesting problem to me though.

This was my 2nd contest solved A,B. Initially afraid of submitting the solution but finally submitted it and got accepted. Aiming to solve more in coming contests. Thanks to the organizing team.

Can F be done with centroid decomposition ?

For each leaf ,go up to root via centroids storing in each centroid ,distance of leaf from it and its index. For each centroid we keep this list sorted on the basis of indices. While answering a query, we go up to root from vertex v via centroids, at each centroid, we need to query min dist of leaf from this centroid whose index lies in L to R(simple RMQ).Take this minimum across all centroids in my path to root .

Is this approach correct ? will it fit the TL ?

Correct, but I guess too slow: 49601147

It is correct and my solution passed. https://codeforces.com/contest/1110/submission/49592451

I think the rating changes for today's contest were not correct. As the "CF Predictor" extension showed something else and the rating changes were different.

Kindly look into the matter.

+1

Problem C was a great question. The only case you had to take care of was when

awas of the form2^x-1. The trick involved wasgcd(a&b, a^b) = gcd(b, a-b) = gcd(a, b). So we need to find the greatest proper divisor ofa. Kudos to the writers!Can D be solved using a recursive approach without running into "Memory limit exceeded"?

Yes, refer to 49598852. Very elegant.

Hey,guys.I have a problem on 1110D — Jongmah. I passed the test 1 locally on my machine,however get WA on the net. Here is running ID #49593747 Help me please!!!QAQ

This is what u can try

When will the editorial be published?

Well,I have already seen it in the Recent actions

Thanks to this contest, I have become a candidate master. I'm so happy.

Congratulations！

KAN, Why 2 links to the same editorial are added to contest page?