Codeforces Round 927 (Div. 3)

9 months ago, # ^ |

← Rev. 2 →

+16

Where are the problems of JetBrains Academy Youth Challenge? Asking for a friend.

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JetBrains Problems with Solutions

9 months ago, # ^ |

+11

Shit, I really found those problems. I feel guilty now. 😟

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edgerunner

9 months ago, # ^ |

nice profile pic and bait)

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Krrishchanchal

9 months ago, # ^ |

I got a youtube ad running lol

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DarkDuel

9 months ago, # ^ |

Can someone plz tell me why my comment got so many downvotes(genuinely want to know,to avoid/learn/improve)

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i_love_edging

9 months ago, # ^ |

Trust me. Some people just downvote for fun. They don't want to clarify your doubts. But they feel annoyed thinking that you don't even know simple things. I hate these egoists.

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abhishekm33

9 months ago, # |

-8

What is the score distribution?

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nskybytskyi

9 months ago, # ^ |

In ICPC rules, all problems are worth the same.

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abhishekm33

9 months ago, # ^ |

-22

So all are problems are rated 1000?

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Xanlar2009

9 months ago, # ^ |

-8

it's based on how many problems you wrote, if equal it's based on penalty

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9 months ago, # ^ |

No, ICPC rules state that you are rated by # of problem solved, then time. There is no point system, however the questions themselves vary in difficulty (so better solve A first for a quick advantage)

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Six_Seconds

9 months ago, # |

+15

Can the timing be changed? It clashes with ARC. Moreover, this time is quite awkward for me, as I have an important appointment at that time.

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9 months ago, # ^ |

-7

Ohh why not sir. Please tell us when you're free and we'll change the timing according to your availability. Thank you.

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Six_Seconds

9 months ago, # ^ |

Great that u understood. I prefer to have it on regular Cf time so that it doesn't coincide with Atcoder Regular Contest and people like me can go on with their regular Sunday Schedule.

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9 months ago, # ^ |

We have a slot free on 30th Feb at regular CF time. Is that alright sir?

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9 months ago, # ^ |

← Rev. 2 →

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9 months ago, # ^ |

This is your first comment, and you've said this? So sad.

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9 months ago, # ^ |

← Rev. 3 →

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9 months ago, # ^ |

What's your main?

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9 months ago, # ^ |

← Rev. 2 →

Cant say sorry. It seems to be your second account as well.

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9 months ago, # ^ |

You do believe that single-account policy exists here right?

For everyone else, don't be this guy. Stick with your accounts.

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9 months ago, # ^ |

← Rev. 2 →

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9 months ago, # ^ |

At least they aren't your level of dumb, literally saying out loud like that. Keep it to yourself mate.

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9 months ago, # ^ |

← Rev. 2 →

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9 months ago, # ^ |

wow, very tough. Wasn't you the guy who said eren to fk off and not do cp because, and quote, "dont try to mess with the tough ones kid"? mind YOUR own business first, then talk to me.

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9 months ago, # ^ |

← Rev. 2 →

i have more than two ids that is not something you should be talking about..i was just saying him not to mock someone...you wont get that so .just go away instead of arguing

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9 months ago, # ^ |

I don't see any mocking done here, more like a joke. But you're right, maybe i should stop arguing and get some problems in. Have fun with your alternative accounts then

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9 months ago, # ^ |

← Rev. 2 →

So glad u understood...yeah i should be going and solve some problems as well..see ya later..best of luck with your journey mate..

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9 months ago, # ^ |

And also sorry for those harsh words..i get your point..see ya

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Erering

9 months ago, # ^ |

-8

Oh yes lets change the contest date because Six_Seconds has an appointment at that time.

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Dominater069

9 months ago, # ^ |

+15

2 of the biggest cp sites should try not to put contests at the same time....especially when one of them has been scheduled for weeks. (Note : leetcode does not count as cp)

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Erering

9 months ago, # ^ |

Yes but I was just joking about the fact that he said he has an appointment like that makes it more reasonable to cancel the contest.

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miHoYo_GenshinImpact

9 months ago, # ^ |

However tourist can still ak both of the 2 contests. That's awful.

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freakout

9 months ago, # |

← Rev. 2 →

Hey boy, are you a semicolon? Because my program is missing you ;)

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9 months ago, # ^ |

← Rev. 2 →

-15

bye

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freakout

9 months ago, # ^ |

← Rev. 2 →

-16

bye

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9 months ago, # ^ |

← Rev. 3 →

-16

bye

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9 months ago, # ^ |

← Rev. 4 →

-16

guys report him she is a he

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Forestmy17

9 months ago, # |

as not a tester i would like to solve this round

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Fork512Hz

9 months ago, # |

Finally after a tedious global round which took away my *1800 it's time to relax. Hope to solve A-F

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Fork512Hz

9 months ago, # ^ |

F in 94:59.

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MKasirloo

9 months ago, # |

← Rev. 2 →

This comment has been deleted.

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miHoYo_GenshinImpact

9 months ago, # ^ |

plz read the rules of extended ICPC mode and stop asking such useless questions.

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xinyu123_AKIOI

9 months ago, # |

Good start time for Chinese.

But I must finish my winter vacation homework!!!F**k

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miHoYo_GenshinImpact

9 months ago, # ^ |

You can say more about it and I will be happy because I have already entered university and I have no homework this winter vacation. LOL

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xinyu123_AKIOI

9 months ago, # ^ |

Originally you also play Genshin

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Md-Merazul-Islam

9 months ago, # |

Back-to-Back codeforces contest. really amazing. Beast of luck everyone. ヽ(•‿•)ノ

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mr.itmo

9 months ago, # |

I have a feeling this round will be amazing.

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WhileTrueRP

9 months ago, # ^ |

Me too.

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SamuelAktar

9 months ago, # |

-10

If problems are already based on jetbrains youth academic challenge, then it means people already know the problems. Not just that, we can find them from internet also.

Therefore, let's not participate in this round.

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ilovenandini

9 months ago, # |

← Rev. 2 →

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2ball

9 months ago, # |

problems of this contest are prepared "step_by_step"

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_SwaRaj

9 months ago, # |

My 1st unrated Div 3 !(;

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synthborne

9 months ago, # ^ |

Nice. Wont hurt to be legitimate this time.

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9 months ago, # ^ |

+16

Cheater

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9 months ago, # |

I would target solving >75% problems.

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Byehttt

9 months ago, # |

Good start time!

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J.A.K.D.Z

9 months ago, # |

Yesterday i became specialist so i hope to reach expert again.

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9 months ago, # ^ |

← Rev. 2 →

I also became a specialist yesterday and I hope not to become a pupil again

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rohitrao835589

9 months ago, # |

I know i will get -ve point in this contest :(

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SAYEM_16

9 months ago, # |

Please note the unusual start time: 18:05 UTC+6

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WhileTrueRP

9 months ago, # |

The timing is great, but it overlaps with ARC's timing.

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Bilal.Coder

9 months ago, # |

Good lucks for everyone!

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rishabhdeepsingh

9 months ago, # |

Good luck to all of my grey friends :D

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cleanser

9 months ago, # |

Redemption time

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_Traniac_

9 months ago, # |

extra registration ??

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AvaraKedavra

9 months ago, # |

← Rev. 3 →

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rohitrao835589

9 months ago, # |

-12

c Problem is very tough :(

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sky_full_of_stars

9 months ago, # ^ |

You just had to iterate string backwards.

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KuR222

9 months ago, # ^ |

you can even use segment tree.(Thanks to n<= 2e5)

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UP84

9 months ago, # |

+34

tourist speedrun the division 3 in 28 minutes to clinch first position,then started Atcoder regular contest 45 minutes late and still came 3rd(Standings ). GOAT stuff.

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https://mirror.codeforces.com/contest/1932/submission/247073793

9 months ago, # ^ |

Tourist teaching nubs how it is done just like Dr Disrespect.

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Spoderman.

9 months ago, # |

I got PTSD because of Problem C

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0gnjen1

9 months ago, # ^ |

Yup, I think I could've done both D and E but just wouldn't give up C because I thought I was going crazy and missing something way too obvious.

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defarias25

9 months ago, # ^ |

Yeah I kept trying to do modular inverse, but realized that only works for certain m.

The observation is to figure out the last value before performing the last operation, then simulate the process backward and compute the cur value with modular multiplication.

My solution is here if it helps: 247081220

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Spoderman.

9 months ago, # ^ |

Wow.good job man

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smit_384

9 months ago, # ^ |

yes , i was not able to solve c , i decided to skip c , and i solved d and e both in contest.

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elseecay

9 months ago, # ^ |

hint: reverse LR string

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I_lOVE_ROMAN

9 months ago, # ^ |

Can you please elaborate?

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elseecay

9 months ago, # ^ |

← Rev. 3 →

calculate where will be left and right indexes after all operations
reverse LR string
calculate values (left moves to the left, right to the right)
reverse result array and print

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9 months ago, # ^ |

Same thing, but maybe more intuitive because doesn't require to calculate starting point is to collect the order of indices (kind of topological sorting), and multiply them backwards.

Code

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tlimitexceeded

9 months ago, # ^ |

Man, i solved C, but couldn't understand b, even after reading it 10 times

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jkulanko

9 months ago, # |

Ugh, I feel so dumb for not being able to figure out C

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9 months ago, # |

+25

Silent moment for ones who tried so hard for bigint in E

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PhuongVN

9 months ago, # ^ |

It was actually pretty easy, and I , of course, didn't think about the optimized solution during the contest. just AC it this morning

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rohitsatpute480

9 months ago, # |

Was F dp + segment trees?

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9 months ago, # ^ |

Yes, although you actually don't really need segment tree

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9 months ago, # ^ |

Yeah, I too only used dp

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pranavvinchurkar

9 months ago, # ^ |

DP: Yes. I did not use segment trees.

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9 months ago, # ^ |

didn't need segment tree, but still can be solved

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arash.

9 months ago, # ^ |

dp + sweep line + multiset

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Anuska

9 months ago, # |

when will i can see the editorial ??

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miniyahirpro

9 months ago, # |

how to solve the E?

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9 months ago, # ^ |

For example S = "12345", then the answer is 1 + 12 + 123 + 1234 + 12345 (hope you can find out why it is)

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miniyahirpro

9 months ago, # ^ |

Thanks I'll try that

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tickbird

9 months ago, # ^ |

observed that 3min before end but couldn't implement on time 😥

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yadavDipesh

9 months ago, # ^ |

same 😥

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castlercodes

9 months ago, # ^ |

In c++, doesnt the sum of these integers overflow long long for a large testcase?

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PhuongVN

9 months ago, # ^ |

yes, that's why you need to implement the addition by yourself, and store the digit in an array (that's my solution, I don't know if you can use string or not)

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castlercodes

9 months ago, # ^ |

thanks

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tenick

9 months ago, # ^ |

you can store in string directly, check my solution https://mirror.codeforces.com/contest/1932/submission/247574018

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Citypop

9 months ago, # ^ |

← Rev. 2 →

A key observation is decreasing 10->0 takes 11 seconds, 100->0 takes 111 seconds...

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miniyahirpro

9 months ago, # ^ |

I did see that, but I didn't know how to implement it.

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Citypop

9 months ago, # ^ |

Think about count 1s for each position. Then add them up.

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Legend_of_Azkaban

9 months ago, # ^ |

According to the statement, going from 10 -> 0 requires time = 11. going from 20 -> 0 requires time = 22. ... going from 100 to 0 requires time = 111. going from 200 to 0 requires time = 222.

Say you are given string S = 12345 First you make 2345 -> 0 (A) then you go from 10000 -> 0 which requires 11111 time. (B) A+B is the ans.

Let's make 2345 -> 0. First you make 345 -> 0, (A) and then you go from 2000 -> 0 which requires 2222 time. (B) A+B is the ans. and so on.

Thus we can observe each i will be the sum of the prefix digits upto i.

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BiggestMohammadyayhater

9 months ago, # |

E and D are just implementation didn't like them that much.

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9 months ago, # |

went good for a first div3 unrated contest

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Cat.Noir

9 months ago, # |

How this code not able to pass C ? link anyway i solve it using Segment Tree link

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dp_16

9 months ago, # |

I can't figure out why this code fails for C: 247057083

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sammas24

9 months ago, # ^ |

Maybe a long long overflow when calculating the product at the beginning.

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9 months ago, # ^ |

Every value of A(i) can go upto 1e4. Now imagine 1e4 values in all the n elements... It will be (1e4)^n and as n can itself go upto 2e5.

So, (1000)^(20005). Do you think It can be stored?

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dp_16

9 months ago, # ^ |

I am stupid

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9 months ago, # ^ |

So am I. Took me ages to figure it out. I literally remember same kind of mistake in My First ICPC in 2022.

Repeating same mistake twice proves me more stupid.

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RPGLILSTAR

9 months ago, # ^ |

OMG, I thought 1e4 * 2e5 at first,so it will not long long overflow

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9 months ago, # ^ |

US moment.

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aditya0610

9 months ago, # ^ |

same

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0gnjen1

9 months ago, # ^ |

This line is definitely the problem:

for(int i = 1; i <=n; i++){
    prefArr[i]=prefArr[i-1]*nums[i-1];
}

Because you could end up with a number as big as $$$10^{8 * 10^{5}}$$$ which just can't fit into a 64 bit integer.

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SearchMeLater

9 months ago, # |

Nowadays Div 3 is harder than Div 2. I don't know if it is only for me or if other users also feel it. BTW happy coding...........

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9 months ago, # ^ |

tourist definitely struggled a lot

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9 months ago, # |

there was no where written about score penalty for wrong submissions? Why penalty has been made?

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MAA1117

9 months ago, # ^ |

Sounds like a bug. I get a score bonus for wrong submissions.

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NoticeMeSenpai

9 months ago, # ^ |

"the round follows the ICPC rules, penalty for an incorrect submission is 10 minutes"

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9 months ago, # ^ |

its written about time penalty only, does it also mean score penalty while calculation of rank?

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NoticeMeSenpai

9 months ago, # ^ |

← Rev. 2 →

time penalty doesn't mean it is gonna remove 10mins of your solving time but it is gonna add 10mins to your solution time (the penalty is time you took to solve problem + 10mins(for every incorrect submissions)).

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9 months ago, # ^ |

ohhh

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J.A.K.D.Z

9 months ago, # |

E got TLE with python but in c++ with making bignum get AC.

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dydx

9 months ago, # |

Problem F was really nice. Thanks authors. :blobheart:

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Mustafa_Elghisha

9 months ago, # |

-23

what the fuck is b and c

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sky_full_of_stars

9 months ago, # ^ |

I had to derive the question by looking at sample test cases.

Such bad problem statements this time.

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tickbird

9 months ago, # |

problem C was quite pleasure to solve

I felt like being Elegia

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9 months ago, # |

How to solve C?

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rohitsatpute480

9 months ago, # ^ |

I used segment Trees to find product of a range % modulo

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9 months ago, # ^ |

Thanks

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9 months ago, # ^ |

← Rev. 2 →

Just do it in reverse order, then you will multiply instead of dividing the current number.

Also, I've seen your code. If you tried to divide the current number, but in modulo form, you need to use modulo reverse rather than divide it directly

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9 months ago, # ^ |

Thanks

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bhoomikacherry

9 months ago, # ^ |

dividing in modulo form might not work in general, as $$$m$$$ might not be prime

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phukaoz

9 months ago, # ^ |

solve from the back

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second_saturday

9 months ago, # |

← Rev. 2 →

Am I the only one who is able to solve C but could not solve any other problem.

what is wrong with me... :(

could anyone tell what is wrong in :

Problem B submission

Problem A submission

EDIT: I found the small bugs in my code both above submissions are incorrect

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Cat.Noir

9 months ago, # ^ |

for A i use recursive it's kinda similair to minimum jump problem code

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9 months ago, # ^ |

For problem A: you will not be able to move further after two consecutive thorns cells. The answer to the problem will be the number of coins up to two consecutive cells with thorns

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nikhil97agra

9 months ago, # ^ |

I didn't clearly understand your solution but let me tell you my logic

Problem A: simply start from left & collect every coin and it any point you encounter two consecutive * (thorns) exit your loop.

Problem B: let A be the year of ith apocalypse, then simply check for the (i+1)th apocalypse if its year is greater then A simply change A = that year, if it is <= a then find the next multiple of that element greater A.

here are my solutions for reference

https://mirror.codeforces.com/contest/1932/submission/246993142 — Problem A

https://mirror.codeforces.com/contest/1932/submission/247002028 — Problem B

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Muhimen

9 months ago, # ^ |

For problem B you can linearly find the multiplier x(i+1) for which ai < a(i+1) * x(i+1)

If you keep doing this for every i, the last element of the array is your answer.

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dsol28

9 months ago, # |

hi guys what means "Exit code is -1073740940"? what should i do with it?

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cleanser

9 months ago, # |

+13

Is G just Dijkstra's but edge weights are computed using some x such that a + bx = c + dx, if the two sides of the equation are 2 different platforms?

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9 months ago, # ^ |

← Rev. 7 →

For all edges $$$(i,j)$$$ we have $$$ l_i + xS_i = l_j + xS_j \mod H $$$. Regroup $$$ x(S_i-S_j) = l_j - l_i \mod H $$$

Find all solutions using the https://cp-algorithms.com/algebra/linear_congruence_equation.html gives you moments when can walk the edge.

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rishabhxchoudhary

9 months ago, # |

I recorded myself live while solving it. I solved A,B,C and D problem. Hope it helps someone: https://www.youtube.com/watch?v=7Sv_tmVLQeQ

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9 months ago, # |

Is there a more effective way to do D other than making an if-else chain?

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mohit458

9 months ago, # ^ |

backtracking

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9 months ago, # ^ |

Hmm, I guess that's one way, didn't click for me

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AvaraKedavra

9 months ago, # ^ |

No, you can't. Each card can have 32 positions. I wasted 2 hours trying to implement it.

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mohit458

9 months ago, # ^ |

Check my submitted solution it is using backtracking

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9 months ago, # ^ |

I imagine you could create a custom comparator and then sort, correct me if I am wrong.

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cleanser

9 months ago, # ^ |

I did the following. I keep the sorted values of the cards for each suit. For the non-winning suits, I can just have the cards beat each other in sorted order, taking 2 at a time until I have 0 or 1 left. If I have 1 left, I'll save it for later. Then afterwards I can just take the winning suit, eliminate all of the remaining ones from other suits, and eliminate the rest in the winning suit the same way as before.

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9 months ago, # ^ |

I see, I believe I did something similar as well. Just couldn't simplify the conditionals effectively, so it almost comes off like a shabby brute-force.

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9 months ago, # ^ |

Store in map

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Citypop

9 months ago, # ^ |

← Rev. 2 →

Sort card for each suit and greedily remove 2 cards at a time, if there is one card left, u need to match it with any trump. At last u do the same thing for the rest of the trump cards.

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fonmagnus

9 months ago, # ^ |

Group each card by suits. Then sort by rank for each suits

Now for each suit other than the trump, take two adjacent card rank. If there are some excesses, keep it and use the trump to beat them

Now use the remaining trump suit to battle each other

The impossible case is only when the excesses are greater than the trump suit

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m_nature

9 months ago, # ^ |

← Rev. 2 →

I used this comparator function to sort all the cards

bool comp(string a, string b) {
    // Both are trump
    if (a[1] == trump && b[1] == trump) {
        return a[0] < b[0];
    }

    // One is trump
    if (a[1] == trump) {
        return false;
    }
    if (b[1] == trump) {
        return true;
    }

    // Both are normal, same suit
    if (a[1] == b[1]) {
        return a[0] < b[0];
    }

    // Both are normal, different suit
    return a[1] < b[1];
}

Then my logic for the ans is just go from left to right

If there are still 2 cards of the same suit -> Choose them
If there is only 1 card left of that suit then get a trump card from the back (If I still has trump cards)
Otherwise, I can't beat the current card -> This game is impossible

    vector<pair<string, string>> ans;
    int rightIdx = 2 * n - 1;
    bool ok = true;
    for (int i = 0; i + 1 <= rightIdx; i++) {
        if (a[i + 1][1] == a[i][1]) {
            ans.push_back({a[i], a[i + 1]});
            i++;
        }
        else if (a[rightIdx][1] == trump) {
            ans.push_back({a[i], a[rightIdx]});
            rightIdx--;
        }
        else {
            ok = false;
            break;
        }
    }

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GlydonNeedsDedication

9 months ago, # ^ |

you can refer my solution Problem D

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MKasirloo

9 months ago, # |

Spent a long time to understand what problem B wants

You too?

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Hemdan47

9 months ago, # ^ |

← Rev. 2 →

i spent about 40 minutes trying to figure out what's required and solved it in like 5 minutes

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ashucoder

9 months ago, # ^ |

Same, I couldn't comprehend what problem B meant spent 40mins there and wasn't able to solve and then when I went to C, D, E all made sense. For problem E solution for python gave TLE, and by the time I wrote the code in C lang time was up. So frustrating could have possibly solved all 5...

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AdityaTakkar

9 months ago, # ^ |

Yup me too, got to know by looking at the sample test cases. Poor writing of the problem statement.

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alpha1215

9 months ago, # |

+14

Ac F at 2:14 i can't believe myself

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codeWithDarshan

9 months ago, # |

A lot of people are cheating . Codeforces please look into this matter. People are uploading their solution on youtube and telegram also. How it is possible that people submited solution of three problem in a gap of three minutes and also in different language while in contest and even after 1 hour suddenly they submit their solution

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pedastrian57

9 months ago, # |

If you need video explanations of C, D, E and F, you can check my video editorials here

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Taejin101

9 months ago, # |

Why long long doesn't work for problem C. Calculating the product in long long then dividing the product by left or right whatever element is deleted?

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Minerva2005

9 months ago, # ^ |

Because long long will overflow

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Taejin101

9 months ago, # ^ |

I accept, I am dumb. Don't know why but thinking it as 1e5 * 1e4 which is (1e4+1e4+1e4+......(1e5 times)). But it is actually (1e4*1e4*1e4....(1e5 times) which will cause overflow.

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Cat.Noir

9 months ago, # ^ |

← Rev. 2 →

it's long long overflow, a[i] can be 1e4, and n can be 2e5 imagine array all the elements are 1e4 of length 2e5 so the product all of them is (1e4)^(2e5) that can't fit in long long

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MKasirloo

9 months ago, # ^ |

Because there is a test case which $$$n=2\cdot10^5$$$ and all $$$a[i]$$$ are $$$10^4$$$, then it will be $$$10000^{200000}$$$ :)

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Fork512Hz

9 months ago, # |

BE CAREFUL OF CARD SIMULATIONS 23min D, 7min E

I did F with linear DP, first finding out bar[x] = which spot is the leftmost viable spot on x's right if x is chosen.

Could someone explain how G works?

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RPGLILSTAR

9 months ago, # |

Can somebody tell me why deque can't solve problem c??? I can't pass test 3.

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9 months ago, # ^ |

p *= a[i]

gonna make integer overflow


  if (s[i] == 'L')
            {
                p /= a.front();
                a.pop_front();
            }
            else
            {
                p /= a.back();
                a.pop_back();
            }

Is not accurate, because you need to use modulo inverse, although I think it is not usable in this problem (because m isn't a prime number sometimes)

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RPGLILSTAR

9 months ago, # ^ |

Maybe i misunderstood the problem? The problem let us get the product of the current array elements,then the let product mod m, lastly remove one element。

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9 months ago, # ^ |

You clearly understand it already

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priy1712

9 months ago, # ^ |

247159936 can you have a look at this one too ,it fails on TC2 , I tried modding it with 1e9+7 because my earlier submission failed on TC3.

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Suplex

9 months ago, # |

Whats wrong with the naive implementation of problem C?

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susus

9 months ago, # ^ |

probably because of overflow

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calebmartim

9 months ago, # ^ |

← Rev. 2 →

If you have $$$2\cdot 10^5$$$ elements and all of them are $$$10^4$$$, your product will be $$$(10^4)^{2\cdot 10^5}$$$ so it overflows.

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Arthas

9 months ago, # |

← Rev. 2 →

D can be solved by using Blossom algorithm, even though it is just an overkill of the problem

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Mu.stafar

9 months ago, # ^ |

Must we use Blossom? I believed the graph wouldn't contain a cycle and is thus bipartite, so I tried to find the matching with max flow. However, I failed on test 3.

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Arthas

9 months ago, # ^ |

← Rev. 3 →

I suppose that it can be solved using max_flow algorithms, but max matching is more intuitive, i think

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q838746907

9 months ago, # |

why greedy does not work for F? :(

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9 months ago, # ^ |

how you suppose to do it greedily?

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q838746907

9 months ago, # ^ |

← Rev. 3 →

sort according to the left points
put those intervals into lists of non-overlapping intervals, like this:

2 4, 5 7, 10 11

2 6, 7 7

4 5

in each iteration, I'll increment the answer by the number of overlapping intervals in the first element of the lists (a little bit confusing description :(, those intervals will form a larger interval (called concatenated intervals) and also removes all intervals contained in the concatenated interval. Then remove empty lists.
output the answer

→ Reply

9 months ago, # ^ |

← Rev. 6 →

Try this input:

7 5

7 7

1 3

6 6

3 7

5 7

The answer is 4 (feed the cat at step 1 and 7)

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q838746907

9 months ago, # ^ |

← Rev. 2 →

A great counter example! thanks!!!

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Waiting_The_Summer

9 months ago, # |

C problem is really hard for Div3 XD.

p.s Who solved this problem without trees? :)

→ Reply

https://mirror.codeforces.com/contest/1932/submission/247080815

9 months ago, # ^ |

I'm curious about the segment tree solution, can you explain a little bit?

→ Reply

Rakesh_898

9 months ago, # ^ |

← Rev. 3 →

Very simple problem if you think using segment tree.find my solution here..

C++ Code

/*ॐ भूर्भुवः स्व तत्सवितुर्वरेण्यं
    भर्गो देवस्य धीमहि
   धियो यो नः प्रचोदयात॥*/

#include <iostream>
#include <algorithm>
#ifndef ONLINE_JUDGE
#include "debug.h"
#else
#define debug(...)
#endif
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define ll long long int
#define ld long double
#define vi vector<ll>
#define forll for (ll i = 0; i < n; i++)
#define tc    \
    ll t;     \
    cin >> t; \
    while (t--)

using namespace std;
using namespace __gnu_pbds;
template <class T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
vector<ll> segTree;
ll mm;
void build(vector<ll> &v, ll start, ll end, ll index)
{
    if (start == end)
    {
        segTree[index] = v[start] % mm;
        return;
    }
    ll mid = (start + end) >> 1;
    ll left = 2 * index;
    ll right = 2 * index + 1;
    build(v, start, mid, left);
    build(v, mid + 1, end, right);
    segTree[index] = (segTree[left] * segTree[right]) % mm;
}

ll query(ll start, ll end, ll index, ll l, ll r)
{
    // complete overlap
    if (start >= l && end <= r)
    {
        return segTree[index] % mm;
    }
    // disjoint
    if (l > end || r < start)
    {
        return 1;
    }
    ll mid = (start + end) >> 1;
    ll left = 2 * index;
    ll right = 2 * index + 1;
    return (query(start, mid, left, l, r) * query(mid + 1, end, right, l, r)) % mm;
}

void solution()
{
    ll n;
    cin >> n >> mm;
    vi v;
    forll
    {
        ll x;
        cin >> x;
        v.push_back(x);
    }
    string s;
    cin >> s;
    ll ans = 1;
    ll ii = 0;
    ll j = n - 1;
    segTree.resize(4 * n + 15, 1);
    build(v, 0, n &mdash; 1, 1);
    forll
    {
        cout << query(0, n - 1, 1, ii, j) << " ";
        if (s[i] == 'L')
        {
            ii++;
        }
        else
        {
            j--;
        }
    }
    cout << endl;
}

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);

    tc
    {
        solution();
    }

    return 0;
}

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ma_da_fa_ka

9 months ago, # ^ |

It’s pretty simple just compute if offline in reverse order. Start from index after n-1 operations. This index is both l and r. in reverse order, if operation is R, append the righter value and increment.

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MAA1117

9 months ago, # ^ |

← Rev. 2 →

yes very simple if you know how to solve it. not so simple if you dont.

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Medeali

9 months ago, # ^ |

basically the reverse approach is hard to think of but it is not the only one you can use segtree and just recompute again for every range thanks to segtree this will be done in n log n instead of n^2

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Atma_

9 months ago, # ^ |

I solved it with sqrt decomposition

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jwacker

9 months ago, # ^ |

← Rev. 2 →

I had no idea that trees could be used to solve this problem :(

I got it wrong when using a brute force method initially as I got a TLE since I was calculating the product again and again for each command giving somewhat O(n^2) solution

Also I realized it was not possible to calculate product just once and remove the contribution of boundary elements in each command...

then it hit me that I could probably start in the reverse order! start from the last element that would be left after all commands and then move reverse in the commands string and keep including the new element in the answer... (Basically not moving from n elements to 0 elements but instead moving from 0 elements to n elements) This way I could do it in O(n) by visiting each element just once. This would give us the answer vector in the reverse order and then I just print it in reverse to get the final correct answer!

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ttirth1234518

9 months ago, # ^ |

Start from reverse. Start from one element that is going to be remaining at last, after removing every other. Then keep multiplying second last, third last… while taking modulo. And then print array of answers in reverse

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quad_wipe

9 months ago, # |

← Rev. 3 →

when you realize taking input in int and str matters just after the contest!-_-

TLE submission

AC submission-_-

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Aditya_Metkar

9 months ago, # |

One MODULO can make all the difference WoW!! (Problem C btw)

Without % -> This
With % -> This

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Hussain_Elemam

9 months ago, # |

I came at the regular contests time and open the contest very seriously and solved the first problem and before starting the second one I looked to see how much time I spend on it, and yeah it was finished :(

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nikhil97agra

9 months ago, # |

Finally I've recovered my -153 rating drop from Good Bye 2023 in this contest.

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aditya0610

9 months ago, # |

I loved Problem E,Great contest

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Welcome24ever

9 months ago, # |

← Rev. 2 →

very good D and E problem,makes my rating down down down.

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freakout

9 months ago, # |

フーヤをもらえますか

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drpav

9 months ago, # |

← Rev. 2 →

+16

Nice contest! Anyway here is my solution for problem F :
Let call cover[i] means the number of segment that cover point i. We can find this by using a multiset.

Iterate from 1 to n. Let's call dp[i] : Consider to position i, what is the maximum answer can we make

If we did not choose a point on i : dp[i] = dp[i-1]
If we did choose a point on i : dp[i] = dp[j] + cover[i] with condition : i and j does not lied in any same segments.
Here's how to find j. For all segments that contain point i, let's call L is the minimum left border of all segments. It's obvious that j = L — 1.

Hope it make sense!

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tOirz

9 months ago, # ^ |

TYSM, that really help me

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Kairm_Zaf

9 months ago, # |

Problems E & F were great! Nice contest! I enjoyed it!

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AvaraKedavra

9 months ago, # |

← Rev. 2 →

Why is this impossible in problem D ?

1
S
7S 3S

Since both belong to trump suit, and 7S has a higher rank, it can beat 3S. ??

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woonder

9 months ago, # ^ |

It is possbile, sixth line of output.

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9 months ago, # ^ |

It's not impossible

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sahil_k_28

9 months ago, # |

can someone explain problem B. Chaya Calendar

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saelcc03

9 months ago, # ^ |

This is the first case: 6 3 2 4 5 9 18 So the first sign occurs every 3 years, so the first time it occurs is in the third year. The second sign occurs every 2 years (2, 4, 6, 8... ), as signs must occur sequentially, the second sign will occur in the 4th year (the 2nd year also occured but the first sign didn't). So the ans will be like ans = 0 for freq in a: ans = ((ans/freq) + 1) * freq Doing this, you assure a year where all the ith signs have occured

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sahil_k_28

9 months ago, # ^ |

thanks

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perlentaucher

9 months ago, # |

Did anyone use DP to solve E? I usually don't use Python and thus had TLE at case 6. Submission

→ Reply

9 months ago, # ^ |

280ms

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saelcc03

9 months ago, # |

qué rico contest // nice contest

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Mehdi_Bachiri

9 months ago, # |

Is there a penalty for unsuccessful hacking attempts?

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Mustafa_Elghisha

9 months ago, # ^ |

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Mehdi_Bachiri

9 months ago, # ^ |

Thank you!

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ByteBattle

9 months ago, # |

How to solve Problem G?

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dartvolley

9 months ago, # ^ |

← Rev. 4 →

+25

Here is my solution:

For each passages, find the times in which the levels of the connected platforms are the same. To do so, you need to solve the equation:

$$$\begin{alignat*}{2} && l_u + t \cdot s_u &\equiv l_v + t \cdot s_v \quad&(\text{mod } H)\\ &\implies& t \cdot(s_u - s_v) &\equiv l_v - l_u &(\text{mod } H)\\ &\implies& t \cdot(s_u - s_v) + k \cdot H &\equiv l_v - l_u &\text{for some } k \in \mathbb{Z} \end{alignat*}$$$

You can solve this equation using the extended Euclidean algorithm. Define $$$C = \frac{l_v - l_u}{\gcd(s_u - s_v, H)}$$$. Note that if $$$t_0$$$ is one solution, then $$$t_0 \pm C$$$ is also a solution.

So, for each passage, you know you can cross it if and only if $$$t \equiv t_0 \left(\text{mod } C\right)$$$.

Then, you can solve the rest of the problem using a simple modified Dijkstra algorithm, where when crossing a passage you update the new distance as $$$t_0 + C \cdot max\left(0, \left\lceil\frac{t_0 - curdist}{C}\right\rceil\right) + 1$$$.

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EulerTour

9 months ago, # ^ |

Can you please explain this, we can find t using the extended euclidean algorithm but how will we get the smallest t ?

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dartvolley

9 months ago, # ^ |

← Rev. 2 →

If $$$d = gcd(a, b)$$$ and $$$(x_0, y_0)$$$ is a solution to $$$ax + by = d$$$, then the general solution to $$$ax + by = dj$$$ is:

$$$\left\{\left(x_0j + \frac{b}{d}k, \, y_0j - \frac{a}{d}k\right)\mid k \in \mathbb{Z}\right\}$$$

Thus, the solution with the smallest value of $$$x$$$ would be $$$\left(x_0j \text{ mod } \frac{b}{d}\right)$$$.

→ Reply

EulerTour

9 months ago, # ^ |

why is it x0j + (b/d)k and not x0j + bk, this will also keep the equation balanced, x0j + bk and y0j — ak

I am sorry I looked at many resources but no one explains this if you could please explain it

→ Reply

dartvolley

9 months ago, # ^ |

Honestly, I just memorized that as a fact without giving it much thought. However, if I have to try to justify it, it would go like this:

If we substitute $$$\left(x_0j + \frac{b}{d}k, \, y_0j - \frac{a}{d}k\right)$$$ in the the equation $$$ax + by = dj$$$, we have:

$$$ \begin{alignat*}{2} &&a\left(x_0j + \frac{b}{d}k\right) + b\left(y_0j - \frac{a}{d}k\right) &= dj \\ &\iff& ax_0j + \frac{ab}{d}k + by_0j - \frac{ab}{d}k &= dj \\ &\iff& ax_0j + by_0j &= dj \\ \end{alignat*} $$$

As for why we choose $$$\frac{b}{d}$$$ and $$$\frac{a}{d}$$$, it is becuase they are the smallest integers which can cancel each other out as above.

For example, note that $$$x_0j + bk$$$ also gives us a set of valid answers. However, they don't give us the full set of answers. More formally:

$$$\left\{\left(x_0j + bk, \, y_0j - ak\right)\mid k \in \mathbb{Z}\right\} \subseteq \left\{\left(x_0j + \frac{b}{d}k, \, y_0j - \frac{a}{d}k\right)\mid k \in \mathbb{Z}\right\}$$$

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EulerTour

9 months ago, # ^ |

Thankyou so much,this was just basic LCM calculation,i completely understood this now

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njmetro_jxzb

9 months ago, # |

Can O(NlogN) solution of problem C be hacked?

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I_need_7.0_IELTS

9 months ago, # |

-6

tourist used dp just to solve a Div3A lol 246990951

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tickbird

9 months ago, # ^ |

-8

ingenious troll, I have 0 clue why and how that even work 😂

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Pirate_King

9 months ago, # ^ |

Spoiler

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njmetro_jxzb

9 months ago, # ^ |

Me too :)

→ Reply

sneki3388

9 months ago, # ^ |

me too...

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Hyrulia

9 months ago, # |

B getting hacked like potatoes!!

→ Reply

amoad

9 months ago, # ^ |

do you know why ?

→ Reply

petertromso

9 months ago, # ^ |

← Rev. 2 →

If you try "brute force", by multiplying ai by 1,2,3... until big enough, you will TLE, for example if a1=1e6, and a2...a100 = 2, t=1000

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MijPeter

9 months ago, # |

Dunno about you all but this is imo very confusing descrption:

description

Why use words sequentially? Why use "strictly after". If there werre not example input/output I would think that you have to find years p, p+1, p+2, ..., p+n such that first sign happens on p, second sign happens on p+1 and so on.

→ Reply

9 months ago, # ^ |

I "solved" only by guessing pattern from samples

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AdityaTakkar

9 months ago, # ^ |

me too.

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AdityaTakkar

9 months ago, # ^ |

Still got hacked though .

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tlimitexceeded

9 months ago, # ^ |

I still didn't understood question B

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Ronak1808

9 months ago, # |

← Rev. 2 →

I tried solving C by calculating product of the range left after each operation using sqrt decomposition (247032959). It was giving TLE. Is there something wrong in implementation ? Because a same problem with similar constraints (243011064) got AC. I know it's not a good approach. It could be solved much simply but that's the approach I thought in contest.

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Atma_

9 months ago, # ^ |

I solved it using sqrt Decomposition and it got accepted. Your Implementation might be messy and slower.

Here is my submission : https://mirror.codeforces.com/contest/1932/submission/247026530

→ Reply

tlimitexceeded

9 months ago, # |

What was B, why was B, didn't understood it from question and explanation :(

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Code_Mystique

9 months ago, # |

If I have submitted 2 accepted solutions in Div 3. Will the later one be considered as my final solution and analyzed in system testing???

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Mahakal_17

9 months ago, # |

Why is B getting hacked?

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Matveyko

9 months ago, # ^ |

← Rev. 2 →

1000000 1 1000000 1 1000000 1... causes tl for solutions that try to linear search next year every time

→ Reply

Farsh_LE

9 months ago, # |

one of the best div 3s ever

thanks a lot

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RahulHarpal

9 months ago, # |

HackForces

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Halym2007

9 months ago, # |

← Rev. 3 →

I solved problem C with segment tree and problem E with lazy propogation.There's has easy solution,but yesterday my brain stopped, silly me.

→ Reply

ChenWenxin2009

9 months ago, # |

I'm joker.I directly used segment tree to solve C……

→ Reply

9 months ago, # |

Editorial when?

→ Reply

sagnn1k

9 months ago, # |

Was this round not rated? I did not recieve any rating

→ Reply

hars_raj

9 months ago, # ^ |

It is rated. Will take some time for the ratings to be updated.

→ Reply

Atma_

9 months ago, # |

MikeMirzayanov I have an optimization for system testing. Instead of testing on whole set of test cases again, test just on new test cases because anyway the submissions which are being tested have passed the pretests. This can potentially speed up system testing.

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JoeSherif

9 months ago, # |

Getting the system test done is like waiting for a snail to finish a marathon!

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xtraglitch

9 months ago, # |

A testcase for A question has been missed in the system testing as well. The case is when the staring point is a goldcoin.Many users dint write the code to cover this edge case.Please look into it.

→ Reply

CodeSmith

9 months ago, # ^ |

Starting or the ending point?

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xtraglitch

9 months ago, # ^ |

← Rev. 2 →

starting (the first cell)

→ Reply

9 months ago, # ^ |

It is guaranteed that the first cell is empty.

Lesson: Read the input section clearly & completely.

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xtraglitch

9 months ago, # ^ |

Ohhhh My bad, Leaving the comment here so that other people will get to know.

→ Reply

sukh_x

9 months ago, # ^ |

Bro your heatmap !! orz...

→ Reply

9 months ago, # |

Why "My submissions" is showing empty when i solved 2 problems yesterday and its showing my my rank in standings but not showing my submissions and contest also not showing in my contest area?

→ Reply

jping2022

9 months ago, # |

When will the rating update?

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Alamin2000

9 months ago, # |

← Rev. 2 →

I solved problem B and It got AC in the contest time. But after system testing it got TLE!! If it says tle in the contest time then I should debug that. But Now this will cause a lot of rating loss.

Very Disappointing

I participated in a virtual contest and didn't know that virtual was unrated. Thank you rahulsingh28082000

→ Reply

rahulsingh28082000

9 months ago, # ^ |

You solved B in virtual contest. How do you expect it to affect your rating ?

→ Reply

Alamin2000

9 months ago, # ^ |

Thank you for your reply. I didn't know that virtual is unrated

→ Reply

mj...

9 months ago, # |

→ Reply

9 months ago, # |

Anyone knows when Editorial will be out?

→ Reply

emanuel.sygal

9 months ago, # |

Why does this C++ segment tree for problem C get a TLE? https://mirror.codeforces.com/contest/1932/submission/247091172

The code copied here:

void build(vector<int> a, int v, int tl, int tr, int m, vector<int> &t) {
    int n = a.size();
    if (tl == tr) {
        t[v] = a[tl];
    } else {
        int tm = (tl + tr) / 2;
        build(a, v*2, tl, tm, m, t);
        build(a, v*2+1, tm+1, tr, m, t);
        t[v] = (t[v*2] * t[v*2+1]) % m;
        }
}
 
int product(int v, int tl, int tr, int l, int r, const vector<int> &t, int m) {
    if (l > r) 
        return 1;
    if (l == tl && r == tr) {
        return t[v] % m;
    }
    int tm = (tl + tr) / 2;
    return (product(v*2, tl, tm, l, min(r, tm), t, m)
           * product(v*2+1, tm+1, tr, max(l, tm+1), r, t, m)) % m;
}
 
void solve(){
    int n, m;
    cin >> n >> m;
    vector<int> a(n);
    for (int i = 0; i < n; i++){
        cin >> a[i];
    }
    string s;
    cin >> s;
 
    vector<int> t(4*n, 1);
    build(a, 1, 0, n-1, m, t);
    int low = 0;
    int high = n-1;
 
    for (auto c : s){
        cout << product(1, 0, n-1, low, high, t, m) << " ";
        if (c == 'L'){
            low++;
        } else {
            high--;
        }
    }
    cout << endl;
}

→ Reply

to_be_or_not_to_be

9 months ago, # ^ |

You should pass vector through pointer in the build function. That is, replace the following:

void build(vector<int> a, int v, int tl, int tr, int m, vector<int> &t)

with:

void build(vector<int> &a, int v, int tl, int tr, int m, vector<int> &t)

→ Reply

moonpie24

9 months ago, # |

+17

when's the editorial come out?

→ Reply

oneautumnleaf_

9 months ago, # |

When is editorial?

→ Reply

pmaddineedi

9 months ago, # |

Those of you from the US probably know that paragraph 3 of Card Game is pretty political...

→ Reply

Krrishchanchal

9 months ago, # |

When will the editorial be posted

→ Reply

9 months ago, # |

← Rev. 3 →

+17

pashka MikeMirzayanov Publish the editorial Bro.

→ Reply

PhuongVN

9 months ago, # |

← Rev. 3 →

excuse me, but can someone take a look at my solution for D? 247482543

Edit

→ Reply

ahnafrahmn

9 months ago, # |

Where can I get solutions of this round??

→ Reply

Banis

9 months ago, # ^ |

← Rev. 2 →

It haven’t been posted now(though it should have been posted), you can view other’s submission instead.

→ Reply

9 months ago, # |

Hello I think this has been a mistake:

Attention!

Your solution 247092052 for the problem 1932F significantly coincides with solutions Bobsy_is_back/247078909, bdyby10101/247085106, nohellodotcom/247085167, Sahilamin219/247086160, soumyadipdasmahapatra664/247086213, orazalyy/247087989, siddhantbhardwaj234/247088775, lender/247088979, nishkarsh1215/247089048, naked_soul/247091483, 3mk_leader/247091900, _Fer3on_/247092036, bgopc/247092052, orzdevinwangg/247092115, AkshitSharma/247092541, AJ_25/247092808, Ari10/247093633, JOO_91/247095387. Such a coincidence is a clear rules violation. Note that unintentional leakage is also a violation. For example, do not use ideone.com with the default settings (public access to your code). If you have conclusive evidence that a coincidence has occurred due to the use of a common source published before the competition, write a comment to post about the round with all the details. More information can be found at http://mirror.codeforces.com/blog/entry/8790. Such violation of the rules may be the reason for blocking your account or other penalties. In case of repeated violations, your account may be blocked.

That contest was unrated for me and I didn't even got in the contest in the start time and I didn't even care? why would I cheat when it's unrated? and the codes aren't even alike like WHAT?? The algorithm looks alike I agree but the code doesn't like why would I even cheat in an unrated contest???

→ Reply

ahnafrahmn

9 months ago, # ^ |

Sorry !!

Correct me please, Are you talking about anything I did? that solution you mentioned I am not even aware of the question!!

if I have misunderstood please let me know, I am sorry for my wrong understanding then!

→ Reply

9 months ago, # ^ |

→ Reply

ahnafrahmn

9 months ago, # ^ |

Thank you for your quick clarification!!

I am highly dedicated to our codeforces rules.

→ Reply

9 months ago, # ^ |

Idk, why would you cheat in the unrated contest?

→ Reply

9 months ago, # ^ |

I haven't that's the reason I'm appealing

→ Reply

9 months ago, # ^ |

Dude I asked you why, not if you did it

→ Reply

9 months ago, # ^ |

I didn't bro, there doesn't exist a reason for a thing that hasn't happened

→ Reply

9 months ago, # ^ |

Bro, again, I didn't ask you to explain cause-and-effect relationship, I asked why you'd cheated in the contest

→ Reply

9 months ago, # ^ |

I think u're not getting the point I DIDN't CHEAT so I don't have a reason to cheat in an unrated contest

→ Reply

9 months ago, # ^ |

So what is your final answer?

→ Reply

9 months ago, # ^ |

I don't have a reason for something I haven't done

end of conversation

→ Reply

9 months ago, # ^ |

So you cheated because you didn't have a reason to cheat? But that makes no sense!

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9 months ago, # ^ |

sounds like a u problem, end of the conversation

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9 months ago, # ^ |

Well, it wasn't really a conversation since you pretty much ignored my question

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9 months ago, # ^ |

Bro, When I haven't cheated, how can I answer it?

U think I cheated and asking me for a reason which I'd be doing the same in this situation, but I actually didn't cheat

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9 months ago, # ^ |

Ok, I'm done now, I was messing with you, for a reason ofc

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