Hello! On 18.02.2024 15:05 (Московское время) will start Codeforces Round 927 (Div. 3), the next Codeforces round for the third division.

**The round is based on problems from JetBrains Academy Youth Challenge. If you participated in it, please don't participate in this round.**

Problems for this round are prepared by denk, step_by_step, goncharovmike, ikrpprppp, pashka, Vladosiya and MikeMirzayanov.

Thank you very much awoo, BledDest, buyolitsez, EgorUlin, Gojova, GrandFruit, Hello_zoka, petyb, scanhex, senjougaharin, shnirelman, SomethingNew, Toy_mouse, Zandler for testing the round.

As usual for the third division rounds:

- there will be 6-8 tasks in a round
- round duration is 2 hours 15 minutes
- the round follows the ICPC rules, penalty for an incorrect submission is 10 minutes
- round is rated for participants with ratings up to 1600
- after the round there will be a 12-hour open hacking phase

Remember that only the trusted participants of the third division will be included in the official standings table. As it is written by link, this is a compulsory measure for combating unsporting behaviour. To qualify as a trusted participant of the third division, you must:

- take part in at least five rated rounds (and solve at least one problem in each of them)
- do not have a point of 1900 or higher in the rating.

**Regardless of whether you are a trusted participant of the third division or not, if your rating is less than 1600, then the round will be rated for you.**

Good luck to all!

UPD: Editorial

First

Hoping for clear problem statement

I hope*d* too

such a great weekends!

All The Best (Zoom out if not visible)

Did You meant to say " All the best???" not quiet easy to guess in desktop...

SpoilerAll I can see is ??L the BE

Yes, here's the expected look.

ScreenShot"The round is based on problems from JetBrains Academy Youth Challenge. If you participated in it, please don't participate in this round." Does this mean if someone participated in that one ,they would have some advantage in this round?(wanted to clarify it)

Yes, because the problems are the same? It's quite a big advantage if you have solved the exact same problems before

So, isn't it unfair?

Is it unfair if someone knows the problems beforehand? Of course it is, that's why those people aren't allowed to participate

How would you validate that the person hasn't participated beforehand?

but why to validate, the people are very honest here.

They won't participate

Nah, I won't believe that. In Fact, I am not honest. And there are many others just like me.

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

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

JetBrains Problems with Solutions

nice profile pic and bait)

I got a youtube ad running lol

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

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.

What is the score distribution?

In ICPC rules, all problems are worth the same.

So all are problems are rated 1000?

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

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)

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.

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

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.

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

,

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

,

What's your main?

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

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

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

,

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

,

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.

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

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

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..

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

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

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)

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.

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

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

bye

bye

bye

guys report him she is a he

as not a tester i would like to solve this round

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

F in 94:59.

This comment has been deleted.

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

Good start time for Chinese.

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

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

Originally you also play Genshin

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

I have a feeling this round will be amazing.

Me too.

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.

problems of this contest are prepared "step_by_step"

My 1st unrated Div 3 !(;

Nice. Wont hurt to be legitimate this time.

Cheater

I would target solving >75% problems.

Good start time!

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

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

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

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

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

Good lucks for everyone!

Good luck to all of my grey friends :D

Redemption time

extra registration ??

.

c Problem is very tough :(

You just had to iterate string backwards.

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

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 ).

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

I got PTSD because of Problem C

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.

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

Wow.good job man

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

hint: reverse LR string

Can you please elaborate?

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

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

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

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

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

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

Was F dp + segment trees?

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

Yeah, I too only used dp

DP: Yes. I did not use segment trees.

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

dp + sweep line + multiset

when will i can see the editorial ??

how to solve the E?

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

Thanks I'll try that

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

same 😥

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

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)

thanks

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

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

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

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

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.

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

went good for a first div3 unrated contest

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

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

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

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?

I am stupid

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.

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

US moment.

same

This line is definitely the problem:

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.

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...........

tourist definitely struggled a lot

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

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

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

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

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)).

ohhh

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

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

what the fuck is b and c

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

Such bad problem statements this time.

problem C was quite pleasure to solve

I felt like being Elegia

How to solve C?

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

Thanks

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

Thanks

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

solve from the back

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

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

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

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

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.

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

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?

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.

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

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

backtracking

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

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

Check my submitted solution it is using backtracking

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

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.

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.

Store in map

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.

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

I used this comparator function to sort all the cards

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

you can refer my solution Problem D

Spent a long time to understand what

problem BwantsYou too?

+1

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

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...

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

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

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

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

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?

Because long long will overflow

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.

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

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

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?

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

gonna make integer overflow

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)

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。

You clearly understand it already

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.

Whats wrong with the naive implementation of problem C?

probably because of overflow

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.

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

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.

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

why greedy does not work for F? :(

how you suppose to do it greedily?

2 4, 5 7, 10 11

2 6, 7 7

4 5

Try this input:

1

7 5

7 7

1 3

6 6

3 7

5 7

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

A great counter example! thanks!!!

C problem is really hard for Div3 XD.

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

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

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

C++ CodeIt’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.

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

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

I solved it with sqrt decomposition

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!

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

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

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

TLE submission

AC submission-_-

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

Without % -> This

With % -> This

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 :(

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

I loved Problem E,Great contest

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

フーヤをもらえますか

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

Hope it make sense!

TYSM, that really help me

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

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. ??

It is possbile, sixth line of output.

It's not impossible

can someone explain problem B. Chaya Calendar

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

thanks

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

280ms

qué rico contest // nice contest

Is there a penalty for unsuccessful hacking attempts?

no

Thank you!

How to solve Problem G?

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:

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 $$$t_0 + C \cdot max\left(0, \left\lceil\frac{t_0 - curdist}{C}\right\rceil\right) + 1$$$.

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

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:

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

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

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:

As for why we choose $$$\frac{b}{d}$$$ and $$$\frac{a}{d}$$$, it is becuase they are the

smallest integerswhich 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

fullset of answers. More formally:Thankyou so much,this was just basic LCM calculation,i completely understood this now

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

tourist used dp just to solve a Div3A lol 246990951

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

SpoilerMe too :)

me too...

B getting hacked like potatoes!!

do you know why ?

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

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

descriptionAccording to the legends, for the apocalypse to happen, the signs must occur sequentially. That is, first they wait for the first sign to occur, then strictly after it, the second sign will occur, and so on. That is, if the i-th sign occurred in the year x, the tribe starts waiting for the occurrence of the (i+1)-th sign, starting from the year x+1

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.

I "solved" only by guessing pattern from samples

me too.

Still got hacked though .

I still didn't understood question B

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.

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

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

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???

Why is B getting hacked?

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

one of the best div 3s ever

thanks a lot

HackForces

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.

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

Editorial when?

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

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

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.

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

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.

Starting or the ending point?

starting (the first cell)

`It is guaranteed that the first cell is empty.`

Lesson: Read the input section clearly & completely.

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

Bro your heatmap !! orz...

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?

When will the rating update?

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

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

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

wo

Anyone knows when Editorial will be out?

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

The code copied here:

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

with:

when's the editorial come out?

When is editorial?

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

When will the editorial be posted

pashka MikeMirzayanov Publish the editorial Bro.

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

EditI found the part I did wrong, if g[x].size() is odd and i==g[x].size()-1, it should break and not iterate anymore

Where can I get solutions of this round??

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

Hello I think this has been a mistake:

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???

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!

no

Thank you for your quick clarification!!

I am highly dedicated to our codeforces rules.