Hello Codeforces!

I'm glad to invite you to participate in Codeforces Round #665 (Div. 2) taking place on Aug/21/2020 17:35 (Moscow time). The round is **rated for users whose rating is lower than 2100**.

All problems were mainly created and prepared by me, and I'd also like to thank:

adedalic for coordinating and reviewing the round,

AmShZ, growup974, dorijanlendvaj, Osama_Alkhodairy, Mohamed.Sobhy, awoo, kdjonty31, TechNite, BledDest, Saurabh_yadav, _cherry_, errorgorn for testing the problems and giving feedback,

MikeMirzayanov for Codeforces and Polygon platforms,

All users for participating in the contest!

You will be given **6 problems** and **2 hours** to solve them. The score distribution will be announced closer to the start of the round.

Good luck, and have fun!

**UPD1** : Score distribution is **500** — **1000** — **1500** — **1750** — **2500** — **2500**.

**UPD2** : Editorial

**UPD3** : System testing has finished. We hope you enjoyed the contest :)

**UPD4** : Congratulations to the winners!

Div.1 (Out of competition)

- Proszek_na_ludka
- Egor
- jiangly
- Geothermal
- dlalswp25
- antontrygubO_o
- I_Remember_Olya_ashmelev
- Maksim1744
- Heltion
- SSRS_

Div.2

adedalic coordinating round will be great!

35 hours before the contest and this still isn't on the home page. Hmmm...

That's the charm of secondthread.........lol

The blog was running on SecondThread ;)

Finally kdjonty31 become tester. congrats ! :3 link

Yay! Thanks. LMAO.

ᕕ( ՞ ᗜ ՞ )ᕗLove it. I keep my fingers crossed for you.

6 tasks for 2 hours and testers from different departments |=> well-prepared competition.

Thank you to everyone who took part in its preparation!

Successful writing!

Hopefully, the problems are as balanced as the color of testers.

will be great round definitely :D

As a Tester, I would like to say that the problems are really interesting, also the statements are short too. Please upvote me now! :)

Love Short Statements

Spoiler![ ](

Hope for strong pretests and not shitty problems from Lemonade255!

Hard to hope for the latter

I cracked so hard and wanted to upvote your comment. But me being a man of culture did not want to destroy the auspicious number of upvotes on your comment :)

it went a little above. Had to downvote it. Sacrifices have to be made for greatness

Underrated comment.

<Mandatory tester comment./>Also, the problemset's really nice and concise.Congratulations..

Thanks lol!

＼(^o^)／As a tester I would say that a good strategy will lead to positive rating.

Emphasize on "good strategy" part please

i think he meant to say A and B are tricky so try out with C and D.

August:

In India — the month of holidays

In codeforces — the month of contests

We already have six contests and three more to go.

Thanks Mike and codeforces.

Augustforces

I thought the next competition was in three days, but suddenly there was a competition tomorrow.I'm looking forward to it!

As a master, I hope you good luck and high rating!

As a master, give me contribution, please.

I gave you contribution.

As a master,saying something is better than contributing something.

WEll, How many contest holds in every month, at least ??

As a tester i advice you to read all problems :)

is it rated?

???

The round is rated for users whose rating is lower than 2100.Can you read the announcement ?

This round is gonna be an amazing

I can relate xD

I can feel you bro, you can check my profile I've reached 1590+ many times but never crossed 1600, 1600+ rating is still a dream for me

I hope You’ll reach blue after This round

Are you a digon?

Oh boy! so we have a monogon here too it seems

Where do you learn these weird polygons ? I never came across these in my school/college.

But can monogon be also belong to polygon? :P

You are at 1599 right now, coincidentally, your profile picture also has a point on the border of the circle.

I'm in dilemma whether to wish you luck or not. If you cross 1600, your graph will lose the charm.

This same thing happened to me also

I can relate bro, I was closer than you xD

Hope for strong pretests.

protests are really strong these days

what is this meme ?

It's Pretest, not Protest. — the first comment hoped strong protests.

Русские поймут =)

Понять-то они поймут, но здесь не политический митинг.

As a tester, this contest made me lose my will to live.

That makes me worried!

I hope that doesn't happen to the contestants after the contest..

But it did not kill your craving for contribution.

Glad to see specialists are also being accepted as problem setters. :)

*testers

You were a tester yourself weren't you?

YES Contigo KHALIYAR

Can I ask a question? What type of problems are you best at solving?

I wish I could say "relationship"

![ ]()

Korean contest! I love it! 반가워요 :)

오 한국말이 되네요

south or north ?

South! i've never seen North Koreans in codeforces yet

There are north Koreans in the ranking though.

As a tester I recommend you to read all problems and wish you big positive deltas!!!

Thanks, really did helped as I skipped B to solve C first

hey bro nice nick

Let's hope the problem statements can be as short as this announcement.

Good luck everyone ;-)

Am I the only person who finds the username Lemonade255 weird? Its like saying "Hi, I am Anus No. 1373"

P.S No offense intended though

Lets just hope that shit doesn't come out.

All hail to you brother! Keep rocking! kdjonty31

LMAO, Thanks!

;)[..]

wow! there aren't too many testers...

Seeing the handle name anus, I first thought it's an Indian round! The name anus sounds like Indian though!

Not as a tester, give me contribution

I had to post this, for 71 minutes the number of comments hadn't changed, a conspiracy.....???

Haha funi sex number i laughed, came, and had an epileptic seizure upon seeing this post. It is in fact so hilarious that after I showed it to my entire family they nearly died from asphyxiation because they couldn't stop laughing at this quirky post. I am currently at my mother's funeral, and the priest is unable to keep a straight face when he reads about the cause of her death.

I'm just happy my mom died with a smile on her face :)

Just asking, what's the point of the 'x' button if I can't remove them?

picJust asking, Why do you want to remove these?

Why not? It's good to have a choice and I'd personally feel better after removing some or all.

You could remove them if you didn't make any submission.

I am scared about 5 page long queue . I hope MikeMirzayanov will fix it before the contest .

UPD — It is fixed now !!Not 5 pages but hundreds of pages .. no submissions are judges for last 4 hours(which i saw ).

Where can you see the queue?

In the NavBar above — Problemset->Status.

Got it. Thanks.

As a tester, this contest will be interesting and entertaining. Thanks, Codeforces.

Donald Trump.

Only two hours left before the contest,

score distributionshould be published. Lemonade255wishing all coderforces family luck and fortune . May god bless you all @Erica

I feel most of the downvotes are from newbies and pupils

I also guess so , bcuz Good coders always enjoy everything ... They are cool people

.

This meme isn't new(actually, it's common), so it's likely to get downvoted.

okay thanks , I removed it. I didn't know that, As I am new on Codeforces.

WOW! E=F, WILL TRY BOTH!

Hell yeah..

Yeah!

speedforces

I suppose everyone here wants to learn. So the test cases must help us to think regarding the problem.

Very nice set of questions. :-)

This is how you make a round.

One of the most stable and decent question sets.

When I'm sure my logic is correct but WA on pretest 2 (x4) @@

Problem 4

Problems are good but somehow I choked so hard on B

even i felt C was easier than B

I made B & C, but couldn't do A. laughing after looking A's solution

At least you're not choke hard on "D" *IYKWIM

is it educational round or what? why problems havent any idea

What do you mean?

It's most possible boring math problems.

Problem D which decided the rank of majority was standard re-rooting. And I guess problem F was some modification of segtree but I haven't looked very seriously into the problem.

D was greedy and not re-rooting.

Of course it was. But at least in my solution, I used re-rooting to find edge weights.

You could have done it like this-

Ah yes of course.

#Always_ready_to_overkill

+1, I understand that maybe this problems can be Educational for Div2 but they seem uninteresting.. :(

you come back please !!

I prefer this problem set. Only one I really thought was too boring was D, but better than having non-diverse round. ABC were even an ur style problems...

For me A was nearly as difficult as D

Will have to check the editorial to understand it lol

2b+k=n find b

Imo C was easier than both A & B.

can you give a hint ?

Just sort the original array and check to see which ones are out of order, if all the ones out of order are divisible by the minimum element the answer is yes otherwise no.

I dont think sorting the array fits under the given constraints ( N<= 10^5 , T <= 10^4)

There is another constraint about max sum of N over all testcases.

SHIIIIIIIIIIIIT

What is wrong in this https://codeforces.com/contest/1401/submission/90583034

You need to compare objects using equals, not !=, even if it is Integer objects.

Thanks bro, But other answers using != are accepted

Maybe they use int[] instead of List ?

Yes

Just curious to know why it runs well on 7 5 2 2 4 but not on 1120 707 16

I don't know. Numbers smaller than 128 are somewhat special in Integer class, because Integer.valueOf(int) will return same Object for int<128. Maybe somehow that comes into play.

Make a copy of the given array and sort it. Iterate from 0 to n-1 and check if a[i] != b[i], if it is different and a[i] % minValue != 0 then there is no solution.

and I can't solve C, I'm too bad :(

After getting 4 WA in D, here are the cases to take care of.

Don't take mod of last greatest values when m > n — 1 and insert in list because mod might make the value smaller. Instead take product of those and add to our answer ASAP.

While computing how many times edge can be part of various paths, don't forget to take it as long long and also list as long long. And don't mod because again value might become smaller.

oh fuck I think I fell victim to both of these... RIP

Oh wow, I think now I know why my solution get WA

Shit..Too bad I couldn't figure out during the contest :(

for point 1, how about sorting p array first, then assigning the values, that way the array is always in sorted order, and no need to care of mod making the value smaller.

Can D be solved by the greedy method?

Yes

Yes, I think the author's solution bases on Greedy approach

Yep, greedily assign the maximum weight to the most used edge (found using dp to get size of subtrees)

I am also interested to know.

Thanks, so it must be some silly mistake

hint —

DFS + SORT

Don't give hint , either tell full solution or don't , we can wait for editorials .What you are telling everyone knows.

My approach —

find number of times an edge is coming in a path.

(n1 elements)--(edge)--(n2 elements) => number of times = n1*n2 (I found n1 = number of elements in subtree of n1 and n2= n — n1)

and then fill max factor in maximum edge.

my sol — 90611208

I have used iterative DFS, because I thought I might get TLE.

Sorry to hurt you feeling Jastin

My feelings are already dead . Just see my contribution . Any way i found the solution and it's same as yours .Thanks for helping me to verify .

Completely exhausted after the contest. I got stuck in the 4th problem. There may be some property that will be used. A lot to learn this from contest. A NICE CONTEST with short problem statements!!

How to solve D :(

Wasted all the time with problem D pretest 5, finally I found that i am sorting after taking the numbers module mod :(

Same here :(

At least you realized it during contest ._.

Holy shit, I'm dumb af

LOL so glad to know I'm not alone

I had the same problem too but wasn't able to notice this. I even solved E mentally and would probably have had time to implement it if it weren't for this mistake. :(

I should have quit it too, but it's really hard to leave a problem solved by about 2K users and go to other harder problem :(

I've done this but still got WA5 90613994 :(

`z.push_back((n - s[u]) * s[u]);`

Overflow will happen here, as both $$$n$$$ and $$$s$$$ are declared

`int`

.lol, thank you!

Thanks for this comment, wasn't able to find the bug in my code until i saw this

ahhhh thanks! after 22 times getting WA on test 5 i just saw your comment. god damn it :(

Video Tutorial for D. Maximum Distributed Tree

Problems, at least problem statements, were too mathy/formal to my taste

It was pure math homework contest. I am surprised as how much people seems to like that kind of problems.

WitchOfTruth and spookywooky if you want to be good at competitive programming , trust me soling math problems will help you a lot .

Math problems per se may not be that bad, but here we had A-B-C three math problems in a row, And then D, which was not math, but the problem statement was so dry and full of formulae that it felt like one.

Great contest...

A round with two data-structure problems!!

And both of them will most likely have difficulty 2000+. So for most of the real div2 participants there were no ds problems.

Explain solution for d please?

F — press segment tree to win

But how to do? Lazy-tag can't work!!

Maintain left and right childs and a tag for each level if they have to be swapped. Reverse operation is same as swapping at all levels below current level.

Observation — order of the operations doesn't matter, so just write amount of each operation for each k. So when pushing from vertex, just look at the number of operations for the current level k. Swap is just swapping children, and Reverse can be done lazily.

Alternatively, notice that all operations are xoring the indices and segment tree works really nicely with xor.

Is it obvious?During the contest I doubted whether the order matters .:(

What was the logic behind C

The numbers which are misplaced must have their common gcd equal to the smallest element in the array. Edit: Just have to check that all misplaced numbers are divisible by the smallest number or not.

Consider the sorted list. Compare to actual list. In the places which differ, check if modulo min of the list.

you can always sort the array as long as all elements which you want to move is multiple of min element of array. because if that's not the case, gcd(element,min element)!=min element. so you can't change its position.

Shit!!!Got where my logic was wrong.Btw Thnax!!

Could E be solved using a linesweep and avl tree?

Today for the first time I had to implement Segment Tree from scratch lol.

lol!

Yes problem A took me 11 minutes!!

Solved B & C, couldn't solve A :P

I'm curious that how one would solve C? if the gcd of two element of array is not needed to be the minimum element of array.

Consider the sorted list. Compare to actual list. In the places which differ, check if modulo min of the list.

I was stuck with this logic. But say array is 2, 8, 4 and you want to swap (8, 4), you can just swap (2, 4) (8, 2) and (4, 2) again.

Basically 4, 8, 2 -> 4, 2, 8 -> 2, 4, 8.

if gcd can be anything (i.e. no conditions on gcd) then according to me its always possible to sort the array, just like swapping every possible pair.

so true..then the problem will not even be considered as prob A

no I didn't mean that gcd can be anything it's like that the gcd should belong to the array but it's not need to be the minimum element of the array.

if you have one element whose gcd is always equal to min with others (like — minimum number) and those numbers if gcd != min are in right position than you can always arrange that array.

ex — 2, 6, 5, 4

if min = 2, and you want 4 to be in 2nd position, but gcd(6, 4)!=min, then you can

swap —

2-6 => 6, 2, 5, 4

2-4 => 6, 4, 5, 2

2-6 => 2, 4, 5, 6

what's wrong in this? https://codeforces.com/contest/1401/submission/90616241 (problem D)

Test case:

Correct answer: $$$1555$$$

Your answer: $$$95$$$

Why the answer is 95? we give the following value: 11*7*5 in the edge between 2-3, 3 in the edge between 3-4, 2 in the edge between 1-2; then answer should be 4*(11*7*5) + 3*(3) + 3*(2) = 1555; Correct me where I am wrong!

Yes, the correct answer is 1555, but instead of do 4*11*7*5 i did 4*11+4*7+4*5

you should consider the cases in which m is larger than n-1.

For C,

I made a vector of elements which had gcd(minOfArray, element) = minOfArray. sorted them and then placed them again in the array in the places where the above condition holds in increasing order. Then checked if array is sorted.

Can't understand the fault in the logic!!

Yah EDIT: Actually logic is correct, I declared the minElement globally. I was doing it for the first case. Forgot to shift it inside the loop.

Like if you sorted after taking mod

I even didn't do that stil got WA. Please help to debug.

UPD:No need i got what i did wrong.SpoilerWhat if m>n-1

Thanks, I wish I would have realized this during the contest.

I kept on getting RTE on test 6 of D. Thanks a lot to python!

https://codeforces.com/contest/1401/submission/90611250

Can anyone tell who user SorryNotSorry (who got the first rank) might be by his coding style?

GCD is resounding in all directions ——when i solved D but not C and only 30 mins left

I am beginner here on codeforces can anybody tell me if my submission will be counted, I submitted my file at the last moment and it ran on the pretests and showed 'pretests passed' after which the time ended, will my code be further evaluated so as to get accepted?

yes it will be further evaluated, if you submitted before the contest ended.

thank you!

Yes

thank you!

Solved F first and didn't have enough time to debug E :(

have you left a question because you just don't want to type that?

solution of C https://codeforces.com/contest/1401/submission/90579190

Without Reverse and Swap last problem is easy to solve)

Thanks for the good samples in F though

Can anybody tell me why my solution for D is failing on pretest 5? It looks like the logic is correct, comparing it to people who got it accepted. Submission link

Alright, so I realised my mistake, one of the arrays I had sorted after taking modulo. I crie. TT

What is the correct approach for problem E? I guess there would be some magical trick.

wingardium leviosa

are you really Dr Vishal Gupta from pilani?

Yes indeed. Why should students have all the fun.

indeed.I was just little surprised to see you here after so long, also because you are only prof I know here. enjoy :)

Please help!

Can anyone tell me why I'm getting TLE on problem D? I switched to c++ from python recently and I really thought these things won't happen anymore :<<<

I really feel my solution is perfect.

Is it about memory allocations? Should I have created an array for the adjacency list instead of creating a new vector or something?

One of the submissions: 90617571

Main part of the solutionI think you should put vvi &adj instead of vvi adj in dfs()

You think that one copy made all the difference? Oh.. As I'm writing this comment, I realize every dfs copies it, because it's recursive :|

Well it was the first dfs problem I solved since switching to C++. There have to be some bumps in the road I guess.

You need to declare vectors given as parameters as references. Here, you copy the array adj each time dfs is called.

For problem D, do we have to greedily arrange the m factors on each edge based on which edge is most visited?

Yes!! But how will we know which edge is visited most? I was stuck in this part only!!

It will be equal to number of times a particular edge xan be used in a path from u to v which will eventually the product of number of elementsin the subtree and other the nodes As that edge will come only if you have a node other than the nodes in the subtree and the other node is in the subtree.

What's wrong with my solution for C? 90603416 I created a copy of the given array and sorted it. Next, for each index, if the elements in the original array and the sorted one aren't equal, I checked if their gcd was equal to the minimum element. If it is different, I print "NO", else "YES".

even I did the same :(

If the arr is 2, 8, 4, then the sorted array is 2, 4, 8. GCD of 4, 8 isn't 2, but it's still possible to fix this. Swap 2 with 4, swap 2 with 8, swap 2 with 4.

Try the test case consisting of 3 elements, 16 2 8. Your code will fail. Correct answer is "YES". __

You need to check if it is multiple of min element, because that is enough. Every two multiples can be triple-swaped with each other using the smallest element.

I felt like a math exam, didn't like this contest at all

Amazing round ! Clear statements with no stories, hopefully strong pretests. This has to be one of the least constructive rounds I have seen in recent times on CF :)

EDIT: And fast editorial as well :)

Great contest :) No shitty stories

what a contest!!

In C, i just took all elements which were not in the correct place after sorting in another array arr2, and found the gcd of arr2. whats wrong in this??

If we have the array = [2,8,4]. Your gcd array will be [8,4] and its gcd is 4 which isn't equal to 2. But still, the array can be arranged in non-descending order.

First: Find minimum element.

Second: Find elements that not in their place.

Third: Notice that we can swap any 2 elements using minimum element as a buffer, BUT, we wont be able to do that, if one of this two elements not divisible at minimum element.

So, u just need to check for every elements that not in his place, if it divisible by minimum element. If all is divisible: "YES", otherwise: "NO"

Query Party!

Stuck in Problem A. Orz

how to solve c????????/ please help

I won't tell you how to solve it directly, but I can give you 3 similar problems to today's C:

1) 432C - Prime Swaps

2) 1365B - Trouble Sort

3) 1148C - Crazy Diamond

my solve: we can never move elements, which are not divisible by min_of_array, so we should check if they are in their "sorted" place. Lets call other elements divisible. after that we can always sort the array with this logic: gcd(min_of_array, any divisible) = min_of_array, with that we can swap for ex min_of_array with the last element and then with min_of_array with max, with that iteration we decreased size of our array to sort by one (we sorted biggest divisble in its right place) and can continue with that logic.

also good problem to practice this kind of thinking is: Replace by MEX — https://codeforces.com/problemset/problem/1375/D

First: Find minimum element.

Second: Find elements that not in their place.

Third: Notice that we can swap any 2 elements using minimum element as a buffer, BUT, we wont be able to do that, if one of this two elements not divisible at minimum element.

So, u just need to check for every elements that not in his place, if it divisible by minimum element. If all is divisible: "YES", otherwise: "NO"

By "swap any 2 elements using minimum element as a buffer" I mean this. U have a1 a2 ... aN . A1 is minimum element (aM), GCD(aM,aX) == aM. So, if we want to swap aX and aY. We can do like this:

[aM,aX,aY]->[aX,aM,aY]->[aX,aY,aM]->[aM,aY,aX]

Could anyone help me,why is my solution for D is TLE. I think my answer is perfect. 90615633

pass the "tre" by reference in dfs function.Because in your code the vector is copying itelf everytime dfs is called

Thank you.

Had the exact same problem. Need to pass 'tre' as a reference (or do like all the experiences ones and declare it as a global variable).

Every time you call 'dfs' you copy the whole vector.

This was the first time I solved 2C and still am able to get a +3 according to predictor. :(

Can someone please tell me why this is giving RTE. link

In dfs function you are using int val = sub[x] * (n — sub[x]); but you shoud use long long int as it (sub[x] * (n — sub[x])) may exceed the limits of int

int is already defined long long in macros :)

Okay, I didn't see that earlier. But now I have found the mistake. for(int i = n — 1; i >= 0; i--){ cal[i] %= mod; factors[i] %= mod; int temp = (cal[i] * factors[i]); temp %= mod; ret += temp; ret %= mod; }

In above for loop you should start i with (n-2) not with (n-1)

okay got it...Thanks. But why my code didnt fail on pretest 1?

I am also wondering, why it was not failing on pretest1. But I didn't came up with any answers.

Can anyone detect why this code gave WA on pretest 2 in problem D , I am not able to understand the issue in it. 90570026

PS:- Finally done , I was missing case when m>n!

Off 5 blue participants in my friendlist 4 did not submit any solution. And the one who did won't be blue next contest.

how to check who took part and didn't submit?

On the contest page you can click the number which shows the participants to get the list of all registered ones. Then there is a friends button.

Same friends button is in standings table.

Yeah and they are the people who will fail when time comes due to lack of courage. I have many a times solved first problem in > 1 hour and then B,C,D in < 1 hours . If you really love CP you won't see al these things.

solved C but got stuck at B :(

Any small hints for problem E?

The editorial is the smallest hint as well as the complete answer .

problem c: n=5. 8 8 4 4 2 why the sample is correct? (this smaple printed "yes" in some codes)

Because 8, 8, 4, 4 are all divisible by 2 (the minimum), so they can easily be swapped to create a sorted array.

ok,i think so.I get it.Thank you.

For C , If input is n = 5

2 12 8 32 24

min = 2, gcd = 4, My ans : "YES" passed, actual ans should be "NO"

Answer is YES. cause using 2 you can swap any element .you have to find just 2 element gcd not all..

you can swap if two elements if their gcd is equal to min element.

Did anybody else feel this contest was so smooth? I mean.. I got my "Wrong answer on pretest 2" the second after I made my submission. Kudos to the platform! Jokes apart, it was nice problem set!! Thanks overall!!

SO, what does the "1-s" mean ??? I didn't get it, so I lose Problem D. Can't you just write directly " 1s " or " '1's " ?

By the way, the other problems are good. Thanks for your contest.

Thank you very much for uploading editorial this fast!!

To not keep you waiting, the ratings updated preliminarily. In a few hours, I will remove cheaters and update the ratings again!

Those who are getting WA in D like me m can be greater then n , it have to be considered! I got WA in 2 facepalm!

During the contest, I submitted solution for F, but quickly realized that it works in $$$O(2^n \cdot q)$$$, though it was easy to fix so I resubmitted. But then I submitted initial slow solution after the contest again, and it passed systests! So I quickly hacked it 90619845 But I'm quite disappointed that systest didn't manage to break my initial slow solution(

when my rating was increased in last contest they updated it afer 6 hours or more and today my rating was going to decrease they updated it inside one hour

why?

Hello I've passed pretest 1 of first problem of this round, but failed on pretest 2. May I know where was I wrong? ~~~~~

## include<bits/stdc++.h>

using namespace std;

int main(){ int no_of_times; cin>>no_of_times;

for(int l = 0;l < no_of_times;l++){ int n,k; cin>>n>>k; if(n == 1 && k == 1){ cout<<1<<"\n"; } else if(n == 1 && n > k){ cout<<1<<"\n"; } else if(n > k){ if(n%2 == 0 && k%2 == 0){ cout<<0<<"\n"; } else if(n%2 == 0 || k%2 == 0){ cout<<1<<"\n"; } else{ cout<<0<<"\n"; } } else{ cout<<(k-n)<<"\n"; } } } ~~~~~

What becomes problem C if instead of having the rule "you can swap elements if their gcd is equal to the minimum element of the array" you have "you can swap elements if their gcd is equal to x" where x can be any number (in particular x may not be in the array) ?

My screencast of the round, where I start to give up on templates, horribly screw up D, continue my curse of never full-solving, and explain solutions for A-E

I have a new record !!! 2000+,I am waiting for your congratulations!!!!

Congrats!

Congrats!

congrachulasunzz

No announcement for any problems in the round is the best evidence to prove an expilcit and clear round.

Amazing Problem statements. Loved problem A very much. Fantastic round!!! Kudos to authors and Mike for fantastic round.

It was unrated for me — rating change 0. lol

Same as 72 other people

Problem C has nothing to do with the greatest common divisor. We only need to consider whether each number not in the correct position can be divided by the smallest number in the sequence.

You are saying that it has nothing to do with gcd just because you figured out how to solve it without explicitly finding any gcd?

for problem A

2∗x=a+k

For what minimum value of a , x<a ? k is constant (can be either even or odd). x , a , k >= 0

does the above equation has any mathematical formula?

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