Hi everybody,

This Sunday there will be a Moscow Olympiad for Young Students. This contest is prepared by Moscow Olympiad Scientific Committee that you may know by Moscow Open Olympiad, Moscow Team Olympiad and Metropolises Olympiad (rounds 327, 342, 345, 376, 401, 433, 441, 466, 469, 507, 516, 541, 545, 567, 583, 594, 622, 626, 657, 680, 704, 707, 727, 751, 775, 802, 829).

The round will be held at Feb/12/2023 11:35 (Moscow time) and will last for 2 hours. **Note the unusual time of the round.** The round will be held according to the Codeforces rules and will be rated.

Problems of this competition were prepared by Tikhon228, TheEvilBird, Ormlis, vaaven, Gornak40, Honey_Badger guided by vaaven, grphil and Helen Andreeva.

Thanks to Artyom123 for the round coordination, and also thanks for MikeMirzayanov for systems Codeforces and Polygon, which was used to prepare problems of this olympiad.

Thanks to BucketPotato, KseniaShk, NemanjaSo2005, MagentaCobra, kuertov, prvocislo, 4qqqq, vsinitsynav, FynjyBath, qualdoom, ivanlarin, didedoshka, petyb, PBBx0, MIKHANGO, charlyk, RP-1, talant, ak2006 for testing.

Due to the official competition source codes of other participants will not be available for an hour after the end of the round.

Good luck everybody!

**UPD:** Score distribution: 500 — 1000 — 1250 — 1750 — 2500 — 3250

**UPD2:** Еditorial

`The round will be held according to the Codeforces rules and will be rated for both divisions.`

Shouldn't it be rated for Division 2 only?Shhh, I’m desperate for a rated contest

Like me!

The feeling when you get a negative in recent contest and want to bounce back.

As a tester I can say that I am a tester (again). Don't hurt me, I hope this time everything will be fine.

testers are so cool!

Thanks, hope we don't let you down!

if FynjyBath = tester,

round = bado.O

Nice avatar

thanks(

Forgot to switch the account back?

thank you for -20(((

As a tester, I enjoyed the problem set.

What will be the distribution of points?

Sorry, saw your message late. It is not testers who decide that. As you can probably see, the author made an update to the blog, the distribution is: 500 — 1000 — 1250 — 1750 — 2500 — 3250

Aiming for first 3 problems then!

Let's see how it goes :)

I just could not figure out what was wrong with problem A.

This was my solution:

`ll promotions_available = (n * m / (m + 1));`

should be`ll promotions_available = (n / (m + 1)) * m;`

can you explain it?

I have been thinking about it and when drawn out on paper, it is correct.

But we want to get the number of (m+1) groups that can be fitted inside n. That is (n / (m + 1)). Then I multiplied by m to get (n * m / (m + 1)). But we use integers and we want to multiply the number of groups with the cost!

As it is now (n * m / (m + 1)) would be rounded down which is not what we want! That is what helped me to understand why this failed when I retried it multiple times.

TLDR; Math checks out but we deal with integers and roundings. (Correct me if I'm wrong @nor)

I found some errant cases for you.

TL; DR -> a = 6, b = 5, n = 9, m = 4, expected output = 44; your output = 45

You can use the following code for the same.

Testcase GeneratorAC Code (you can use any other as well)Debugging ProcedureRun the generator and save the input in a text file. Use that input to get output from your code, as well as AC code.

Compare the two and debug. Note that this can be done efficiently with scripts. I don't have windows scripts on hand right now, but it's easy with Linux.

Thanks!

Very Good!

As a tester I can say that problems are very interesting. Have a good luck on this round:D

i can say problems are very hard LOL :)

Note the unusual time i.e. 6hrs before general time.

is score distribution going to be announced before the contest?

also can it be delayed ? it clashes with Innopolis finals time

It can't be delayed since it has to be held during Moscow Olympiad for Young Students.

The name of the writers had not been updated yet in the Contest Section.

Note the unusual time of the round.

Fingers Crossed for +delta (◠‿◠)

Don't cross your fingers, it will be difficult for you to type during contest. (◠‿◠)

lol

I will solve this olympiad. Good luck to everyone at the round and the olympics!

Friendly time for Vietnamese like me

And Chinese like me

I'm also Vietnamese!!!

eagerly waiting for a rated contest . Good luck everyone !!

is this china mathforces round or am i safe?

I am looking forward to the problems of this contest, hoping to surprise me.

I hope we can enjoy ourselves in this contest.Make it Sunday, February 12, 2023 at 16:35(UTC+8)

This is at midnight for me. Hopefully I don't brick it :/

I have classes T_T... But still good luck awa

i guess it will be speedforces round

Hoping to have a great round with great learnings!

hope the best for all and to get my cyan color today :)

Edited okay I think I will wait for my cyan for another day :D

same. cant wait to demote to newbie bro

Good luck everybody, hope problems will be interesting!

Is the other contest still ongoing?

Such a permutationforces.

The problem F is the same as 765F.

My code was copied from https://www.luogu.com.cn/problem/solution/CF765F.

Edit: Please don't upvote this comment.

I don't understand why you upvote this. Firstly,if you notice that a problem is similar with another one in the past,keep it for yourself and don't spoil the contest. Second of all,saying that you copied your code from a solution source is against the values imposed by codeforces(ig).

A funnily embarassing contest for me -- A >> B > C for me. I solved the problems in the order C, B, A and got 5 WA on A before getting it. My math solution had a bug I couldn't find, so I tried to binary search the amount potatoes to buy for 'a' coins, which obviously didn't work. Then I thought the function was unimodal, so I tried ternary search on the potatoes, which still didn't work! Finally, I decided to scrap my mess of a code and prove a solution on paper and then implement it. It was less stressful after the 4th WA anyways, since I already hit the maximum penalty for a problem... Finally I solved Problem A, 76 minutes into the contest! I am proud of my persistence and disgusted at my stupidity.

I used binary search for A and it worked. Have no idea how to solve it normally though...

I did the case work on paper: let mx be the most multiples of (m+1) that we can buy without overbuying. Then the answer is the minimum of {n*b, (mx+1)*(a*m), (a*m*mx)+(n-(m+1)*mx)). I wish my binary search solution worked so I could have solved it earlier.

OMG, B is so hard, can anyone explain the solution problem B for me :((

I didn't prove it, but I solved it in a few minutes by greedily going max_sum, max_sum-1, max_sum-2... min_sum, min_sum+1, min_sum+2.. max_sum-1. I was surprised it worked.

Thank you so much. The sad contest, I can't solve B.

Same. I'm beginning to question my ability to make observations :(

Note the length of the array is always

`(x-y)*2`

.I couldn't figure that out, I just had a counter and incremented it by 1 every time I outputted a number lol

if x>y ->

x, x-1, x-2, ... , y+1, y, y+1, ... , x-1

there is only one higher and lower point and always difference 1

what if x = 3 and y = 2? 3 2 ??

Just start from max sum, go down to min sum and then go up to max sum -1. The reason it works is that, only the max/min sum will be local maxima/minima so, all others are unimportant.

n = (x — y) * 2. That's because local minimum and local maximum are alternating(x1, y1, x2, y2, x3, y3, ...(that's quite easy to prove)) which means that the quantities of minimums and maximums are equal(that's unimportant to the solution, it's just an interesting fact). So with this information we can easily count n. Lets look at slice [x1...y1..x2). We can easily count the quantity of all elements in the slice [x1...y1...x2): n = x1 — y1 + x2 — y1. Then add n to answer and do such thing till you return to your first slice. We can see that the final result will be ans = 2 * x1 + 2 * x2 + ... + 2 * xm — 2 * y1 — 2 * y2 — ... — 2 * ym = 2 * (x1 + .. xm — y1 — ... — ym) = 2 * (x — y). Finally, we can easily create an appropriate array with only 1 maximum and minimum which are equal x and y.

How to solve F?

I tried Mo's algorithm, but it won't pass the time-limit. Any nice optimization required in Mo's algo?

Idk if it will pass or not but anyways you should try this optimization: Optimization with Hilbert curve

Bricked B and then couldn't submit D in time.

Seems that last problem is very popular.

last problem is a copy of 765F - Souvenirs

the n = 2 case in B is very stupid

agreed i am surprised that the answer for the case when x — y = 1 is just an array of {x , y} :((

statement so unclear and stupid

Make it unrated, unless you play too much genshin.

Why Genshin out of nowhere...

it's just a joke hhh

Best contest I have ever seen!..

what even is pretest 6 for problem D? i tried to use segment tree type of answer and i still don't know the corner case (that or i somehow fluked the 4th and 5th pretest lol). Also tricky problem A ;)

I mistyped min to max when initializing prefix min of positions. However it also squeezed into the first 5 pretests:(

Segment Tree seems pretty overkill for D. My strategy was to count the #of subsegments such that they share a common MEX by iterating over the MEX value and using 2 pointers to handle everything. I got it to pass the example cases too. Wish I could have submitted it :(

Edit: It got accepted. Here's the code if anyone's curious: 193333734

One tip is to write a brute force solution for D and compare your submission with the brute force solution.

SolutionI started by counting solutions with subsegments not including ones, then includeing 1s and not including 2s then includeing 2s and 1s not including 3s ... I used combination for counting this stuff which you can make O(1) easily, so my approach is O(n) which passes(and passed).

Codenote that i used my template here and used cpbooster to parse and submit the code so there are a lot of unnecessary lines. you can assume that solution starts from void solve()

did u find what is pretest 6? mine is also failing

How to do F ? I tried using Mo's algo with sets, but that doesn't feel to squeeze in the TL

just go to https://codeforces.cc/problemset/problem/765/F and read editorial

Anyway, is it possible with MO? I have TL too.

Yes, it is possible with MO. To pass time limit, you need super fast set for integers and super fast IO. Check out my code.

Can you copy-paste your code into somewhere else, because currently we're not allowed to look into other's submissions.

Oh, didn't notice that. Code

I solved https://codeforces.cc/problemset/problem/765/F with Mo's algorithm: https://codeforces.cc/contest/765/submission/164782195.

However, I'm sure my solution won't pass TL in today's problem.

is it sure that F isn't https://codeforces.cc/problemset/problem/765/F?

yayy I was able to attempt D this time.. hoping system test pass

accepted :)

My solution to $$$C$$$.

For every element $$$l$$$ as left position, there is minimal $$$k$$$, such that it is required to $$$r \ge k$$$ for $$$l$$$ not to be minimum or maximum. Find all these $$$k$$$ in linear time using stack (note, there can be no such $$$k$$$). Let it be $$$left[i]$$$. Then do the same in other direction: $$$right[i]$$$ is rightmost position, at which $$$i$$$ is neither max nor min.

Now we have to find $$$l$$$ and $$$r$$$ such that $$$l \le right[r]$$$ and $$$r \ge left[l]$$$. Fix $$$l$$$. We have to check, if there is $$$r \ge left[l]$$$, such that $$$right[r] \ge l$$$. We can do this using segment tree for maximum and position of maximum.

Whose solution is longer?

I used two pointers -- while(l < r && flag), flag = false, if a[l] or a[r] = min or max, increment l or decrement r, increment min or decrement max, and set flag to true. Then output l and r as the answer,

Same

I did the exact same thing, even the same variable names.

But it feels like there is a better sol

As i understand you try to find

`l`

and`r`

for every element in array, so it is`O(n^2)`

. There is no requirement that length of subsequence should be minimal, so we could just check that left and right elements aren't current minimal or maximumI loved problem C! My solution is even simpler. I solved with also

two pointers, but instead of the segment tree, I used 2RMQs. Starting from the first and last element, check whether the range is good or not with RMQ;if it is good, print and stop. Else, update left and right by equality conditions, respectively. The segment tree method is another solution, but I think RMQ is even shorter and simpler. Btw used the sparse table method, so it gives min and max inO(1).I got AC without segment tree or sparse table. Just notice that at the beggining you have all values from 1 to n in your segment, so when you move your pointers you can just update min and max.

Yeah, that is also correct and even simpler xD. Congrats :)

It was a guess though. I wasn't 100% sure if it would work.

Wow! So simple approach! Thanks for this :)

I have also overkilled! same idea and used 3 sparse table :)

Why is C easier than A and B?

no it isn't

Problem E is just beautiful!

Solution sketch:

First sort array $$$a$$$. Henceforth I assume $$$a$$$ is sorted.

We can prove by greedy exchange that the set of satisfied people form a prefix of $$$a$$$. Now, let's fix the prefix of satisfied people.

Let's maximize the number of distinct books that this fixed set of satisfied people can be assigned such that all have to be read by at-least one personWe can prove again, that:

So for each prefix, we have found the maximum number of distinct books. But the number of books that can be assigned to a prefix is monotonic: if you can assign $$$k$$$ books, you can also assign $$$k-1$$$ books. So if answer for a prefix of size $$$j$$$ is $$$k$$$, update answer for each $$$i<=k$$$ with $$$j$$$. This can be done trivially by maintaining suffix maximums.

193328416

brought a permutation p of length n to the zoo This sentence lol!

What will happen if you try to hack someone's solution with this test x = 1e9 ,y = -1e9 in problem B ?

solution exist only for n<1e5 (given) so above case will not have any solution

Yes, absolutely. I think they should have written that the sum of the differences between x and y would not exceed 1e5

First line of the question

`For his birthday recently Fedya was given an array a of n integers arranged in a circle`

Last line of the question

`It is guaranteed that the sum of n over all test cases does not exceed 2⋅10^5`

Im extremely sad. I just solved question D 5 minutes after contest ended. And the even sadder part is that I joined the contest 30 mins late :((. You can tell because I only solved first question after 35 mins. And solved the next two questions almost immediately.

`Problem C`

just feels lot easier than`Problem B`

.bad problem F

Difficulty order according to me: B>A>D>C

I would swap A and D around, but I agree with this lol. Wasted a ridiculous amount of time on B

what was your idea for D that you put is so low on difficulty?

can we solve question e using binary search like our predicate function will be greedily picking smallest k number of people and satisfying their needs. here I am applying binary search on number of peoples we can satisfy tle solution also works

.

my submission

basic idea is with 3 conditions

price on first day is a and price on second day is b

then

if (a < b) ... you should alway buy on first day

if (a > b BUT you can save with a discount ) then you should get discounted items on first day and remaining on second day

else buy everything on second day

Why is problem B such a troll? It says that $$$-10^9 < y < x < 10^9$$$ and the answer has length $$$2 \cdot(x-y)$$$. In my opinion, you should not put this kind of "traps" in Div2 problems, especially in A and B.

First line of the question

`For his birthday recently Fedya was given an array a of n integers arranged in a circle`

Last line of the question

`It is guaranteed that the sum of n over all test cases does not exceed 2⋅10^5`

This is a wrong use of the phrase it is guaranteed.

See this

Meaning 2: Make it clear that this ‘guarantee’ is an additional constraint on the input.Input is only allowed if X.In the example above: “Only input for which some C exists is legal input.”oh, thanks for sharing this. Something for me to think about

Exactly , I was wondering how would i even print such a huge sequence even if i get one :/

Exactly, It had took my lot of time for optimizing number of operations, and I checked constraints on x and y more than 5 times, Also problem statement of A is not well written, it could be written in 2-3 lines. Even F is taken according to me.

Agreed, perhaps the worst Div. 2 B of all time, or at least in recent memory.

I totally agree with this, just this line alone $$$-10^{9} \le y < x \le 10^{9}$$$ can mislead people to think that $$$x = 10^{9}$$$ and $$$y = 10^{-9}$$$ is a valid input, even though the next line says that the sum of $$$n \le 2 * 10^{-5}$$$

And as such leaving their head scratching that there is a better answer than $$$2*(x-y)$$$ for $$$x = 10^{9}$$$ and $$$y = 10^{-9}$$$ that can fit into the given constraint on $$$n$$$, which can possibly cause them to get stuck forever

The problem A is just way too annoying!!

yaa !did u solved it ? i thought it's easy but took lot of efforts but still didn't got accepeted '~'

can you explain why my code is wrong?

A: Divide n potatoes into floor(n/(m+1)) groups and n%(m+1) single potatoes. The price for a group is min(m*a, (m+1)*b), and for a single potato is min(a,b).

B: Go down from x to y by step -1, then go up from y to x by step +1.

C: Recursion. First, check if [1,n] is a good segment. If not, then a[1] or a[n] will be the max or min value on this segment. If a[1]=1 or a[1]=n, then we need to solve recursively on [2,n], otherwise we solve on [1,n-1]. We do this recursively until we find an answer, or the length of the current segment is <4. Also we need to maintain the max and min value on the current segment.

D: Consider for each possible MEX in [1,n+1]. For MEX=1, we need to find the position of 1 in p and q, and count the number of segments which doesn't contain these positions (which is a subsegment of 3 segments splited by positions of 1), and for MEX=n+1 it's trivial (only [1,n] will have MEX=n+1). For MEX in [2,n], we need to find the minimal segment containing numbers in [1,MEX-1] in both p and q (this segment can be updated by expanding previous segment to contain MEX-1), and find the positions of MEX, we can expand the minimal segment in both directions until it touch these position (also, if there's any occurrence of MEX in the minimal segment, the contribution of MEX is 0).

Can you prove B?

Let the difference between the sum of local maxima and local minima be $$$d$$$, and let $$$n$$$ represent the optimal solution size (note that $$$n$$$ must be even). Assume that $$$d>1$$$ so that $$$n$$$ must be greater than 2.

Consider a local maxima — it must occur as $$$\ldots,x,x+1,x,x\pm 1,\ldots$$$ and now, remove the centre two elements, and we have a new sequence $$$\ldots,x,x\pm 1,\ldots$$$ which has $$$d'=d-1$$$, and a solution with $$$n-2$$$ elements, and reduce.

EDIT: (And a local maxima will always exist)

yaa in first attempt in A i did exactly what u r saying but didn't get accepted so started thinking something else

Great Explanations !! I managed to solve C using 2-Pointers as following: " Don't forget to move the pointer, and if we got "l==r" then -1 "

if "l" or "r" is pointing at min then "new_min = min+1"

if "l" or "r" is pointing at max then "new_max = max-1"

if none of above then print (l, r)

In A isn't it possible that if we buy x number of the potatoes out of n from a and remaining from b so now we vary this x How can we claim that it is best to divide all n in m+1 groups for a and remaining for b Can't there be any intermediate x out n for which we find the best possible minimum I tried this using binary search but W/A

It's not "divide all n in m+1 groups for a and remaining for b". In fact, there are at least n%(m+1) potatoes we can't buy them by favorable price (so we must buy them by price min(a,b), and we can by at most (m+1)*floor(n/(m+1)) potatoes by favorable price (which is min(a*m/(m+1),b)).

Let c=intmax, l=0,r=n,mid what i am saying is that what if we take i=mid and j=n-mid then we can calculate 'a' for i and 'b' for j

if this is less than equal to c we do r=mid-1 and else we do l=mid+1

i have understood that we can give n%(m+1) to min(a,b), but now it is only applicable when we give (n-n/(m+1)) to 'a', so what i am thinking is if we find certain 'mid' which can give a better result by using same distribution formulae?

I can't understand why you think about binary search. But we can consider problem A atart anew:

--Consider we want to buy

onlyk potatoes where k<m+1, what's the minimal cost? Well, since k<m+1, if we buy them on the 1st day, the price is k*a, on the 2nd day the price is k*b. Therefore, the minimal cost is k*min(a,b).--Consider we want to buy

onlym+1 potatoes, what's the minimal cost? If we buy on the 1st day, we need to pay m*a since the promotion, and on the 2nd day we need to pay (m+1)*b. Therefore, the minimal cost is min(m*a,(m+1)*b).--Now consider we want to buy n potatoes, what's the minimal cost? By the previous discussion, we can do 2 kinds of operation:

buy 1 potato, cost min(a,b).

buy m+1 potatoes, cost min(m*a,(m+1)*b).

Since the 2nd operation includes promotion and 1st does not, it cannot be worse than the 1st operation when there're >= m+1 potatoes we need. So we can do 2nd operation repeatly until we can't do more operations (which means, there're <=m potatoes we need to buy). Then we need to use 1st operation for remaining potatoes.

That B problem T_T

Is anyone able to tell me why in problem B, in the first test case where the

local maximum = 3and thelocal minimum = -2the solution cant be just the numbers3and-2?It doesn't seem to break any of the requirements? Am i missing something?

My submission to problem B

Am i missing something? : Yes

consecutive numbers should have a difference of 1 :)

The problem says

absolute differenceand the numbers3and-2seem to have an absolute difference of 1 right?Or is it that

abs(3-(-2)) = abs(3 + 2) = 5?Absolute difference doesn't mean the difference between the absolute values of the numbers.

It is the absolute value of the difference

Ok thank you, it was my mistake and makes sense now :D

it's the latter one sadly

Can someone show testcase 6 in D?

Check this case:

answer = 10

my answer is 10, but thanks

my code passes this and the samples but fails tc 6 lol

Check this case:

6

4 1 2 3 5 6

5 6 4 1 2 3

answer = 6

also correct, Im waiting for open submissions ;d

OMG,D is so hard,can anyone explain the solution of problem D for me :(

FOR（MEX，1，n+1） { Find Range of L in this MEX Find Range of R in this MEX answer += len of Range_L * len of Range_R } you can solve MEX=i by MEX=i-1

The fact that there are people who failed to solve D, but were able to solve F in under 30 minutes :)

Why is this getting downvoted, the round was pretty nice?

Cause F problem was present in previous contests. Most probably round gonna get unrated

This round will not get unrated, Mike did a very good blog post as to why it won't get unrated.

where can I see check it? :) , I need the editorial

I managed to solve A-D and I have to say I quite liked all the problems. Sadly, I did not look at E.

Problem B give me [-1e9;1e9] and so I say: Nani? How I can minus one with range 1e9 LOL? Finally, I solved it by subtracting the value by 1 and it worked XD! Sorry for my english !

I spent nearly an hour on problem F but now it coincided with another problem in the old round. I don't have time to solve D and E. /kk

you can check the video editorial for

problem C: https://www.youtube.com/watch?v=39LiI3MKack

problem D: https://www.youtube.com/watch?v=H-7tJaW7crI

Even without considering that the problem F has appeared in the past round, I think the problems in this round is simpler than usual.

Last two rounds max only maths problems

I spent almost 40 mins for solving the problem B for the given constraints. Finally left the contest. One of the worst contests in recent times.

two-pointer forces :).

The objective of the last problem is fairly simple, (not the solution, just the objective), so it makes sense that a similar problem has already been proposed before.

Imo, this coincidence wasn't the only issue with this round, Speedforces rounds are becoming increasingly common these days; though it was indicated beforehand by the problem ratings, still this isn't ideal.

When will the submissions and testcases open?

When can we see the

source codes of others? I'mDESPERATEto see where did I get wrongIn the standings page you can double-click on their scores

I don't think that this one is open yet.

be patient for the sys test and hacking session to be done

How to solve A? I have tried deperately in contest math, binary search but everytime something got wronged

my submission

basic idea is with 3 conditions

price on first day is a and price on second day is b

then

if (a < b) ... you should alway buy on first day

if (a > b BUT you can save with a discount ) then you should get discounted items on first day and remaining on second day

else buy everything on second day

cost1 = (n/(m+1))*m*a + min((n%(m+1))*a, (n%(m+1))*b) cost2 = n*b

ans = min(cost1, cost2)

Explaination : cost1 will be generated from first day buying potatoes on with set of m+1 at the cost of m and rest will be bought either at price of a or b whatever is min. Cost 2 will be generated from buying all the potatoes second day and min of these two will be your answer.

Bro there will be three cases 1) The purchaser will buy all the potatoes on day 1 without the offer. 2) The purchaser will buy all of it on day2. 3) The purchaser will buy the potatoes on day1 n/(m+1) times and the remaining either on day1 or day2.

Here is my submission. 193334956

Yes your approach is correct but your 1st case is redundant because it's going to be checked in 3 case. but nothing harm in it

193350417

why getting WA? what is wrong with my code can anyone tell please?

got it! i was minimizing two indexes each time but i had to check every index; sad though

when will u update editorial please :o

Hi! this is my first competition.. I was wondering when my rating will be updated ?

Is this round Unrated?

Problem C : Dora and Search video Editorial Link : https://youtu.be/l5GgWZmsZ98 Happy Coding

Good Contest.

Approach for D: For MEX 1: we have to find no. of common subarrays which dont include 1. So lets say of the 2 arrays leftmost 1 is at i, and rightmost 1 is at j then all subarrays forming less than index i, in b/n i and j , greater than j will give MEX=1.

Now for MEX=2 : We have to fix the subarray where 1 is present in both so i to j is fixed. Now check how many common subarrays we can form excluding 2. Here 2 can lie in the opposite direction of i,j or at the same direction. If it lies in b/n i,j so MEX=2, doesnt exist. also expand i,j which will include 2 for next MEX calc

In todays contest i swear to god i done problem C by myself but the system is showing that "Your solution 193304701 for the problem 1793C significantly coincides with solutions....." .For the first time i solved 3 problems, and i was very much happy to be able to do that, but this happend :( .Please can anyone tell me whome i should reach for this issue . This is my first comment on cf so please forgive me for my mistakes if any.

You can do nothing my bro.

isn't it unfair for those who did not copied ?

You are Pupil my bro Congrats!!

Thanks bro :)

In my opinion, C is easier than B.

todays order was a > b > c literally reverse

why is this not working for A? can someone help please

is this contest unrated ?

unfair contestunfair contest

In problem B, the statement "It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2⋅10^5$$$" is misleading since it made me believe that for all possible inputs $$$x$$$ and $$$y$$$ $$$(−10^9≤y<x≤10^9)$$$ the answer will never exceed $$$2⋅10^5$$$ which is obviously not true for $$$x=10^9$$$ and $$$y=-10^9$$$ since the minimal length is $$$2x-2y$$$. This lead me to believe that the expected solution was not optimal enough and that the problem had to be solved with some optimal value that minimized the length in order to not exceed $$$2⋅10^5$$$ by spamming something like $$$\sqrt{x}$$$ and $$$\sqrt{y}$$$ or using binary search.

It would've been much better to either:

Rewrite the statement as "It is guaranteed that the sum of $$$x-y$$$ over all test cases does not exceed $$$10^5$$$".

Lower the constraints to $$$(−10^5≤y<x≤10^5)$$$ (same problem, same solution).

Ask the solver to output -1 if the minimum length exceeds $$$2⋅10^5$$$.

Rewrite the statement and make it clear that you will not

actuallyreceive all possible inputs $$$x$$$ and $$$y$$$ $$$(−10^9≤y<x≤10^9)$$$, but rather all possible inputs $$$x$$$ and $$$y$$$ $$$(−10^9≤y<x≤10^9)$$$ such that the minimum length of the optimal solution does not exceed $$$2⋅10^5$$$.I don't understand what's wrong in my implementation of ternary search for the problem A

Submission — 193300171

For me this is an inspiring contest with excellent B&E.

And F itself is just a good problem at the wrong time.

None of you done it on purpose,that's enough.I hope that it will not destroy your confidence and courage.