**UPD: Pay attention to the changed start time of the competition.**

<almost-copy-pasted-part>

Hello! Codeforces Round #627 (Div. 3) will start at Mar/12/2020 16:05 (Moscow time). You will be offered 6 or 7 problems (or 8) with expected difficulties to compose an interesting competition for participants with ratings up to 1600. However, all of you who wish to take part and have rating 1600 or higher, can register for the round unofficially.

The round will be hosted by rules of educational rounds (extended ACM-ICPC). Thus, during the round, solutions will be judged on preliminary tests, and after the round it will be a 12-hour phase of open hacks. I tried to make strong tests — just like you will be upset if many solutions fail after the contest is over.

You will be given 6 or 7 (or 8) problems and 2 hours to solve them.

Note that **the penalty** for the wrong submission in this round (and the following Div. 3 rounds) is **10 minutes**.

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 behavior. To qualify as a *trusted participants of the third division*, you must:

- take part in at least two 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.**

Thanks to MikeMirzayanov for the platform, help with ideas for problems and for coordination of my work. Thanks to my good friends Daria ZeroAmbition Stepanova, Mikhail pikmike Piklyaev, Maksim Ne0n25 Mescheryakov and Ivan BledDest Androsov for help in round preparation and testing the round.

Good luck!

</almost-copy-pasted-part>

Thanks to Artem Rox Plotkin and Dmitrii _overrated_ Umnov for help with testing the round!

**UPD2**: Editorial is published!

"You will be given 6 or 7 (or 8) problems and 2 hours to solve them."

Does this means that the problems haven't been decided ?

No, Because it is

almost-copy-pasted-partThe numbers of questions has been decides, maybe they will have a little change before contest.

May not. I think it's just a general idea so that almost every contest could use this template. ^-^

Which contest coincided with the original timing?

Reply code challenge : https://codeforces.com/blog/entry/74502

sorry, misunderstanding from sentence!

The time that this contest was supposed to start. The contest was supposed to start 1.5 hour later.

Does that matter?

Edit: Well, I mean, do the time zones matter?

Good luck everyone !!!

Athena Chu <3

haha

The same to u <3

-.- <3

Dont show that corona face

Who else is hyped for Div 3 because can't get good rating from Div 2?

getting a good rating in Div3 is even tougher :(

What does mean by "almost-copy-pasted-part"?

The "almost-copy-pasted-part" is almost equal in the announcements of most contests

So excited for first contest. Good luck everybody!

Good luck and high rating <3

Thank you so much <3

Just solved three problems

How was your Holi celebration ? Was anyone not able to play due to coronavirus threat?

So many downvotes :( . I haven't asked anything wrong :( .

your question was right but on the wrong platform.

2 hours in Div 3 and 4 hours in Reply Code Challenge, it's going to be an exhaustive day.

I think you are highely inspired by TVF Pitchers.

True

Hope that I can solve A-B-C.

Good luck guys <3

stfu

well，is there anyone know what's the name of the other contest（1.5hours later） after this contest？

Reply code challenge : https://codeforces.com/blog/entry/74502

Thank you so~much

You're welcome

Hope to become expert after this round

No, you won't

actually I'm tzc_wk's fake account (surprised)?

I think ... He will :)

Would you like to share your secrets... with us :)

I hope to get blue in this contest

I don't think i can solve over two problems...

if you think you can't, so you can't.

Believing in yourself is the first step to success.

you were wrong

Hope problems will not be indirect like this

Well, you can jump through the centre.

Ya I can for that I require wings of Advanced mathematics and data structures. Anyways Good Luck to everyone for the contest.

Bad luck for me becoz i can't attend this contest.

Good luck Everyone

Hope i will be green after this round :))

same

I use this site during contest :)

really that's i waited for

I got logged out 2-3 times during contest. Did it happen to anybody else or problem at my end?

Same here :(

Glad to see I wasn't the only one who faced the same issue.

Made me wonder if my account got

hackedor something :/Well, the thing you had also happened to me, so I felt unhappy during the contest:(

Yes,this happened with me also. Maybe a coincidence, but it happened mostly when I clicked submit button.

Me too. I thought it was my browser problem or something like that and erased cache and it happend again... :(

same here, maybe it's the browser's problem or something else, idk

Same to me ((

I feel it difficult to understand some of the problem statements,does anyone feel the same?

me!!!

Yes :/

me，especially problem a

F too

Me too ,I got such penalties!

statement of A was simple for those who played tetris before.

yeap, understanding Problem A was harder than solving it

Couldn't agree more

True that

Still don't get it ;/

[100] step1) replace ai with ai+2 = [102] step2) replace ai with ai-1 then how [102] = [0] ?

If different between any two ai's is odd then it's not possible to achieve them at equal level(0) ..

I took maximum time on A

problem E is difficult to understand

E was the most difficult to understand. I understood A maybe because I had played that game before.

Exactly!

E was really confusing, in the examples they didn't even specify the mod h part

me

Good to see that vovuh is the writer.

The problems are too easy so that more than 500 participants solved all the problems

Guys,I have a problem!

do not have a point of 1900 or higher in the ratingIs that means standings in contest is different from your rank which is the final standings?The contest will be rated for those participants who have a rating below 1600. However, participants with rating greater than or equal to 1600 can participate unofficially. Separate standings are available for official and unofficial+official participants.

I get it.Thanks!

You're welcome

During the contest, I automatically got logged out from codeforces. I had two tabs of codeforces open in firefox. When I tried to submit I got to know I am not logged in. This has happened thrice. Is it normal/minor bug or there is some other problem ?

Even I faced this issue around 4-5 times.

I didn't count but it was infuriating.

For those who cannot understand what is subgraph and subtree. Click :(

Screencast Explanations

Faster than Lightening! Thanks!

Thank you

.

There's an easier solution for C BTW. Just count max length of continuous L's in the string.

Thank you very much for the rating friendly contest

How to solve D ;-;

My approachMy source codeFound Mistake: (wrong formula)

`cout << 1LL * n * (n + 1) / 2;`

->`cout << 1LL * n * (n - 1) / 2;`

Found Mistake: (integer overflow)`ll res = q * (q - 1) / 2 + q * zero;`

->`ll res = 1LL * q * (q - 1) / 2 + 1LL * q * zero;`

AC Code (Spoiler warning)You can use binary search. create a diff array such that diff[i] = teacher[i]-student[i]. sort this array. at a particular point i calculate how much you'll need to compensate for when you choose j. e.g if teacher[i]-student[i] = -2. then we know we need diff[j] >= 3, so as to make teacher[i]+teacher[j] > student[i] + student[j].

you can easily binary search this value in diff array.

I had the same logic, but failed somewhere in implementation i guess. Did you get AC with this logic?

Yup ... I had the same logic... ACed

I have implemented same logic in java but getting TLE in 24 testcase. but it is taking nlog(n) time. Your text to link here...

Change your int c[] to Integer c[] My submission of your code: 73104414

Thanks a lot, buddy. What is the reason?

Yes, I had AC

Let $$$c_i = a_i - b_i$$$. Now you have to count the number of pairs of $$$c_i$$$ with the positive sum. It can be done in $$$O(n log n)$$$ with binary search.

Can I ask why my approach failed ? T_T

Create an array of differences of a[i]-b[i] then use lower bound to find good pairs for negative numbers and zero numbers and for positive use binomial coefficient (ncr).

What formula should I use ?

I used PBDS.

how?

Let's just iterate over the array from the start, and since we need to find unordered pairs, so for a particular i, we need all elements before it such that: ai — bi > bj — aj or Let's denote bj — aj as val, then we need all such elements where : val < ai — bi So PBDS on pair of elements acts like a multiset with additional feature ,giving count of elements less than x too.So use that function for each i and get count of elements less than {ai-bi-1,inf}, and add it to answer and insert pair {bi-ai,i} to the pbds.

Can u please explain — "st.order_of_key(mp(arr[i]-arr1[i]-1,inf))"--- In this part of your code,why you used inf?

Use PBDS as a multiset (comparator less_equal). Traverse from the end of the array and for each index, increment answer by order_of_key of $$$a[i]-b[i]$$$ (basically index of upper bound). Then insert $$$b[i]-a[i]$$$ in the PBDS.

Yes, yes, yes, shooting pigeons with a cannon — that's my style! 73043201

Can you share your code? I tried using pbds, but it gave wa on test 6.

You can look it up in my submissions and if are not permitted, then add me as a friend and then look it up in standings submission. Sharing code here will scribble the comment section.

It can also be done with sorting; if you have a sorted array $$$c$$$ and you want to know for particular $$$i$$$, how many $$$j$$$ satisfy $$$c[i]+c[j]>0$$$, then you can decrement $$$j$$$ from $$$n$$$ down to the first index where $$$c[i]-c[j]<=0$$$, and count $$$n-j$$$. Then if you iterate over increasing $$$i$$$, this $$$j$$$ can only decrease.

What is wrong with my approach ? ;-;

problem is similar to inversion count, so used BIT tree to solve it, after compression.

I'm glad i'm not the only one who used BIT :)))))

I to used BIT to solve this. But I complicated it as hell. ACed

Let c[i] = a[i] — b[i]. Now we have to count the number of pairs with positive sum. Sort array c.

Set two pointers, one at index 0 (pointer j) and the other at index n-1 (pointer k). In each iteration decrease k by 1 and find the corresponding index j, by running a separate loop within the previous loop. Exit when j >= k.

Initialize ans = 0, and keep adding (k — j) in each iteration, i.e. ans += k — j

Note that j keeps on increasing in each iteration. We

do notrestart j from 0 each time.Time Complexity: O(n)

You have to sort array c as well, otherwise this 2 pointer technique will not work.

thanks for taking your time to write an explanation.

Thanks for saying that <3

There's no one later than me to solve A! I like playing with the tetris game,but I

do not likethis problem!Can any one give a hack on the following approach for D?

Sort them by $$$a_i-b_i$$$, then use binary search to find last okay one.

Gradually sort?

Can you explain it further please?

I think simply sort will TLE. What's your method?

Pasting my code here: (ignore the

`stable sort`

part, using`sort`

fails also) LinkWe need ai-bi+aj-bj>0 instead of ai-bi>aj-bj

sorting them will change the order of the indices

and in the problem you need $$$j$$$ to be $$$>$$$ $$$i$$$

That doesn't matter, if you are counting all

distinctpairs $$$(i,j)$$$ which satisfy some property, then this will equal the number of distinct pairs $$$(\sigma(i),\sigma(j))$$$ which satisfy the same property, for any permutation $$$\sigma$$$deduce condition ,suppose if ai-bi=di; then we need all di+dj>0 (j>i)pairs (counted once)

for this make a array d=ai-bi you can divide array into three parts — positive(np) ,negative(nn) ,zero(n0); solve separately for zero , positive and negative;

we can get our condition satisfied in three way , 1. positive di pairs 2. zero paired with any positive di 3. positive negative pairs where magnitude of positive is bigger than negative;

a1=np(np-1)/2 (Ist cond); a2= n0*np(IInd cond) ; and for a3(IIIrd cond) ,you can loop in (O(nn+np)) time and at last ans=a1+a2+a3;

Make an array of Ai — Bi, sort it and binary search the right one for each. That's the solution I think.

Maybe you can use the Binary Indexed Tree in value.

First discrete all $$$a_i - b_i$$$ and $$$b_i - a_i$$$.

Then for every $$$i \in [2, n]$$$, calculate all for $$$j \in [1, i)$$$ which $$$b_j - a_j < a_i - b_i$$$

The time complexity will be $$$O(n \log n)$$$.

Lightning-fast system testing

There is a 12 hr hacking phase, after which system testing will begin.

Codeforces has large servers which can carry out fast testing.

What is the 8 test case for problem D?

Why, i kept logging out after few intervals ,throughout the contest??

Me too!

Same here

Same here !

How to solve E? Also, what is the name of the concept used in E?

I believe it's called dynamic programming.

Dynamic programming. $$$dp_{i, j}$$$ is the number of good times when we consider the first $$$i$$$ times ($$$0$$$-indexed) and did this $$$-1$$$ "operation" $$$j$$$ times. The answer is $$$\max\limits_{j=0}^{n} dp_n$$$.

what's the meaning of the first $$$i$$$ times，I can't get it

More correctly, $$$dp_{i, j}$$$ is the number of "good times" when we consider first $$$i$$$ values $$$a_i$$$ and used $$$-1$$$ $$$j$$$ times. Transitions are also pretty easy: if the sum of the first $$$i$$$ values $$$a_i$$$ is $$$s$$$ then $$$dp_{i + 1, j} = max(dp_{i + 1, j}, dp_{i, j} + [(s - j) \% h \in [l; r]])$$$ and $$$dp_{i + 1, j + 1} = max(dp_{i + 1, j + 1}, dp_{i, j} + [(s - j - 1) \% h \in [l; r]])$$$. $$$\%$$$ is modulo operation ofc.

I understand that now, thank you very much~~

state are index and modulo

Submission:73073024

in problem D , solution of 10^10 was giving TLE for test case 12,althought time limit was 2 seconds, http://codeforces.com/contest/1324/submission/73076815 i thought 10^10 solutions can run in 2 sec

I think a bound of 10^(7-8) is good for 2 secs, 10^10 is taking it too far.

Okay..

Downvoted because the sentence in problem E is completely unreadable.

The sentence "The i-th time he will sleep exactly after ai hours from the time he woke up" and "Vova can control himself and before the i-th time can choose between two options: go to sleep after ai hours or after ai−1 hours." sounds inconsistent.

for problem D, i Stored two arrays one array has (

`ai-bi)`

values and the other has`(bi-ai)`

values and i ran a nested loop from`i to n-1`

&`j=i+1 to n-1`

and counted the number of elements that are greater than current element. what could i have done better?In fact, you can first discrete all $$$a_i - b_i$$$ and $$$b_i - a_i$$$.

Then use a Binary Indexed Tree to count all the values.

The time complexity will be $$$O(n \log n)$$$.

Use a binary indexed tree.

Insert $$$a_i - b_i$$$ into it after checking how many elements there are before $$$i$$$ which satisfies $$$a_j - b_j > b_i - a_i$$$. Discrete all $$$a_i - b_i$$$ and $$$b_i - a_i$$$ first thing first.

UPD: Sorry for the comment. I didn't think much, it's my bad.

Appologize for the round, sorry.

Do you write is serious? The "subSTRING" problem, as you say, is not "more interesting" and can be solved in $$$O(n)$$$ considering $$$3$$$ or $$$4$$$ consecutive characters.

D — data structure problem? The problem "sort the array and do (lower bound or binary search)" is data structure problem?

Sorry for that. I didn't think too much.

I am terribly sorry for the words.

How to "sort the array and do (lower bound or binary search)"?

It seems very interesting.

I used binary indexed tree after discretion.

See this

Can't the substring problem be solved in O(N)?, I did that only after I misread the problem! :( Just check all "3 consecutive characters substring" and "4 consecutive characters substring".

Interesting fact for myself: Everytime people celebrate a rating-friendly contest, I lose ratings; Everytime people complain about toxic problems, I gain ratings.

You did not think that the tasks are too simple even for div 3 ?)

Speed is really need. lol

vovuh made them easy to please div3 ppl

here

Haha, that's funny. I don't know how to prepare "good" round. I prepare slightly hard round, people say "omg that's div.1, not div.3!". I prepare slightly easy round, people say "loool 2ez4me that was not even a round". I don't know what to do.

Lol. You should ignore comments made by candidate masters on div3 rounds. A candidate master saying div3 is easy is like a pupil / grey guy saying div1 is hard.

actually, vovuh, you did well, i really appreciate as most of div.3 contests are made of you

Don't coordinate anymore divs 3 rounds )

Yea, yea, I remembered you. You wrote the same about pikmike. That's all I need to know.

axaxaxxa

You look like Radewoosh

I'm his brother

Blue blood in ICPC?

Twin brother.

who knows who knows

Actually, vovuh, you are the best. You prepare quality Div 3 contests, and I enjoy participating in your contests. Also, I participating out of the contest; I get to learn new concepts and techniques from your contests. I hope you keep up this great work and keep contributing to more such awesome contests.

I really liked D problem of this contest. I've been focusing on pbds/fenwick last 2 weeks and this problem really put my knowledge to test. I like how I saw 3 different approaches on this problem(bin search, pbds, fenwick and treap). The problem really managed to capture the creativity of the community. Well done, our Div.3 King

Great job vovuh, don't give in to the hate. As it is, it's hard and time consuming to create problems and test cases, let alone deal with criticism such as this.

Personally, I think this is a good level of difficulty, otherwise how do we differentiate between Div 2 and Div 3 problems. Once in a while, problems in Div 3 are easier compared to other Div 3 contests — but I think many people in Div 2/Div 3 enjoy the opportunity to solve a few problems (that are in reach and where we can apply what we've learnt recently) and then learn from the other unsolved problems.

I think that this is the perfect difficulty level for Div 3. Myself being an Expert (Rating: 1734), I was only able to solve A, B, C, D. So, I believe that it justifies my claim.

Speedforces

Tried E for the first time in my life. Got wrong answer test 40! I think I made a silly mistake but couldn't figure out yet. My submission: 73079617

It failed for me as well, test case is involving l = 0 I guesss, and I had counted t = 0 case, which is wrong. Removing that gave me AC.

Why these two submissions using PBDS for problem D got different verdicts Vovuh please look into the matter.

Correct SOlution:-73082017 Rejected Submission:-73079109

It's because, in your case, it couldn't contain same elements.

thank you for reply oh I got that it is property of sets

During contest I was logged out several times. Does someone know why?

The same happened with me 4 times as well.

It's quite easy. Even my i got more rating, but i think i wouldn't satisfy with those.

What's wrong with this submission for D? https://codeforces.com/contest/1324/submission/73081429

Why I am getting ILLEGAL CONTEST error while pushing "hack it"? Please fix

you must go to submission link and then press hack it there instead of hack from status

TY. It works

Hey guys Can anyone tell me

Why this submission fails :https://codeforces.com/contest/1324/submission/73068802

and this one passes: https://codeforces.com/contest/1324/submission/73082736

I'm not able to find the problem can anyone point this out with an example.

Thank you!

Please guys :)

ll cn1=pos.size()*(pos.size()-1)/2;

pos.size() => returns (int value) not (long long value) you must work casting.

Thank you so much it worked now. Didn't work during contest so sad :(

Hello? I'm a bit curious about how to solve F, can someone help me? Thanks

can someone please explain the reason for

TLEin my submission of E my code -linkBecause at last where you have done dp[id][sum][f]=ans, your value of sum is changed, but your dp[id][sum][f] should use original value of sum not the updated one.

1 5 2 2 2 2 4 why this will give yes in A?

NO, since all elements must have same parity (either odd or even), then only answer can be "YES"

can you show the steps?

Add 2 with each 2 then reduce 1 by 1 in 4 steps

Is it a full test or a single case? If it is a full test(i.e. 1 represents number of cases,5 represents number of columns) then simply put blocks in column 1-4 If it is a case(i.e. there are 7 columns),the answer is NO.

+104 -> +8, thx for B pretests, goodbye expert:(

THIS HAPPENED WITH ME TOO :'(

exactly same problem.... :/ why would they create such weak tests?

Actually, it means we should be more careful, and find out all special cases.

the thing is , the hacks arent even special cases, the hack could have easily been in the example test.

If you didn't consider something, for you it was a special case I suppose.

how did you figure out your rating change?

Chrome extension "CF-predictor"

Oh thanks. I can't see the prediction, I guess I was supposed to have it before the contest too?

Sorry, I don't know.

Hack for B?

!!NEED HELP!!

My solution for problem-D : 73076688

According to me,it's complexity nlogn, but i m getting time limit. Can anyone explain why it is happening or where m i wrong??

TIA

`sum+=(ll)distance(it,ms.end());`

has linear complexity -> O(n^2logn)If I hack my own solution, then will that be an advantage?? like my scores would not reduce

No, you'll just hack your own solution instead of someone else.

Seppuku, the honourable way.

So, I made that:)

quite sad that my solution for B got accepted in the first place , was it intentional to make the test cases all weak for only an odd length palindrome / 2 different numbers only? would've solved 5 problems in one contest for the first time.

Yeah i mean i totally forgot about that test case.. I mean i knowingly did this when i should have not done it and it would have passed.

Hi, can anyone explain why this solution 73084554 is giving WA. While this solution works 73037149 by tmwilliamlin168. Thanks in Advance!

Problem B$$$O(N)$$$73087070

I liked this contest vovuh. Good job. I'm interested to know your thoughts on the difficulty levels of $$$D$$$ and $$$E$$$ if this was a Div $$$2$$$ contest. Do you think $$$D$$$ is harder than $$$E$$$ ?

How to solve $$$F$$$ ?

I think D is between the div2B and the div2C, while E is between the div2C and the div2D.

However, you may just wait until when these problems are recorded in PROBLEMSET and the difficulty of them will be visble for you.

I think segment trees are too difficult to be asked for $$$B$$$. It might be a $$$C$$$.

Can you please tell me how to solve $$$F$$$ ?

In fact, you don't need segment tree for D since the binary search is enough for it.

For problem F, I apply DFS twice on the trees.

In the first DFS, I regarded no.1 node as root and regarded the trees as a rooted trees. In this DFS, I get the correct answer for no.1 node.

In the second DFS, I started my DFS from no.1 node and calculated the answer of other nodes in my traversal.

For more details ( since I'm poor in English but good at C++ ), you can look up my code.

Nice ! I never thought of binary search for counting inversions ! :)

My main doubt was in the second DFS. Thanks !

I implemented the same algorithm in Java, but failed test case 3. I couldn't figure out why it fails. Can you help me? My submission is: 73110580

It seems like ts3 is just a random tree whose node is all black. So you can just make a small test to see what happens in your program.

My tree building logic was wrong, I need to build a 0-rooted tree recursively, thanks anyway!

Root the tree arbitrarily and let $$$dp_x$$$ be the answer for $$$x$$$'s subtree. We can do this with a single dfs.

But then $$$dp_x$$$ won't necessarily have its full answer (because we didn't consider the path through $$$x$$$'s parent). So do a second dfs and for each parent node $$$y$$$ update each of $$$y$$$'s child's $$$dp$$$, (child may benefit from path via their parent $$$y$$$), but be careful to not to over count that child's contribution from $$$dp_y$$$.

for clarity you can see my submission: 73067648

Ah ! The overcounting is tricky ! We should subtract only if $$$f(c) > 0$$$. The reason is that if $$$f(c) > 0$$$, then we never added $$$f(c)$$$ to $$$f(v)$$$

Thanks !

it is the first time i tried in E

but i got WA on 40

any one can help me why it is wrong

https://codeforces.com/contest/1324/submission/73078402

Corner Case:

1 3 0 0 1

You miss this case.

yo boy!

Can someone please explain to me the DP state for problem E?

Before considering sleeps #$$$i,i+1,\ldots,n$$$, suppose you last slept at time t. Then under these conditions, dp[i][t] is the maximum number of good sleeps among $$$i,i+1,\ldots,n$$$. The question asks to find dp[1][0].

Can

Bbe solved inO($$$n^2$$$)will it pass?i think. YES, as n=5000

Yes, but if you have a constant factor $$$a$$$ you will have runtime $$$O(a * n^2)$$$, which may TLE because worst case scenario you will do something like $$$a * 5000 * 5000$$$ operations which may not be executed in time (I hacked a couple ppl with this idea)

Yes, since sum of all n is <= 5000.

Can someone explain why the order (i<j) won't matter in problem D ??

i<j simply means that pair (i,j) and (j,i) are same. so you can ignore this statement and divide it by 2 at last to consider it only once.

It is another way of telling that pairs (i,j) and (j,i) are the same. And they have to be counted as 1.

Is problem F a classic problem? Because i notice that many people write similar code in their dfs. If it is, please tell me. Thanks is advance :)

i think you are talking about F

Thanks! fixed now.

i think because its straight forward.

Yes it's a classic problem based on dp on trees. You can refer to rachitiitr youtube videos regarding in-out dp.

Can you provide the link please :)

Can you suggest links for rerooting technique and in-out dp?

Could someone help me figure out my mistake in Problem E 73091842. Thanks in advance.

It's because you don't take maximum of dp[i][j]s any where in your for loops.

Ok

Are you guys mad? I have missed the contest due to change in timing, now I can't attend div1 challenges.

Vovuh you are great author your problems are amazing

Hack case of B . 1 3 2 2 2

how is this O(n^2) solution timing out on B for 2 secs??.Link

It seems like you are using

`map<int, int>`

which takes $$$O(\log n)$$$ time for adding and looking for element. Also,`std::map`

is especially slow (big constant factor) so your solution has time complexity $$$O(n^2 \log n)$$$ with big constant factor.Ah, wow. i always thought map was constant access and insertion. thanks btw

I think what you want is unordered map which uses hash. It has constant access time on average (but can be up to $$$O(n)$$$ worst case). It also is quite slow since internal structure is somewhat complicated. I mean, don't expect something like a array-access speed.

Also, since it uses random, it can be hacked if some experienced coder tries to craft specific testcase to kill unordered map.

I think there was a good blog on CF about this (and also how not to get hacked from that) but I can't find it now.

Neat info thanks. i guess using a boolean array in this case would have been better. my loss

You might be talking about this.

coderanant orz

So this should time out too? 73028508

I don't think so. Your code is performing $$$n$$$ map access operations which make the whole complexity like $$$n \log n$$$. The code originally asked was using about $$$n^2$$$ map access.

Thanks. I was misinformed about the access time of maps, now its clear.

I think map takes O(log n) time for searching thus complexity goes to O(n^2logn)

thanks

When it's my first time to solve 4 problems and I've a new high rank:HACKS:Nice Contest! My solutions: A. Check the parity of all elements. If the parity is same, then the answer is YES, else NO B. Its always enough to check for palindrome sequences of length 3. So considering each element as center, check if the right and left parts contain atleast 1 element in common. C. its always beneficial to use R and not L. (why? bcoz with L you can only reach back to reach some R, which anyway you would have reached before reach the cell with L). so find the largest gap in reaching a R. D. Compute ci = bi-ai and sort this array. Then for each ai and bi count the number of elements in array c less than ai-bi. Also make sure to consider the edge cases where you have counted the same element again. E. Use dp. The state is (i, time), where i is the day (or the ith sleep) and time is the current time. Write transitions to each of ai and ai-1. F. Assuming you run bfs from some initial root node, Compute max(wi-bi) for each node i in its subtree rooted at i. Now we need to compute the value in parent part. This can be done using tree rerooting technique.

Could someone help me with my problem E submission. I am not able to figure out the mistake.Thanks in advance[submission:73096035]

Corrected code 73127806

Explanation$$$dp[i][j] = 0$$$ makes it so impossible situations (states of $$$dp$$$) can change the result.

Setting $$$dp[i][j] = - inf$$$ fixes it.

Example of simple hacktest:

2 24 23 23

1 1

Correct answer is 0.

Thanks

Can anyone tell me why this 73097950 passes and this 73098017 does not ?

they will give different output when v contain same value more than once.Check this 3 6 6 6 2 2 2

Can somebody hack this?

PS: I find similar solutions getting hacked.

what's wrong? After submitting the (A) question of Round #627 Div. 3, in test case #2 it is showing this: "wrong answer Answer contains longer sequence [length = 100], but output contains 28 elements ". Can I know the reason?

You should print 100 output not 28, That is, 1 output for each test case.

I think my code is doing it. can you please check it? solution no: 73099642

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Edit: now i know that these are the number of hacks.

I got TLE error during contest instead of WA! even after the contest when I was submitting correct answer I was getting TLE. I got frustrated and then copy pasted solution of another user then also I got TLE. Also, I was logged out multiple times automatically during the contest!! solution link: https://codeforces.com/contest/1324/submission/73104474

use pypy 3 instead of python 3 as your language on codeforces.

It worked, thanks for the help! do you know reason/line where its failing in python3.7?

There are answers in stackoverflow.com

I am getting error in 6th problem too, for 36th test case!

can you help me in that too? https://codeforces.com/contest/1324/submission/74307112

Will there be any system testing after the hacking round is finished??

Yes

All of the AC solutions will be rejudged with original tests and successful hacks once the hacking phase is over.

Why I want to hack a solution and it shows Illegal contest ID ？

Try to hack solutions via the Status page instead of the Standing Page

Ok I get it. Appreciating your help !

Can Somebody please explain why this solution is not giving compile time error on pretest cases and arbitrary cases on offline compiler Solution for Problem 2

When system test will start?

What's wrong with my solution for E problem? 73075794

I am wonder that when the system test will begin.

Anyone who solved problem D using two pointers? Please explain your approach