Codeforces Round #464 (Div. 2)

6 years ago, # ^ |

+24

Thanks, corrected!

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6 years ago, # |

+10

Hope that the problem statements are as short as the announcement.

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Ali_Adelkhah

6 years ago, # |

high rating for every one :D

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6 years ago, # |

+24

4 days 4 amazing rated contests!!

i'm feeling lucky!! 4 days doing favorite thing! XD

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DenXman111

6 years ago, # ^ |

4 rated contests?

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6 years ago, # ^ |

It's including tomorrow's contest! I'm sure that it will not go unrated! And ya 4 rated contest for div 2 XD

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Alhelal_01

6 years ago, # |

How many problems in Round #464 ?

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starboy_jb

6 years ago, # ^ |

I think 5.

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Alhelal_01

6 years ago, # ^ |

-7

If it is 5,generally saw that div2 C problem is solve more than 1000. if it is more than 5,C is solve besides 1000.

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_HossamYehia_

6 years ago, # |

← Rev. 3 →

if someone will become div. 1 in this round .. Is (Educational Codeforces Round 38) will become unrated for him ?

UPD: Sorry i didn't see that Hack is opened 12 hours only in this round xD

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Usu

6 years ago, # ^ |

No, because hack period for this educational is only 12h length so the rating will be updated until the cf 464 will start

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_HossamYehia_

6 years ago, # ^ |

yeah , i didn't see that xD .. Thanks

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justadlet

6 years ago, # ^ |

← Rev. 2 →

+26

Hey, your expectation was justified!

Participiant MiteshAgrawal became Div.1 on Educational Codeforces Round 38 (rated for Div. 2) and got +139 on Codeforces Round #464 (Div. 2).

Here is the link to standings: click.

made at imgflip.com

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racsosabe

6 years ago, # ^ |

-14

"Just take advantage of bug :v"

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justadlet

6 years ago, # ^ |

+21

Oh, there is also somebody who became Div.1 on Educational Codeforces Round 38 (rated for Div. 2) and fell to Div. 2 on Codeforces Round #464 (Div. 2).

Look!.

made at imgflip.com

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killer_god

6 years ago, # ^ |

sadly even I did.

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shanto_kundu

6 years ago, # |

← Rev. 2 →

-25

It is rated for Div.2.4 days 4 amazing rated contests!!

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-_-_-_-_-_-_-

6 years ago, # ^ |

-35

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ZZZZZR

6 years ago, # ^ |

It is rated for Div. 2. XD

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MonkeyKing

6 years ago, # |

Will the scoring distribution be announced soon?

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1.618

6 years ago, # |

Look at the first line of the announcement — It should be "Hi, Codeforces". I'm afraid that the author should be a litle more careful.

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MonkeyKing

6 years ago, # ^ |

He forget to thank KAN for preparing the contest either :)

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6 years ago, # ^ |

the "He" you are referring to is KAN XD

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MonkeyKing

6 years ago, # ^ |

Yeah, it's only a joke

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6 years ago, # ^ |

i Understood bro, just pulling your leg XD XD!

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starboy_jb

6 years ago, # |

-28

Can you please extend time for half an hour.

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rvalue

6 years ago, # |

+41

Perfect time for Chinese users!

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1.618

6 years ago, # ^ |

-9

But it could be better! ---- I'm having dinner during this time.

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SrO

6 years ago, # ^ |

It is very good to have a contest when I am working, not sleeping!

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Assaduzzaman_Sohan

6 years ago, # |

Back to back contests. Passed a great weekend.

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Zindow

6 years ago, # |

What a pitty! I hava to have dinner when the contest starting.

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samyakjain

6 years ago, # |

+19

4 days 4 rated contests for div.2... perfect ending of valentine's week...xD

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Nibir_Hossain

6 years ago, # |

When will be the rating updated for "Codeforces Educational round#38"???

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KUMAR_PRANJAL_RAI

6 years ago, # |

Probably it is the first time for Div-2 users when their ratings will be updated twice a day.

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IIIIIIIIIIIIIIIIIIIII

6 years ago, # ^ |

← Rev. 2 →

But in December ratings of Educational round 35 updated at the same time with Goodbye 2017.

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thienlong1511

6 years ago, # ^ |

This because the Goodbye 2018 is held for two division (div1 + div2 combine). However, this contest is only for div 2.

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Notali.haidar

6 years ago, # |

to newbieeeeeee

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P___

6 years ago, # |

Could you please update the information about the number of problems and their scoring distribution?

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xht13127

6 years ago, # |

Seldom to see so friendly a round time for East Asia and Southeast Asia participants!

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Bill_Yang

6 years ago, # |

How many problems are there in this round?

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6 years ago, # |

+78

For technical reasons, system testing of Educational Codeforces Round 38 (Rated for Div. 2) is delayed until the end of this round. After the round, the following sequence of events will happen: system testing of Round #464, system testing of Educational Round #38, rating updates for the Educational Round, rating updates for Round #464, in that order. Ratings for Round #464 will be updated for participants who were in div. 2 prior to the Educational Round.

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C137

6 years ago, # ^ |

-10

So rating updates for Round #464 will be ready by tomorrow???

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rvsmr

6 years ago, # |

For problem A, do A,B,C need to be distince planes?

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killer_god

6 years ago, # ^ |

it doesn't matter because f[i]!=i

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hulk_man

6 years ago, # |

← Rev. 2 →

When local time in the 1-st timezone is 1 hour, local time in the i-th timezone is i hours.

How 3 hours (in first timezone) is equal to 1 hour (in the second timezone) and 2 hours (in the third timezone)?

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6 years ago, # ^ |

Please ask questions with "ask a question' button on the page with problem list.

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hulk_man

6 years ago, # ^ |

I wasn't aware of that.

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Dazzler

6 years ago, # |

+53

Did someone notice? They added the feature to copy the sample test cases of a problem. Thank you Codefroces, keep going!

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and800

6 years ago, # ^ |

+36

a bit buggy, but still very wonderful feature

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rmm_0304

6 years ago, # |

How to solve E?

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MoreThanANoob

6 years ago, # ^ |

Ternary Search!

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6 years ago, # ^ |

← Rev. 2 →

My solution is this:

It's easy to prove that we will always pick the max element in the sequence, and then we can ternary search for which prefix should we use.

Hope I won't FST. (UPD: AC in system test)

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mental.cat

6 years ago, # ^ |

+20

It can be done faster: notice that if we pick new max element, amount of elements in answer can't decrease, so we can use 2-pointer algorithm

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6 years ago, # ^ |

Great idea, I didn't found that in the contest, lol.

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IWillBeRed

6 years ago, # ^ |

Why ternary? I'm fine with binary search:)

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6 years ago, # ^ |

← Rev. 3 →

My implementation is doing binary search on the slope.

However, I am used to call it ternary search because I'm ternary search on the value.

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6 years ago, # ^ |

How can we update prefix' sum fast?

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6 years ago, # ^ |

Things are given in non decreasing way, so we can simply maintain prefix' sum like this:

int x; cin >> x;
pre_sum[n] = pre_sum[n-1]+x;
n += 1;

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6 years ago, # ^ |

Is n the size of multiset?

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6 years ago, # ^ |

yes(maybe size of multiset+1), my bad

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6 years ago, # ^ |

← Rev. 2 →

Omg, dude. I've just looked through the statement and noticed that:

1 — Add a positive integer to S, the newly added integer is not less than ANY number in it.

During a contest I thought that the newly added integer is not less than minimum number in it.......

Now I understood how easy the problem is :(((

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6 years ago, # ^ |

I don't see why it's true that we always pick the max element. Also why is there only a single peak for the prefixes?

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6 years ago, # ^ |

I figured out why there must be one peak. But still don't see how it's always optimal to pick the overall max.

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6 years ago, # ^ |

← Rev. 2 →

Let’s assume that we have an optimal choice, where max element is m₁ and sum of other elements is S.

So we can conclude that $\text{[math]}$ (M is the max element in the original set)

nm₁ - S + m₁ ≥ nM - S + M

m₁(n + 1) ≥ M(n + 1)

m₁ ≥ M

We know that nothing could greater than M, so m₁ must be M

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6 years ago, # ^ |

← Rev. 3 →

I still don't see it. For m₁ it's possible that the optimal choices for S and n are different from the optimal choices for M. (so we can't assume that the S and n on the left side are the same as the S and n on the right)

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6 years ago, # ^ |

← Rev. 3 →

m₁, S and n is optimal, so any subset won’t greater than it.Then, I simply replace m₁ with M trying to get another subset, but we can prove that it’s the same subset containing m₁

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6 years ago, # ^ |

Understand it now, thanks!

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SrO

6 years ago, # ^ |

← Rev. 3 →

You needn't use binary search or ternary search. You can first choose the biggest number, and then choose some small numbers to make the average value as small as possible. If you use C++, you can use priority_queue. (upd. AC in system test)

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marshal4world

6 years ago, # ^ |

Thanks , your solution was helpful . pity that i missed the trick during contest

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6 years ago, # ^ |

You don't need a priority_queue, too. You can just use an pointer pointing the last element you choose. (This one)

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Danylo99

6 years ago, # |

How to solve F?

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6 years ago, # ^ |

Wondering too.

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6 years ago, # ^ |

← Rev. 4 →

+21

Just a DP with segment tree is ok. dp_i, j stands for the minimum times of flipping, if the face now is cooked j seconds at the end of interval i (i.e. r_i).

It's obvious that we don't need to flip more than two times in an interval. So just consider how long every face get fried, this yields a Θ (n²·k) dp.

Notice that the points that a state affects is always an interval, so we can use a segment tree with interval-minimum modify and single-point query(or single-point modify and interval-minimum query).

This yields a $\text{[math]}$ solution, which suits the time limit. (Actually it runs very fast, because there are lots of useless states in the dp, ignoring these can make this solution run in about 200ms)

Hope this will help.

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6 years ago, # ^ |

Sorry, what's minimum times meant for? It's time for what?

我怕我英文說得不好，這裡用中文寫一下我的問題：可不可以跟我解釋一下反轉魚的最小時間（也就是DP的意義）是什麼，我的理解他是j，可是這就很奇怪了

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6 years ago, # ^ |

← Rev. 2 →

Sorry, the time means the number of operations. (I don't know a better explanation) It doesn't mean the time when you flip it.

我怕我英文說的不好，這裡用中文回答一下你的問題：DP的意義是翻轉魚的最小次數，不是時間。英語中time做不可數名詞的時候是時間，可數的時候是次數（可能還有其他用法）。j是當前被烤的那一面，被烤的時間。dp就是說，考慮了前i個區間，被烤的那一面烤了j秒，所需要的最小翻轉次數。

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6 years ago, # ^ |

← Rev. 3 →

Oh!!!! Great Thanks.

My English is so poor, I didn't notice the 's' after "time".

Thanks for your wonderful explanation.

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nutella_waxberry

6 years ago, # ^ |

← Rev. 2 →

An alternative solution:

Still, use dynamic programming, and dp[i][j] stands for stands for the minimum times of flipping, if the face now is cooked j seconds at the end of interval i.

It's easy to get dp[i][j] = min(dp[i — 1][j — r[i] + r[i — 1]], min(all elements within dp[i — 1][(j — r[i] + r[i — 1]) ... (j — l[i] + r[i — 1])]) + 2, min(all elements within dp[i — 1][(r[i — 1] — j)...(r[i — 1] + r[i] — l[i] — j)]) + 1).

Notice that: 1. dp[i — 1][(j — r[i] + r[i — 1]) ... (j — l[i] + r[i — 1])] is an interval. 2. As long as j0 < j1, (j0 — r[i] + r[i — 1]) <= (j1 — r[i] + r[i — 1]) and (j0 — l[i] + r[i — 1]) <= (j1 — l[i] + r[i — 1]) (Either of the terminals of the interval 0 is on the left of the corresponding terminal of the interval 1).

So it's not hard to see that two pointers + deque trick can be used to optimalize this algorithm.

More detailed: 35414333

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teja349

6 years ago, # ^ |

+11

Try to use the fact that in each l[i] to r[i],whatever possible can be achieved with atmost two flips. So use seg tree(or sparse table or multiset) to update dp values of each level based on previous level,
Now we need to find what all states contribute to i using len length segment we can get from i+len%2,i-len%2,i+2+len%2,i-2-len%2, and so on till within limits. So segment tree on odd and even positions should be maintained.
My solution took 2500ms but some solutions took 61ms. So I believe some other nice way exist.Can someone explain???

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georgerapeanu

6 years ago, # ^ |

you can update the dp in O(N) using a deque,so it becomes O(N * K)

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teja349

6 years ago, # ^ |

Yeah.Nice So was that the optimisation that gives 61ms??

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georgerapeanu

6 years ago, # ^ |

i think so,mine runs in 31ms

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6 years ago, # |

Hope that I handle the output criteria of C more careful. Half an hour wasted with 4 wrong submissions because of the mistake.

Anyway, the problems and pretests are good. Thanks :D

P/s: Is problem E solvable with a greedy approach and two (but technically one)-pointer(s)?

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teamskiy

6 years ago, # ^ |

I have written such solution in E and it have passed pretests

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6 years ago, # ^ |

Same with mine. However I feel quite uncertain — like Is it that easy, or there is a trap, or it was just me thinking that the problem is easy?.

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teamskiy

6 years ago, # ^ |

I am quite confident that it is a solution, even if mine will fail sys.tests, cause I am bad at coding without bugs :D

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6 years ago, # ^ |

Well just give yourself a faith. Good luck. ;)

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sdssudhu

6 years ago, # ^ |

Yeah I wrote such a solution and it passed system tests. There is a proof sort of for it.

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6 years ago, # ^ |

Can you elaborate the proof? ;)

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sdssudhu

6 years ago, # ^ |

Yeah. Initially when you consider adding all occurrences of a number and you find the difference between the current and the previous one, and if it is positive then you can take that number into account.

Consider n/sum — (n+occ*ai)/(sum+occ)

If difference is -ve you can stop.It is beacuse:

a/b<(a+occ*x)/(b+occ) for all x>1

Hence the greedy approach works.

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6 years ago, # ^ |

It's clear enough ;) also by reading this I know that my thought during contest time was equivalent ;)

Thanks! :D

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Ahmad

6 years ago, # |

Was F just dp but instead of having an O(n) transition you change it to O(1) by choosing whether or not to flip at some time (if valid) and return the number of seconds one side is cooked for (and then keep the minimum number of moves in a different dp table)?

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6 years ago, # |

What's pretest 5 in problem C, if someone was stuck there?

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6 years ago, # ^ |

← Rev. 2 →

+11

I think pretest 5 involved the wraparound condition. Where people of last x timezones and first f-s-x timezones participated Edit: f-s-x, not s-f-x

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JoaquinManuel

6 years ago, # ^ |

I think is something like that:

Input: 3 3 1 5 1 3

Output: 2

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6 years ago, # ^ |

gotcha !

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MonsterInMe

6 years ago, # |

no chances for problem hack a and b...."pretty straightforward"

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mkagenius

6 years ago, # ^ |

showoff

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Dishant_18

6 years ago, # |

Any idea what was the pretest 5 in C??

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http://codeforces.com/blog/entry/57819?#comment-414901

6 years ago, # ^ |

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Dishant_18

6 years ago, # ^ |

← Rev. 2 →

Thanks :)
Our comments have 2 minutes time difference xD

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rahulkhairwar

6 years ago, # |

I solved E literally 2 seconds after the contest got over :(

I hope it doesn't get accepted when I submit later.

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JasonBaby

6 years ago, # ^ |

+14

I solved that 5 minutes after the contest lol

Don't feel upset if it gets accepted, though. Whether or not we are able to solve a problem is more important than "the ratings we could've gotten," in my opinion.

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6 years ago, # |

+12

When I realize that only one digit '1' needed to be changed to '0' in the code to get AC...

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ujzwt4it

6 years ago, # |

← Rev. 2 →

problem E would have been a much better problem had the numbers been not in increasing order.

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6 years ago, # ^ |

How to solve it? Use some Data Structure to maintain it?

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ujzwt4it

6 years ago, # ^ |

← Rev. 2 →

+11

The only way I can think of is binary search +segment tree

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6 years ago, # ^ |

Maybe some kinda ternary search tree?

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Gipsy_Danger

6 years ago, # ^ |

binary search and Prefix And，I thinnk (前缀和加二分） (倍增法也可以）

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mouse_wireless

6 years ago, # ^ |

There are data structures which allow you to emulate an array with the possibility of insertion in the middle (complexity changes to O(logn) instead of O(1)). Treaps are an example of such a structure. So the problem wouldn't have changed much. Just had to use a data structure instead of an ordinary array.

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contest_dhuye_pani_khai

6 years ago, # ^ |

whoa, I totally didn't notice that numbers were given in increasing order. I overkilled my solution :')

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nima10khodaveisi

6 years ago, # ^ |

same :D so during contest I implemented ternary search + segment tree but time wasn't enough :D

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JonnyDavies

6 years ago, # |

Does failing a pretest during the contest affect your rating? I've looked through contest rules etc couldn't see that part discussed.

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6 years ago, # ^ |

I don't get exactly what you meant, but this is what I think.

If you haven't submitted any solutions, and then you submitted a "WA/TLE/etc. at pretest X", your rating would start to be affected, since from this point you're set as really participated in the contest.

If not, then it's similar to any failed submissions — just a minus 50 points of penalty.

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JonnyDavies

6 years ago, # ^ |

Apologies, I see what you mean the question wasn't clear enough.

That makes sense. Is there a difference in the penalty between a "WA/TLE/etc. at pretest X" and failing on system testing? Or are they both just considered a failed submission -50 points?

Thanks for the reply.

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6 years ago, # ^ |

← Rev. 2 →

Well, technically, they both mean -50. But that penalty is considered only when you managed to have a correct solution after.

And in this case, system test failures are assured to not gaining or decreasing your points at all (since you have no chance to re-submit oficially and get points).

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JonnyDavies

6 years ago, # ^ |

Oh I see! I think this is what I was missing. I thought the penalty was -50 from your overall points for the contest, but it is a penalty on the total points that you can gain from submitting the correct solution for that particular question.

Thanks Akikaze

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aiden_pearce

6 years ago, # |

← Rev. 2 →

WA in D(pretest 6). can anyone check whats the problem?? link to my solution of Problem D

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VTifand

6 years ago, # ^ |

Using this test case

3
aaa
bcc

your code gives an incorrect answer. I'm not sure what's wrong though.

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E_N_D

6 years ago, # ^ |

return dfs(graph[i][j], fd); This line of code gives you the error. You are not considering all the adjacent nodes and just returning the value from the first adj node you visit. Eg. the graph consisits of: (a, b) (a, c) if you search for a c again it would return false as first 'b' is visited in adjacency list of 'a' and thus it returns false

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interestingLSY

6 years ago, # |

+23

Light-speed System test. Amazing!!!!

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I_D0nt_Kn0w

6 years ago, # |

what could have been the pretest 4 for problem C?

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bsgnl952

6 years ago, # ^ |

I failed on that one too (several times). From my observations, it's the first test with s>1

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6 years ago, # |

+22

Before the announcement on B, I submitted B and locked it.

For any case where we can not take any box, I printed -1 0 because I have no idea what to print in such case as the question is unclear.

But my solution failed in System Test.

This should be got AC because, at first the statement was not clear.

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prakhar17252

6 years ago, # ^ |

Your solution probably failed somewhere else. My code also printed 1 0 in case of using no box and I have got an AC.

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6 years ago, # ^ |

My solution printed -1 0

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6 years ago, # ^ |

I think that was clear even without the announcement. It's technically not different from any other case.

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6 years ago, # ^ |

but, what is the meaning of taking zero box of a specific type? nothing logical here before the announcement!

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prakhar17252

6 years ago, # |

The fastest system testing ever!

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6 years ago, # |

For D, was there a better method than finding the edges of the minimum spanning forest of the graph?

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latvian

6 years ago, # ^ |

You can find the connected components of the letters. In each component, all letters have to become the same. So a single DFS or BFS is enough

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6 years ago, # ^ |

Oh! That's clever. Thanks!

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we7d

6 years ago, # ^ |

+11

dsu p[26]

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6 years ago, # |

After how much time of the system testing can we submit the solutions in Practice? Cant now

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6 years ago, # ^ |

After 13:30 UTC as mentioned in the update.

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6 years ago, # ^ |

Hadnt seen the update. Thanks

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learner_321

6 years ago, # |

How to solve C?

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an6285

6 years ago, # ^ |

I used a sliding window kind of approach.Passed the pretests but my solution failed on last system test due to silly mistake. It was just because of using < instead of <=.

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skpro19

6 years ago, # |

What's wrong with my solution of C?

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skpro19

6 years ago, # ^ |

an6285 what's wrong with my approach?

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starboy_jb

6 years ago, # |

something going wrong in test case 20 in the problem B.

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sirjan13

6 years ago, # |

When will the contest be available for virtual participation?

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6 years ago, # ^ |

After 13:30 UTC as mentioned in the update.

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madn

6 years ago, # |

Can anyone prove this? 35412837

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6 years ago, # |

What is pretest 13 of problem A?

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6 years ago, # |

As its not possible to view the test cases currently did any one else get a runtime error in test case 19 of B? If yes then do you know why?

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latvian

6 years ago, # ^ |

I managed to hack a few solutions by forcing RE with the following tests:

19999999999999 1
362143000000

and

3 1
4

For the first case the typical mistake is that maxValue is initialized to INT_MAX or LONG_MAX instead of LLONG_MAX or the equivalent, which is >=n.

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6 years ago, # ^ |

This is my code. http://p.ip.fi/67F5.

It gives an answer of 1 9313 on the input

19999999999999 1
362143000000

No run time error though

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6 years ago, # ^ |

← Rev. 3 →

It should give 1 55.

P.S In your code array a should be long long type

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6 years ago, # ^ |

Alright thanks :(

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Master_coder

6 years ago, # |

is this contest is rated?

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Snickeen

6 years ago, # ^ |

Link

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6 years ago, # |

System test is finished and still we can't submit ?!

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Ahmad

6 years ago, # ^ |

We probably need to wait until yesterday's educational round finishes its testing.

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bharath

6 years ago, # |

What would happen if a participant with rating above 1900 participates in div2 contest? Does it effect his/her rating ?

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6 years ago, # ^ |

No. It won't affect his/her rating.

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ghoshsai5000

6 years ago, # |

Can someone please post the questions asked in the Nizhny Novgorod Olympiad, that this contest was based on ?

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6 years ago, # ^ |

← Rev. 2 →

-6

Problems were same as of Codeforces Round 464 (Div. 2).

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ghoshsai5000

6 years ago, # ^ |

It is written that the problems are not identical in the announcement thread.

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6 years ago, # ^ |

+12

The problems are in Russian, so it won't work if I just post them. Let's go through problems of Codeforces Round:

A: prepared specially for the round,
B, C, D: same as on the olympiad,
E: on the olympiad time limit was 1 second.
F: on the olympiad the constraint on n was 500 000.
there was one more easy problem on the olympiad.

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ghoshsai5000

6 years ago, # ^ |

Thank you so much for your reply !

Can you tell what the easy problem was ?

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6 years ago, # ^ |

+13

So why in the Codeforces round, the constraint on k of problem F was 100, but in the test data it never exceeded 20? I think if it became 100, a lot of $\text{[math]}$ solutions will get TLE(but mine will not :P). Can you explain it a little bit? Is it a bug or a feature? Thx a lot.

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cyz666

6 years ago, # ^ |

← Rev. 3 →

Why will nklogn solutions get TLE? The time_limit is enough for nklogn when k is 100.

but.. I still don't know why my O(nk^2 log32 /64) solution can pass the testcase within 62ms.

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6 years ago, # ^ |

But the segment tree runs a little bit slow... Those who runs more than 1s will likely get TLE, actually there are a lot of them.

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Sharon

6 years ago, # |

+20

"The round is based on XIV Nizhny Novgorod Olympiad in Informatics for high school students named after V. D. Lelyukh, which will take place on Saturday in Nizhny Novgorod."

Wow, there is an Olympiad for high school students that are all named after V. D. Lelyukh? That's amazing!

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6 years ago, # |

What's wrong with my code of problem A?

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gatecse2018

6 years ago, # ^ |

if (n==2){ cout<<"NO"<<endl; } probably, i printed YES for that case.

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Art

6 years ago, # ^ |

There have to be "NO", triangle with two elements can't exist

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gatecse2018

6 years ago, # ^ |

← Rev. 3 →

my comment was incorrect.

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Art

6 years ago, # ^ |

your solution is right, it works correctly with two elements. And remember that plane can't love himself:)

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Amaryllis3

6 years ago, # ^ |

It's supposed to be NO. I think the mistake is because for the case n=2, the line NO would be printed twice.

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6 years ago, # ^ |

Thanks a lot.

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9646516

6 years ago, # ^ |

3 1 2 2 try this input data

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Amaryllis3

6 years ago, # ^ |

Since the problem states that f_i != i, I don't think that would be a valid test case.

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E_N_D

6 years ago, # ^ |

You should return 0 from the if condition after printing NO.

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6 years ago, # ^ |

Thanks very much.

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ehsanoo

6 years ago, # |

It's 13:45 UTC now and we still can't submit solutions.

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Ahs_58

6 years ago, # |

My B is hacked :( ...

can anyone give me some test case...

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ehsanoo

6 years ago, # ^ |

← Rev. 3 →

Is 99999999999 sufficient?

Edit: N = 10¹⁸, K = 1, a₁ = 4756234141534.

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Ahs_58

6 years ago, # ^ |

oh shit....

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latvian

6 years ago, # |

Apparently the succesful hacks have not been added to the final system tests. What is the reason for this?

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Diegogrc

6 years ago, # ^ |

Hacks are only added to final system tests in Educational Rounds.

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PrakharJain

6 years ago, # |

Enable upsolving please!

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6 years ago, # |

Why is the right answer 4 in fourth C test? 10 7171 2280 6982 9126 9490 2598 569 6744 5754 1855 7 9 If we begin round at 4, only participants from 4, 5 and 7 time zones take part in it, and if we begin round at 10, participants from 1, 8 and 9 time zones take part, so in first case we have 9126+9490+569=19185 participants, and in second case 7171+6744+5754=19669 participants. ???

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6 years ago, # ^ |

← Rev. 3 →

The time zones which has time f will not participate. It's said in the problem statement.

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6 years ago, # ^ |

But I consider it. In first case 6th zone has f time, and in second one 10th zone has it.

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6 years ago, # ^ |

So why does time zone 7 take part in it in the first case? $\text{[math]}$ !

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6 years ago, # ^ |

The round start at 10 and finish at 1. It starts later than 7, finish earlier than 9 and it doesn't start at 9. What's wrong?

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5555555555

6 years ago, # |

For E Can we pick max and min element get their average and then pick all less elements than Average?

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6 years ago, # ^ |

I don't think so. As you add more small elements the overall average decreases. At a certain point adding new elements will start to increase the average, even if it's less than the original average.

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lmn0x4F

6 years ago, # ^ |

← Rev. 2 →

I solved it using this idea but changing the average as I insert the elements... This way is very simple and doesn't need ternary/binary search.

Take a look: my submission

We always use the max and a prefix of the elements, and this prefix always increases, so it's very easy.

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6 years ago, # ^ |

← Rev. 2 →

thnx,your idea is very well!!

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ilovekidneystone

6 years ago, # |

When will the ratings be updated?

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royappa

6 years ago, # ^ |

I have a tremendous fear of kidney stones. Why on earth have you chosen such a handle, may I ask??

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ilovekidneystone

6 years ago, # ^ |

Because I actually have them in my kidney.

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6 years ago, # |

How to solve C?Anyone please explain .

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6 years ago, # ^ |

Let's find the max sum on each segment with the size of (f-s). First segment will be s...f-1, second s-1...f-2.

Ok. But there is 1 detail: the beginning of the segment can be in the end of our array and the end of it can be in the beginning of our array.

I hope my code'll help you to deal with it

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6 years ago, # ^ |

10

7171 2280 6982 9126 9490 2598 569 6744 5754 1855

7 9 Why here answer is 4? Please explain :)

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6 years ago, # ^ |

What is the correct answer you think?

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saihemanth9019

6 years ago, # ^ |

This problem can be solved in O(n). Initialize an array with value 0 of size n. Consider 1st timeline as the starting point and can be represented as 1 2 3 4 ... n. While, 2nd timezone can be represented as 2 3 4 ... n 1 (wrt to 1st timezone) and so on... Replace the indices with the participants. And then do an array addition from s to f-1 for each timezone in the initial array. The index corresponding to max value is the answer.

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Sugardorj

6 years ago, # |

Before this round my rating in div 2 But after by education round my rating was 1954. Is calculate my rating this round?

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Sugardorj

6 years ago, # ^ |

Oh no. Now my rating 1816 :(

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lashavi

6 years ago, # ^ |

I think rating has been updated without considering performance of educational round. This should be fixed.

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isaaclo97

6 years ago, # |

← Rev. 2 →

In problem C, why this test case is 4 and not 5

10
7171 2280 6982 9126 9490 2598 569 6744 5754 1855
7 9

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6 years ago, # ^ |

Because （9126 + 9490） is maximum

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6 years ago, # ^ |

Explain,he knows that is maximum.

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6 years ago, # ^ |

← Rev. 4 →

 4    5    6    7    8    9   10   1    2    3
7171 2280 6982 9126 9490 2598 569 6744 5754 1855
 N    N    N    Y    Y    N    N   N    N    N

I think he can understand

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6 years ago, # ^ |

← Rev. 2 →

Thank you brother :) got it properly . But when there are many answers,how do i select the best? Upd : Solved

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6 years ago, # ^ |

Congratulations, I believe you can do better next time contests.

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didenko.ilia

6 years ago, # ^ |

Obviously the maximum is 9126+9490. So we want the contest to start at 7 hours in the region with 9126 to cover the two. Now we need to find at what hour it will start in the first timezone.

Timezone / People / Hour
1 7171 4 <-
2 2280 5
3 6982 6
4 9126 7
5 9490 8

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6 years ago, # ^ |

In 8th hour,answer is being maximum which is 5 in timezone.So,why the answer is 4 ?

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didenko.ilia

6 years ago, # ^ |

If we start when it is 5 in the first zone, we will actually get 9490 + 2598 which is not maximum. Because in the zone with 9126 it will be only 6 hours the time which is too early to participate (<s)

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Fullmetal_Algorithmist

6 years ago, # |

+18

Some people who became div-1 in the Educational round was rated in this round. If you go to the rating changes of this round you can see some purple guys got their rating updated which shouldn't happen as this round is for div-2. This should be fixed.

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6 years ago, # ^ |

+12

List:

MysteryGuy2, RCG, Timsel, WORLD_OF_TANKS, winner, andrew, faustaadp, furys, Chefer, levapnilirbuz, smusmu, killer_god, Trote_w, hands_yhy, Sugardorj, Gorix, win11905, kzyKT

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6 years ago, # |

← Rev. 2 →

35423572 can someone debug this solution for problem C?

EDIT : My approach for the problem: I made the array of twice the size so that round or wrapping condition for the end & beginning timezones is considered. I calculated prefix sum for 2*n +1 size array. Then, I searched for maximum diffrence of (a[i+f-s-1] — a[i]). Then calculated corresponding value for the time in the first time zone.

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anant_k_singh

6 years ago, # ^ |

imalgoholic, I had this exact problem. In case there are multiple subarrays with same (max)answer, you have to output the one which leads to minimum start time in timezone 1.

For everyone failing on Pretest 9 of problem C, Try this: 6 1 3 4 1 3 4 1 3 correct ans : 3 your ans : 6

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6 years ago, # |

← Rev. 2 →

+10

More than 20 Candid Master have there rating change in this Div 2 round.

This should be fixed.

MikeMirzayanov

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6 years ago, # ^ |

+15

KAN has mentioned it in his comment here : http://codeforces.com/blog/entry/57819?#comment-414872 . He said those who were in div. 2 prior to the Educational Round will be considered as div. 2 and has rating changes.

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skpro19

6 years ago, # |

← Rev. 2 →

Can someone tell me, what's wrong with my approach in C?

I used range updates to add a[i] to (s + i — 1 , p — 3 + i), and then checked the maximum.

Here's my code.

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6 years ago, # ^ |

← Rev. 2 →

The only thing you needed was to find the segment with max sum in.

First segment is [s,f). Second — [s-1,f-1); If segment number two has the max sum, answer is two. If segment 3 has the max sum, answer is three, and so on.

Look through my solution. At least it looks easier than yours ;)

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Yazan_Khatib

6 years ago, # |

In problem A, why starting the loop from 0 instead of 1 gives a wrong answer !?

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6 years ago, # ^ |

← Rev. 2 →

If you start from 0 you should read a[i] and then decrease it (a[i]--).

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Yazan_Khatib

6 years ago, # ^ |

yeah I know, but I don't understand why what's the difference

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6 years ago, # ^ |

← Rev. 3 →

"There are n planes on Earth, numbered from 1 to n, and the plane with number i likes the plane with number fi," Here's the sample:

3 1 2

You consider that person number 0 likes person number 3. But if you start from 0 there is no person #3. (a[0],a[1],a[2])

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6 years ago, # |

Can somebody explain what's wrong with my solution of C? 35398638

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manu1311

6 years ago, # |

Will never use cin and cout in my life again! Problem E. :/

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6 years ago, # ^ |

std::ios::sync_with_stdio(false)

Do you add this?

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6 years ago, # |

← Rev. 2 →

When the universe just sided with you..

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6 years ago, # ^ |

← Rev. 2 →

？？What is the difference between Educational Codeforces Round and Codeforces Round?

I always thought Educational Codeforces Round wouldn't change the rating......

so after Educational Codeforces Round24 i didn't take part in it any more

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6 years ago, # ^ |

It used to be unrated. But now it is rated for div2 users.

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