Codeforces Round #438 by 2nd Hello Barcelona Bootcamp with Sberbank

7 years ago, # |

Typical concern: will the drain be adjusted :)?

→ Reply

Errichto

7 years ago, # ^ |

+52

And please make differences between point values more geometric. The last problem should be worth a lot lot more than a middle one.

→ Reply

Taan

7 years ago, # |

-11

![ ]()

→ Reply

darth_coder

7 years ago, # ^ |

+13

Each Data Structure is Special :P

→ Reply

rajat1603

7 years ago, # ^ |

+64

you can't just use rage comics in 2017.

→ Reply

GreenGrape

7 years ago, # ^ |

+42

2k17 actually

→ Reply

Decayed

7 years ago, # |

← Rev. 12 →

-45

→ Reply

_S1MPLE_

7 years ago, # |

← Rev. 2 →

-23

bad timing :(

→ Reply

OneClickAC

7 years ago, # |

-36

Why this unusual time? :(

→ Reply

7 years ago, # ^ |

-64

I don't think you used good meme. This fits better IMHO:

→ Reply

OneClickAC

7 years ago, # ^ |

-46

Sorry. It was my first time. Probably this tutorial from you will improve me.:)

Meanwhile the community (for you)

→ Reply

kostka

7 years ago, # ^ |

← Rev. 3 →

-36

How about no?

→ Reply

Gassa

7 years ago, # |

← Rev. 2 →

+76

Out of the recent Division 1 contests, more than half are longer than two hours:

AIM Tech Round 4 (Div. 1)	2:30
Codeforces Round #431 (Div. 1)	2:00
Codeforces Round #432 (Div. 1)	2:30
Codeforces Round #433 (Div. 1)	2:00
MemSQL Start[c]UP 3.0 — Round 1	2:30
Codeforces Round #434 (Div. 1)*	2:00
Manthan, Codefest 17	2:30
Codeforces Round #437 (Div. 1)	3:00
Codeforces Round #438 (Div. 1)	3:00

*Round 434 was intended to be 2:00.

Personally, I like when an online contest takes at most two hours, be it a TopCoder SRM, a Codeforces Round, an AtCoder Round, etc. As they get longer, I'm usually unable to fully take part, and so have no love for them.

So, here goes: are there plans for shorter Codeforces Division 1 Rounds as well, for a change? An 1.5-hour experimental round sounds fun.

→ Reply

7 years ago, # ^ |

+29

There are few things often associated with longer contests:
1) more tasks
2) merging divisions
3) harder tasks
4) not adjusted drain
5) bad scoring distribution

4) is always a tragedy, but easily fixable if coordinator cares.

5) is also bad, but this is a more controversial subject, discussed few times. I prefer sth like 250-500-750-1000-1500-2000-2500 and I do not like "standard" 500-1000-1500-2000-2500-3000-3500, because Div1A is worth 1500 and Div1D is worth 3000 pts whereas ratio should be higher than 2 times and if drain is not adjusted then probably submit to Div1D will be worth less than submit to Div1A.

I also do not like 2), I end up with mostly green/cyan guys in room and have nobody to hack on >=Div1B problem and there's nobody to hack my solutions if they contain a tricky case. Moreover if I do a stupid bug in some easy problem I lose a lot of points and I may end up in very far position losing more rating than appropriate.

I basically don't care about 1), but it is kinda a cause for 5) which I don't like.

And I like 3), but this is a rare case that we get more time because of harder tasks compared to reasons like "we wanted to merge divisions and added Div2 tasks to solve for Div1 users"

→ Reply

7 years ago, # ^ |

On the other hand, I actually like longer rounds because they allow me to worry less about working fast and take more time to think through things carefully :)

→ Reply

7 years ago, # |

+14

Any lecture videos available?

→ Reply

claporslap

7 years ago, # |

+44

Explain?

→ Reply

bazsi700

7 years ago, # ^ |

+20

It's a typo, that is supposed to be #439.

→ Reply

haleyk100198

7 years ago, # |

+14

Ah, at last, finally an early Asian contest for me to raise my ratings.

SILVER SCRAPEs STARTS PLAYING

(League of Legends Worlds starts today at 5 mins before as this round)

→ Reply

E869120

7 years ago, # |

Hope problem statement is short... !!!

→ Reply

dush1729

7 years ago, # |

← Rev. 2 →

Unable to view submissions in my room after locking.

EDIT — Works fine in chrome

→ Reply

RohanRTiwari

7 years ago, # |

+64

This is just a suggestion, please ignore if deemed unfit by you.
In longer contests, the points drain for problems is adjusted, which is ideal. However, shouldn't the penalty for a wrong submission be adjusted too? Usually, there are cases when one contestant submits his code after some other person, but still ends up with more points on that problem, because the other person has a wrong submission on the same problem. This makes sense if the difference in their submission times is not too large. However, in longer contests, like today's, the points drain is slower than usual, and hence, a wrong submission is actually far worse than it normally is. For example, there are cases when a person has got his solution passed much after some other person, but still has more points, owing to the other person's incorrect submission(s). All I am saying is that, there is far lesser incentive to finish off a problem quickly as compared to solving a problem in the first go.
PS. This is just a suggestion, and not a complaint or a rant. I wasn't even affected by this, I just happened to notice some others who had been.

→ Reply

starbot

7 years ago, # |

+28

RIP rating

→ Reply

Trath

7 years ago, # |

Anyone else also did C using case-work ?

→ Reply

Len

7 years ago, # ^ |

lol

→ Reply

Trath

7 years ago, # ^ |

← Rev. 2 →

+15

Nah, this is just sad. :(

I really think that my one is worst

→ Reply

I_am_Heisenberg

7 years ago, # ^ |

+19

I did.. :p 31042120

→ Reply

eviebot

7 years ago, # ^ |

That is the biggest submission i have ever seen.

→ Reply

I_am_Heisenberg

7 years ago, # ^ |

Ha ha.. I tried every possible case. Took 40 minute to code.

→ Reply

AshishYadav

7 years ago, # |

+12

How to solve D ??? I spent 2 hours on it alone :(

→ Reply

Sophie

7 years ago, # ^ |

It seems that answer cannot exceed 8 but I couldn't proceed more from that.

→ Reply

ChickenInKitchen

7 years ago, # ^ |

really ? why the length of the strings can be very long (2 ^ m * 100)

→ Reply

VastoLorde95

7 years ago, # ^ |

When you combine two strings, you create at most K-1 new substrings of length K

→ Reply

VastoLorde95

7 years ago, # ^ |

You only need the first K and last K characters from every string where K is the maximum possible answer.

→ Reply

Vicennial

7 years ago, # |

Hack cases for B and C?

→ Reply

7 years ago, # ^ |

← Rev. 7 →

6 5 59 1 2, answer is NO

6 5 59 12 1, answer is YES

11 59 59 12 11, answer is YES

most frustrating is when you see a bunch of solutions using doubles, and you know they will fail system test but you don't want to hack them yourself because you know you won't be able to create a good test case to fail them

I'll admit that reading most B solutions was extremely painful, so I actually typed most of them out and just tested them against strong test cases. Now my hands really hurt and I think hacking should allow copy-paste :P

In all honesty I think a problem like B should have strong pretests because nobody likes hacking problems such as these :P

→ Reply

send_nodes

7 years ago, # ^ |

+14

Why will double solutions fail?

→ Reply

7 years ago, # ^ |

← Rev. 3 →

I think most of them should fail simply because it's unreasonable to use double to check if two numbers are equal and expect the result to be correct a large portion of the time

→ Reply

ChickenInKitchen

7 years ago, # ^ |

what about long double

→ Reply

Gassa

7 years ago, # ^ |

The general notion that one shouldn't compare doubles for equality is too vague, and doesn't apply probabilistically, since doubles work in a deterministic way.

In this problem, doubles don't seem like much of a problem. For one, distinct positions of clock hands are too far apart to compare wrong. Also, integers are stored with full precision as floating-point numbers, and dividing the same integer by the same constant produces the same result — unless the result is stored in different types (double and long double) due to, e.g., optimization.

→ Reply

Sasa96

7 years ago, # ^ |

for first test case 6 5 59 1 2 doesn't that mean that the minute hand will coincide mesha's starting position?

→ Reply

7 years ago, # ^ |

← Rev. 2 →

No, it won't coincide because the minute hand will be between the 5th minute and the 6th minute, extremely close to the 6.

→ Reply

Cro-Marmot

7 years ago, # ^ |

if input is 12 0 0 12 1, what is the answer 'Yes' or 'No'? and 12 0 0 1 12

I don't quite understand the meaning of the step over

→ Reply

7 years ago, # ^ |

In problem statement it's guaranteed that one of the query hands won't coincide with any of the hour, minute, or second hands, so this test case is invalid. (In this case, the hour hand is exactly on the 12, so it coincides with the query)

→ Reply

Cro-Marmot

7 years ago, # ^ |

oh. thx!

→ Reply

Dazzler

7 years ago, # ^ |

You should see my solution then.

→ Reply

SureYeaah

7 years ago, # ^ |

← Rev. 3 →

For C:

2 3
1 0 0
0 0 1

Ans is YES

→ Reply

7 years ago, # ^ |

← Rev. 2 →

What about this test case?

→ Reply

SureYeaah

7 years ago, # ^ |

→ Reply

ainta

7 years ago, # |

+33

After I solved six problems, I realized the hardest problem(G) is unsolvable to me, so I started to hack other source codes. But when I attempted hacking, there was nothing but never-ending loading bar. Why there were problems although the server was working?

→ Reply

Vicennial

7 years ago, # ^ |

+14

This is mostly related to the browser.
It happens to me on chrome regularly.

I use firefox for hacking and I haven't faced this issue.

→ Reply

.o.

7 years ago, # ^ |

Why use Flash even on the submitting form? After preparing the hack test, I had to waste some time waiting for the form to pop up.

→ Reply

Len

7 years ago, # ^ |

Maybe it's to make harder to copy & paste solution(I think it's still possible with some ocr)

→ Reply

.o.

7 years ago, # ^ |

Of course you're right. But if that is only the case, I think using Flash only on showing codes is enough.

→ Reply

7 years ago, # ^ |

Why don't we want people to copy-paste solution :P

I think for problems such as B, it would be very beneficial to the sanity of most people to allow copy-pasting for hacking :P

→ Reply

-Morass-

7 years ago, # ^ |

Good day to you,

In my opinion hacking with copy-pase (vs.) hacking with "flash" might pretty different. Hacking flash is mostly about observation & finding a mistake (somewhere). If copy-paste would be allowed, one could easily create stress-tests and hack the solution without even looking at the code :)

Yes, this have some drawbacks, yet I guess this might be the main reason.

Good Luck & Have Beautiful Day! ^_^

→ Reply

JustasK

7 years ago, # |

Before system tests, it seems that 40 people solved E and 70 people solved F, yet only 14 people solved both. This is weird.

→ Reply

SirirNicheBirirDokan

7 years ago, # |

← Rev. 2 →

What's the idea behind G? We repeat making $\text{[math]}$ continuous sweeps and then work out the formulas?

→ Reply

SirirNicheBirirDokan

7 years ago, # ^ |

← Rev. 3 →

I think I solved the case $\text{[math]}$ . Considered all n cyclic sweeps. I overlooked that when $\text{[math]}$ we can make less sweeps and realized it too late. :(

→ Reply

7 years ago, # ^ |

It's not like key difficulty of this problem lies in reducing to case gcd(n,k)=1 by dividing by common divisor and then there are just few simple formulas...

→ Reply

Los_Angelos_Laycurse

7 years ago, # ^ |

I think there is O(log(n)) solution, we should compute

sigma((i*n/k)/2^(i+1)) 0<=i<=k-1. this can be solved by Euclidean algorithm

→ Reply

ksun48

7 years ago, # ^ |

Hey, could you elaborate? I don't see how to calculate that, and I also don't see how that suffices to solve the problem.

→ Reply

Jeffrey

7 years ago, # |

I don't think my D should pass, but it passed...

→ Reply

http://codeforces.net/contest/868/submission/31028937

7 years ago, # ^ |

I think that in pretests operations did not create anything more than what was in original strings.

→ Reply

Jeffrey

7 years ago, # ^ |

← Rev. 2 →

And it really passed system tests XD

Is it correct? 31024676

→ Reply

WildUtah

7 years ago, # ^ |

That's hilatious. Mine was about 10x as slow, long, and complex as yours and also passed.

→ Reply

Jeffrey

7 years ago, # ^ |

So seems like my solution is wrong, just change the maximum preserved length of string may result in wrong answer. I'm just lucky.

You can check here: 31053045 (change L to something else, my solution in contest equals L = 500)

Set it to 500, 550, 700, 1000, 567, 543, 510, 100000 will get AC.

However, set it to 400, 475, 499, 501, 600, 456, 511, 512, 10000 will get WA in various tests.

→ Reply

xuanquang1999

7 years ago, # |

I need one more second to submit my solution for problem D...

Btw, is the solution just iterate the length from 1 to 15, brute-force all binary string of that length and check using DP?

→ Reply

Satyam22

7 years ago, # ^ |

I did the same and did for 1 to 10 only dont know will it pass systest or no

→ Reply

7 years ago, # ^ |

No need for dp — just brute everything.

→ Reply

Satyam22

7 years ago, # |

What is upper bound on answer. I have taken 10 and it has passed pretest not sure about systest

→ Reply

newCEA

7 years ago, # ^ |

← Rev. 4 →

I could prove it for 20 (still should work without TLE but couldn't code in time) but I think it could be smaller than that. There are at most $\text{[math]}$ substrings of length k that can be made. Finding maximal k with 100 × (100 - k) + 100 × 99 × k ≥ 2^k yielded k = 20 for my very approximate calculations.

→ Reply

Satyam22

7 years ago, # ^ |

For 20 i think it will get tle as it will 100*2^20*log100

→ Reply

newCEA

7 years ago, # ^ |

It's basically $\text{[math]}$ so it may pass, but very close to TLE without a lot of arithmetic optimizations.

→ Reply

newCEA

7 years ago, # ^ |

← Rev. 2 →

I just got myself a calculator and this analysis gives k = 17 as the greatest possible value.

→ Reply

Sarkin

7 years ago, # ^ |

← Rev. 2 →

+32

For a fixed length k, there are at most 100 different substrings initially, and each move can add at most k new substrings, so we have 100 + k * 100 ≥ 2^k, and so k ≤ 10 is enough.

→ Reply

adamant

7 years ago, # ^ |

Isn't it k ≤ 11?

→ Reply

Sarkin

7 years ago, # ^ |

For k = 11 the inequality doesn't hold.

→ Reply

adamant

7 years ago, # ^ |

oh, got it :)

→ Reply

Anadi

7 years ago, # ^ |

+18

I used 15. You cannot create more than k new substrings while merging two strings. So the max answer cannot exceed what we have at the beginning plus what we add (m * k), it's something like 20 000

→ Reply

skpro19

7 years ago, # ^ |

Why can't we create more than k new substrings while merging two strings?

→ Reply

Anadi

7 years ago, # ^ |

If we merge string s1 and string s2, new substring have to contain part of string s1 and string s2. So we can place it only on the k — 1 last positions of string s1 (otherwise it would be completely in string s1 or string s2). So we can't create more than k — 1 new substrings.

→ Reply

kingofnumbers

7 years ago, # ^ |

I've just got AC using 6 as upper bound (and I believe 6 is very easy to achieve so upper bound can't be less)

→ Reply

7 years ago, # ^ |

Tests to this task were very weak...

→ Reply

7 years ago, # ^ |

← Rev. 2 →

Why downvotes? I wasn't fu**ing with you guys. Post above + this prove that tests were weak.

→ Reply

7 years ago, # ^ |

← Rev. 2 →

+10

What the heck ._.? Setters should get familiar with DeBruijn words. My bound was 12, but if I'm not mistaken we can get even 13 if concatenation of words on input is DeBruijn's word of order 13.

EDIT: Ahhh, I see that there was a limit of total length up to 100. My upper bound of 12 works. That makes it harder to construct cases, but 6 is still easily breakable.

→ Reply

HBK_Wasi

7 years ago, # |

Systest started.

→ Reply

eat_code

7 years ago, # |

Can someone explain the clear logic behind B problem. Also, if someone understands this submission 31012140 I_love_Tanya_Romanova

→ Reply

I_love_Tanya_Romanova

7 years ago, # ^ |

+12

There are two possible paths from start to finish: clockwise and counterclockwise. For each hand I'm checking which of these two paths it blocks — it will block exactly one of them since it is guaranteed that start and finish are different from position of any hand.

There is a solution if and only if one of two paths is unblocked.

→ Reply

GyojunYoun

7 years ago, # |

+22

Could you tell me why considering only two problems set is enough (in Problem C)? Is it related to the range of k? If k = 20 then the solution does not work??

→ Reply

getSchwifty

7 years ago, # |

← Rev. 2 →

How to solve C ? [ Hope you pass System Tests ]

→ Reply

yaksha

7 years ago, # ^ |

this problem C of today's contest looks like the type of problem where system tests totally crush the solutions..

→ Reply

getSchwifty

7 years ago, # ^ |

People On Codeforces are good in Sense of humour

→ Reply

yaksha

7 years ago, # ^ |

→ Reply

MiteshAgrawal

7 years ago, # ^ |

← Rev. 3 →

Consider 2 problems, which are exact same in terms of teams who know it. Adding both never make sense, hence we can consider only 1 occurrence of it. Now as there are only 4 maximum teams, there are only 16 different possible combination of question type.

Now consider all 2¹⁶ possibilities of question set, and check validity.

code

→ Reply

7 years ago, # ^ |

← Rev. 3 →

I (and many others I know of) submitted a solution where we checked:

If there's a problem that no teams know, answer is YES.

If there is a pair of problems such that no team knows both of them, answer is YES.

Otherwise, answer is NO.

I haven't formally proved this, but it seemed very logical when I tried some smaller test cases, and it passed the system test. Does anybody have a proof of this?

→ Reply

voover

7 years ago, # ^ |

I have one, but it's not pretty.

Otherwise, answer is NO.

This is the only non-obvious part. Let's assume the previous cases are not true, but the answer is still YES.

If there is a problem that is known by exactly one team, all problems have to be known by it (otherwise we'd have answered YES in the previous step).

Now, all the problems are known by at least 2 teams. It follows that k=4 and we need to compose the problemset from problems that are known by exactly 2 teams (otherwise, the number of team-familiar problem pairs would be too large).

Now it's some case-ology to show that no such problemset can exist. W.l.o.g. let's assume two of the used problems are known by teams 1,2 and 1,3. Next one would have to be 1,4 or 2,3 — in both cases we cannot further expand the set.

→ Reply

yaksha

7 years ago, # ^ |

-9

Wow, that's a very elegant solution! how did you think of that,I was thinking of something a little greedy like but coudnt come up with something.. Please tell me your train of thought , how did you come up with this solution, Have you done problems using this technique earlier or it striked you during contest?! Anyways,brilliant technique!

→ Reply

7 years ago, # ^ |

+39

It's not clear why adding both doesn't make sense, but I think that for k<=4 we can always find a solution of <=2 questions so in fact it is true, but I think your claim will become false if we allow bigger values of k

→ Reply

Gassa

7 years ago, # ^ |

← Rev. 2 →

+51

Yeah, here is a counter-example to the statement "size of answer <= 2" if we allow k = 6:

4 6
1 1 1 0 0 0
1 0 0 1 1 0
0 1 0 1 0 1
0 0 1 0 1 1

Each two distinct teams have a common known problem, but if we take all four, each problem is known by exactly two teams.

I can't figure out what's with k = 5 though.

→ Reply

ksun48

7 years ago, # ^ |

It's actually still true for k=5.

Assume there is a problem used on the test which is unknown to exactly three teams. (the four and five cases are easier.) Call them teams 0, 1, 2.

If some other problem is unknown by teams 3 and 4, then the answer is YES and we can use these two problems.

Otherwise, each of the other problems will have been seen by at least one of team 3 and 4. Then we cannot make a test where teams 3 and 4 have seen at most half the questions using the original question. (and we can remove it)

→ Reply

OptxPrime

7 years ago, # ^ |

Why it never makes sense to add two same problems in terms of teams who know it? Maybe adding that problem will give us additional 0 to one team and ones to others which wont make problem.

→ Reply

satyaki3794

7 years ago, # ^ |

+15

You can prove that if the answer exists, the size of the subset of chosen problems is atmost 2.

Code

→ Reply

dwivedi

7 years ago, # ^ |

← Rev. 2 →

What is the problem subset in the following case:
4 4
1 0 0 1
1 1 0 0
0 1 1 0
0 0 1 1
I think we have to consider all four problems for a valid problem set.
Edit: Choosing (1st and 3rd) or (2nd and 4th) problem is sufficient

→ Reply

Aemon

7 years ago, # ^ |

+17

The first and the third problems are sufficient. No team will know more than 1 problem.

→ Reply

dwivedi

7 years ago, # ^ |

Oh Yes! Sorry. My bad.

→ Reply

Ishan.nitj

7 years ago, # ^ |

any proof?

→ Reply

towrist

4 years ago, # ^ |

Lets convert the binary matrix into a vector of binary numbers and arrange them in increasing order of highest bit.

If the minimum maximum bit is 1, then the last team should know every problem.

If the minimum maximum bit is 2, then you can use inequalities.

If the minimum maximum bit is 3, then there is a balance between bit-4 and bit-3 which will be broken after cross-matching.

If the minimum maximum bit is 4, then team 1 knows every problem

→ Reply

imalgoholic

7 years ago, # |

+25

when they do system testing manually...

→ Reply

7 years ago, # |

← Rev. 2 →

how to solve B ? i have tried it but failed on test 56 (:

→ Reply

7 years ago, # ^ |

← Rev. 2 →

Then you did not solve it.

I just divided the clock into 12*60*60. Hour hand moves by 1 such unit every second. Minute hand moves by 12 such units every second. Second hand moves by 720 such units every second.

Then just compare these values with t1*3600 and t2*3600 and that's it.

→ Reply

7 years ago, # ^ |

Okay, just to be clear on this — so the whole question depends on the system being dynamic, not static right? Is that correct? My attempted solution tried finding opens paths in the clockwise and anti-clockwise directions consider the point in time in which t1 and t2 are considered. That worked for the sample cases, but failed miserably in the pretests.

Could you please go into a bit more detail about your logic? That would help us beginners a lot. Thanks!

→ Reply

http://codeforces.net/contest/868/submission/31017176

7 years ago, # ^ |

← Rev. 2 →

Every second, all hands move in clockwise direction. In order to measure this precisely I divided the clock into 12*60*60 units (the total number of seconds in the whole turn on the clock — 60 seconds in minute * 60 minutes in hour * 12 hours in total). Now you can observe that with every second hour hand moves by one such unit. Minute hand completes the whole turn within an hour so you divide 12*60*60/60*60 = 12 units with every second. Second hand completes the whole turn within a minute, so you divide 12*60*60/60 = 720 units.

Once you scale the values, you have precise integer values of hands positions. You can then compare these values against t1 and t2 (you have to scale them too by *= 3600). The comparison is simple — just check if none of the hands block one way or another:

→ Reply

7 years ago, # ^ |

Thank you so much for the explanation! It makes so much more sense now. The worst part is that it is a very simple concept as well, isn't it? I really must stop overthinking things. Thank you again for taking the time to lay it out so well — it really helped me understand the solution.

→ Reply

7 years ago, # ^ |

You're welcome. I'm glad I could help.

As regards simple concepts — that's almost always the case with short contests. I used to call these tasks tricky as they really require to get a trick — sometimes an observation, sometimes just restating the problem in a much simpler way, sometimes reducing to something simpler. I hope that finding such ideas will be easier with practice :)

→ Reply

OptxPrime

7 years ago, # ^ |

-20

You can divide clock like this guys said, but it wasnt necessary. I only converted hours to seconds: hours = 5*hours + minutes/60 + second/3600. and minutes to seconds: minutes = seconds/60 You must work with doubles,but suprisingly it wasnt problem, i got AC without having precision troubles.

→ Reply

7 years ago, # |

+10

Tests to D are very weak. My buggy code got accepted after the contest.

→ Reply

7 years ago, # ^ |

can you please explain this one also !

→ Reply

7 years ago, # ^ |

← Rev. 4 →

+10

My code was checking initial values in strings and then was setting prefixes and sufixes only if the concatenated strings were remembered as whole strings. If they were not remembered as whole strings (it means their length was greater than 109) it just checked new strings on the concatenation of sufix and prefix of these strings and do not set new prefix and suffix. Which means that if this string will be concatenated in future, it will present empty prefix and sufix.

It fails this test:

4
0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111
1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111
4
1 2
3 4
5 6
7 1

BTW, it is also interesting that you could have stored all the generated strings up to 109 length and it would fit within memory limit: 254500 KB

Edit: I corrected that bug during the contest and lost 180 points. It showed up that it was not worth it. Very mean as usually when I do not notice such bugs my solutions fail and when I noticed them, they are not caught by systest...

Edit2: Test exceedes input size but you can get the idea of the problem. Submission: http://codeforces.net/contest/868/submission/31030570

Edit3: When I fixed the bug and I tried 109 upper_bound it exceeded memory limit but I'm pretty sure something around 90 would fit. Don't want to check it though.

→ Reply

rsazid99

7 years ago, # |

B 46 test case: 10 50 59 9 10 is it ok ??? bcz in statement it i found Misha's position and the target time do not coincide with the position of any hand. so how it becomes a valid case ?

→ Reply

eviebot

7 years ago, # ^ |

The hour and minute hands will be a bit ahead of 10, so it is a valid test case.

→ Reply

nukeykt

7 years ago, # |

Sent same solution with other compiler and it got accepted.

→ Reply

rajat1603

7 years ago, # |

+11

The pretests seem too weak in D , I didn't memoize my DP hence my complexity should have been something like 2^Q * (Something related to K) and Q was upto 100.

→ Reply

7 years ago, # |

how to solve D?

→ Reply

mDziewanowski

7 years ago, # |

Guys i dont get these statement with B problem...

4th pretest : 10 22 59 6 10 ANS = YES

Ok, there is a path between 6th hour and 10th hour, even if our point is directly on the hour hand we can get it

But 5th pretest : 3 1 13 12 3 ANS = NO

But why? Here there's the same situation in my opinion.. there is a path between 12th hour and 3rd hour, but we just have to go backwards and our point is also directly on the hour hand just like in 4th pretest...

why answers are different? cannot we go backwards? The task statement says that "he doesn't have to move forward only: in these circumstances time has no direction"

What do not i understand?

→ Reply

eviebot

7 years ago, # ^ |

Because seconds hand is at 13, minute hand is little ahead of 1. Because minute hand is a little ahead of 1, hour hand is a little ahead of 3. Hence answer is NO.

→ Reply

Rafaaaaaaaaaaaam

7 years ago, # ^ |

← Rev. 2 →

Just imagine a clock. At exactly 3 o'clock the hour hand is at the position of 3. When one minute and 13 seconds has past, the hand is a bit further than position 3. So it is between 3 and 4.

→ Reply

Nson

7 years ago, # |

How to solve F?

→ Reply

7 years ago, # ^ |

http://codeforces.net/blog/entry/8219 Divide and Conquer.

→ Reply

anonymousbuggy

7 years ago, # ^ |

← Rev. 2 →

I too used Divide and conquer optimization but got TLE ... Can you check this once? 31025909

→ Reply

7 years ago, # ^ |

for(int i = m; i > cap + 1; i--){ tmp += arr[a[i]]; arr[a[i]]++; }

This makes your code O(n^2) for some case. Check the go function here to optimize it: http://codeforces.net/contest/868/submission/31021372

→ Reply

-osdajigu-

7 years ago, # ^ |

Why go function doesn't change the complexity of your solution?. I realized that D&Q could work but I couldn't come up with an idea to calculate the cost..

→ Reply

7 years ago, # ^ |

← Rev. 3 →

Let's see the movement of both pointers:

Right pointer: when it gets to [l, r], it comes from r + 1, goes to the mid, solves [l, mid — 1] ending in mid — 1, solves[mid + 1, r] ending in r (so when it comes back, it is in mid — 1 from the previous call). So for each range [l, r] it will move O((r — l) / 2). If you take the sum of every range it will be O(n*logn)

Left pointer: when it gets to [l, r] assume that it starts on optr, it will move on [optl, optr] to find the answer, move back to the optimum point, solve the left part (going back on the optimum point after the call), go back to the optr and solve the right part. So for every range you take the 2 * (optr — optl) and that's O(n*logn)

The idea is the same as parallel binary search for the complexity of moving the pointers.

→ Reply

-osdajigu-

7 years ago, # ^ |

Wow thanks, man. It's very clear now.

→ Reply

mochow

7 years ago, # ^ |

Isn't it similar to what we do in Mo's Algorithm?

→ Reply

7 years ago, # ^ |

It is but at the same time it isn't because the way you divide the queries you want the answer is:

online
not in buckets of size sqrt(n) for the left pointer

→ Reply

Cro-Marmot

7 years ago, # ^ |

your go function is awesome! you're so smart

→ Reply

thedark

7 years ago, # |

Can someone explain how to solve D?

→ Reply

7 years ago, # |

← Rev. 2 →

Hey all,

In the contest, I got hacked by the following test case (Problem C — Qualifications Round):

The answer for this should be "YES", isn't it? Question 3 only has 2 teams which know the solution, and that is less than or equal to half of k=4. Am I missing something here? (The judge says that the correct answer is "NO").

Thanks!

→ Reply

7 years ago, # ^ |

← Rev. 5 →

You misunderstand problem.

The number of known problem for each team should be less than or equal to half of total number of problem in problem set.

For given example, total number of problem in problem set ( T ) is 1.

The second team know 1 problem, but this value is greater than T/2.

So given example is invalid.

→ Reply

7 years ago, # ^ |

Thanks for the reply, ainch96. I think there was some problem with the formatting — fixed it. The given example has 3 problems with 4 teams in total. Can you please explain this test case (or have I interpreted your answer incorrectly?)

→ Reply

7 years ago, # ^ |

I will give simple example for given test case.

For given problem A = {1,2,3} you can choose some of them.

Let assume that we get two problem S = {1,2} ( |S| = 2 )

Then, Team 1 know problem 1,2, Team 2 know problem 2, Team 3 know problem 1,2, Team 4 know problem 1

For this, Team 1 know 2 problem, which is greater than |S|/2. So this is invalid case.

For all nonempty problem set, {1},{2},{3},{1,2},{2,3},{1,3},{1,2,3} ,they are invalid. Get it?

Sorry for my English. I'm not the native, so i guess there are so many grammar error and word error.

→ Reply

7 years ago, # ^ |

Ahhh! Now I get it — stupid mistake from my side in understanding the problem. So, basically, the problem set size keeps changing depending on how many we are considering at one time, but I was checking against a fixed number (one of my several mistakes in this problem). Also, this explains why some solutions I had checked were enumerating subsets. Hmmm.

And, no, your English is absolutely topnotch. Thank you for the explanation! It helped a lot! :-)

→ Reply

7 years ago, # |

Can anyone give me hint for problem E?

→ Reply

Jakube

7 years ago, # ^ |

+11

It's easy to see, that the criminals will hide at the leaf nodes. And so the policeman will catch each criminal at leaf nodes.

Perform dynamic programming: dp[c][v] = time to catch c criminals, if the policeman is currently at a leaf node v.

Is that enough? Computing these values is not trivial.

→ Reply

7 years ago, # ^ |

I get it with reading your code. Thanks so much.

→ Reply

satyaki3794

7 years ago, # ^ |

Could you elaborate on the DP relations? What are the transitions from dp[c][v]? I tried exactly this state but the transitions got messy. :\

→ Reply

Jakube

7 years ago, # ^ |

← Rev. 3 →

+25

You can compute the value dp[c][v] using a binary search. Criminals want to be found as late as possible, therefore you look for the biggest time that is possible for the criminals to archive (using a good spreading).

Checking if a specific time >= M is possible, you simply place as many criminals as possible at each leaf, such that either choice of the policeman ends up in a time >= M. (placing x criminals at leaf u takes cost[v][u] + dp[c-x][u] time).

Here is my implementation: 31030667 (variable naming is quite messy, but I hope you get the idea)

→ Reply

nimom

7 years ago, # |

Is bitmask dp required for solving C?

→ Reply

7 years ago, # ^ |

You can solve it just by brute force with bit. You don't have to use dp.

→ Reply

AshishYadav

7 years ago, # |

← Rev. 2 →

How to arrive at the maximum number K in the question D ???

→ Reply

Borisov

7 years ago, # ^ |

← Rev. 2 →

I used such idea. For K you should get 2^K different substrings with length K. Let's denote that we have string S with L length.

Count of different substrings with length K for string with size L are equal L — K + 1.

Then we have next formula: 2^K <= L — K + 1 => 2^K + K — 1 <= L

maxL ~ 10^4 (each step you add string with size 100)

=> maxK ~ log2(10^4) ~ 14

Correct me, if I'm wrong.

→ Reply

misael

7 years ago, # ^ |

MaxL isn't 10^4, each step you can add the last two strings, then maxL is 2^m

→ Reply

Borisov

7 years ago, # ^ |

Yes, you are right it was my mistake)

→ Reply