Codeforces Round #532 (Div. 2)

6 лет назад, # ^ |

← Rev. 2 →

+41

why not?

Rather than complaining about it, maybe it would be more beneficial to get better at the topic, because it's not going away.

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6 лет назад, # ^ |

-107

why not?
because it is completely pointless and rather than wasting time "getting good at it" for no reason, they might as well give us better and more useful problems overall.

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http://imo-official.org/country_team_r.aspx?code=IRN

6 лет назад, # ^ |

+35

Do you have any reason why other problems are “better” and “more useful” besides just your opinion?

It’s not pointless if your goal is to get better at cp. You’ll end up doing better in contests because you’ll know the topic.

And if you don’t want to learn geometry either, that’s perfectly fine. Just don’t start complaining about the questions when you can’t do it, since you made the choice not to learn. There’s 5 other questions for you to try.

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eisen_shinzou

6 лет назад, # ^ |

← Rev. 2 →

-34

I'm neither with any sides nor saying geometry is bad, but you know, memorizing( or even searching ) a formula and coding an O(1) program is not really fun.

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Um_nik

6 лет назад, # ^ |

+87

All you need to memorize is the definition of sin function. If you can't memorize 8-th grade math, then I have bad news for you.

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OmarKimo

6 лет назад, # ^ |

How does the definition of sin function help in solving this problem?

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Um_nik

6 лет назад, # ^ |

+75

You have to connect center of inner circle, center of outer circle and point of tangency between two outer circles, In this triangle we know two sides, one acute angle, and also we know that angle between tangent and radius is right.

$\text{[math]}$

I'm not saying that this problem is innovative or interesting, but in no way that's because it is geometry or math in general.

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eisen_shinzou

6 лет назад, # ^ |

-14

OK, I'm too stupid, and I know it!!!

BTW in our country they teach it in 11th grade, and I haven't reached there yet.

And again; BTW I know what sin func is.( But still stupid )

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M_H_H_7

6 лет назад, # ^ |

in your country you learn it on the second lesson of the 10th grade math!

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Mikaeel

6 лет назад, # ^ |

+15

Your school is very ghavi!

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VadymKa

6 лет назад, # ^ |

jfyi Iran is quite good at math:

Good luck studying more advanced mathematical concepts like tanh or derivatives.

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6 лет назад, # ^ |

-51

exactly, it isn't fun nor useful.

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6 лет назад, # ^ |

+14

It can be boring if you can instantly come up with the correct formula and code it up (but it's also good, since it will give you more time to work on other problems).

If you don't know the formula, you can spend some time deriving the solution using what geometry knowledge that you know.

And isn't going through the process to reach a solution the fun in cp?

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6 лет назад, # ^ |

-66

so in conclusion, it is pointless. period. glad we could all agree.

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6 лет назад, # ^ |

-11

Sorry if you didn’t see it before.

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6 лет назад, # ^ |

-67

i saw it but because you have anime profile picture i have automatically skipped it so i didn't count it.

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Um_nik

6 лет назад, # ^ |

+28

What is useful?

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eidan

6 лет назад, # ^ |

+42

So geometry is useless, but wasting time to create fake accounts is useful?

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6 лет назад, # ^ |

-19

how can an account be fake?

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eidan

6 лет назад, # ^ |

+27

You didn't have the balls to post that comment with your actual account because you feared downvotes, so you created this one. Do you honestly think we don't know that?

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6 лет назад, # ^ |

-12

THIS is my main account. you can ask the codeforces administration about this.
your downvotes don't affect me, they are worthless because i know what i say is true.

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neel71

6 лет назад, # ^ |

+13

You yet to do a contest in your "main account" and you are complaining about problemset. WOW!!!

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6 лет назад, # ^ |

-11

? what are you talking about?

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neel71

6 лет назад, # ^ |

← Rev. 3 →

+10

Before trying to make the community fool, you may not notice your color is still black now :P :P And you may be forget people can go to your profile and see the number of contest you attend till now.

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6 лет назад, # ^ |

-25

you imbecile just because it glitched for you doesn't mean that's how it is for everyone.

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IMRED

6 лет назад, # |

+47

Интерактивка без предупреждений незаконно.

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Noche_3

6 лет назад, # |

Illuminati wants to know your location.

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Tigutor

6 лет назад, # |

+131

There is a drastic difference between A-C and D-F, so the contest turned into speed coding.

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6 лет назад, # ^ |

True. but problem D was not that tough. But don't know why people couldn't solve it.

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TO_OO_EndGame

6 лет назад, # |

I hope you not to repeat it

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6 лет назад, # ^ |

+20

That's not constructive criticism.

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TO_OO_EndGame

6 лет назад, # ^ |

-29

I am really sorry but , the contest went into wrong direction and I hope the next contests that prepared by him to be more good and avoding the mistakes that has happend at this contest . hope good luck for all .

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6 лет назад, # ^ |

-15

you’re right it was trash

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6 лет назад, # ^ |

Was that your another fake ID or something else?

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6 лет назад, # ^ |

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AZ01

6 лет назад, # |

+53

Worst Problem statement with worst Explanation. Wasted 10 min in just understanding the problem Statements

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6 лет назад, # ^ |

+40

Also boring A,B,C and unbalanced problemset. Slight error in implementation and you are very low in ranklist.

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6 лет назад, # ^ |

That happened to me :(

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slappy

6 лет назад, # ^ |

i am still trying to understand D. LOL

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AZ01

6 лет назад, # ^ |

← Rev. 2 →

-11

....

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slappy

6 лет назад, # ^ |

← Rev. 2 →

-24

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gfonn

6 лет назад, # |

Why D, E, F so hard? Massive AC-gap between C & D :(

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SlavaSSU

6 лет назад, # ^ |

+27

For me E is easier than C.

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TheStory

6 лет назад, # ^ |

That's right.But D is too hard to put in a Div.2 D I think.

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6 лет назад, # ^ |

C was pretty straight forward. If you know Trigonometry, It is easier than A , else you wouldn't be able to do it :( How about E ? Didn't read that. Was it a data structure problem ? Or something else ?

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slappy

6 лет назад, # |

A, B, C = A and D, E, F != DIV 2.

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6 лет назад, # |

← Rev. 2 →

In problem F, are we supposed to answer the queries for maximum subarray xor? Probably somehow offline?

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aviroop123

6 лет назад, # ^ |

+13

No, we have to choose some numbers from l to r s.t. their xor is maximum ... It's a problem similar to https://www.spoj.com/problems/XMAX/.

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6 лет назад, # ^ |

← Rev. 2 →

I am aware of that, let me reformulate my question. Is that problem equivalent to finding maximum subarray xor on segments of the given array?

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aviroop123

6 лет назад, # ^ |

No, it's related to finding independent basis vectors of l to r. It can be done using gaussian elimination. But my solution was N*logN*20^2 which is not enough to pass t.l.

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IceKnight1093

6 лет назад, # ^ |

In general, maximum subset xor and maximum subarray xor aren't going to be the same — subarray is a fairly restrictive condition when you need to pick things.

For example, take a = [5, 6, 2]

Maximum subset xor is 7, maximum subarray xor is 6.

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6 лет назад, # ^ |

Isn't subarray a subset of an array?

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IceKnight1093

6 лет назад, # ^ |

If that's the definition you're going with, then yes, the answer is maximum subarray xor.

I'm going with the more commonly used definition for competitive coding, where a subset is your usual subset and a subarray is a contiguous segment of the array. Similar to how a substring is a contiguous segment of a string.

Something like this, basically.

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6 лет назад, # ^ |

Can it be done without Gaussian elimination?

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IceKnight1093

6 лет назад, # ^ |

Maximum subarray xor can be done with a trie — see here for a tutorial.

Maximum subset xor, I don't think so. I'm not aware of any other method at the very least.

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mkisic

6 лет назад, # |

How we were supposed to locally test D?

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hypomaniac

6 лет назад, # |

How to solve D?

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sniffleheim

6 лет назад, # ^ |

+26

Bring the king to the center. Then count the number of rooks in each quadrant, and go diagonally in the direction opposite to the quadrant with the least number of rooks.

Verify as an exercise that this works.

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RedNextCentury

6 лет назад, # |

+26

How to solve F?

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6 лет назад, # ^ |

+49

Take yesterday's G -> TLE with seg tree -> use divide and conquer that you can use just prefix/suffix -> done

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RedNextCentury

6 лет назад, # ^ |

+11

Yeah I got TLE with segment tree, very nice solution, thanks!

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aviroop123

6 лет назад, # ^ |

Is it possible to merge two independent vector sets in better than 20^2 time?

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6 лет назад, # ^ |

← Rev. 2 →

+26

I don't think so. My solution computes suffix/prefix when dividing [l, mid) [mid, r) (to add a number to a prefix/suffix is O(bits) time) and then solves the queries with l < mid and r >= mid in O(bits^2).

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aryanc403

6 лет назад, # ^ |

Your Gauss struct is black box for me.
If you can provide some intuition how those max and min are working?
First time I have seen inserting an element in O(bits).
My implementation in O(bits^2) was giving tle.

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6 лет назад, # ^ |

You can add number by number. So instead of thinking of a N*BITS matrix when you have N numbers, think of a BITS*BITS matrix where you insert one value at a time. Also, don't use vectors in code like this that uses a ton of memory, it'll use a lot more memory and it'll have more cache misses. For max, you have the basis so just go from the number with highest bit to lowest bit and activate the bit whenever you can.

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aryanc403

6 лет назад, # ^ |

← Rev. 2 →

+10

Got it. Thank You.

What I have understood is that each no in your table has a bit (the most significant bit) associated with it. Similar to the basis vector which has at least one dimension with them.

And that no is responsible for activating(max)/ deactivating(min) that particular bit.

P.S. After reading I got core idea of your implementation. But was bit confused with just go from the number with highest bit to lowest bit So, I added prt function in your implementation. And then I came up with the above explanation.
Thank you. :)

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6 лет назад, # ^ |

+10

In your submission (linked below), the Gauss::add() function confuses me.

I think that the invariant for the Gauss class is that if table[i] > 0 then the $i$th bit of that needs to be set. But if you don't make that check in your add(), this invariant could be violated.

Your submission did get AC, so what am I missing here?

48339644

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6 лет назад, # ^ |

← Rev. 3 →

Edit: True, let me think how that worked rofl. So... I got careless and think that code shouldn't work, systests were weak?

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pranjal.ssh

6 лет назад, # ^ |

+34

You can split your array in chunks of size 20 and use sparse table on array of size n/20. Query can now be answered online in 20*20 time by sparse table query + adding elements partially belonging to a chunk one by one.

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Ari

6 лет назад, # ^ |

Nice! I thought of sparse table but I didn't think of dividing into blocks, that's a cool idea.

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6 лет назад, # |

+43

Does E have the following solution?

Sort all the edges according to c. Binary search over this to find the edge so that no cycle is present. Compute the topological order of this graph without cycle and add reverse remaining edges which violate the order.

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Determination

6 лет назад, # ^ |

Sorry if I understand your idea wrong, but don't you mean to reverse all the edges that have a lower cost than the middle in your binary search ? I thought the same but there might be some edges that you can reverse that should not be reversed which made it so much harder

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6 лет назад, # ^ |

You cannot reverse all the edges which have lower weight because you might end up getting cycles in new graph. Instead what you do is reverse those edges which violate the topological order as a directed graph has no cycles iff it has a topological order.

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6 лет назад, # |

+33

The problems were actually pretty good, but the difficulty gap was pretty bad. Hope you can balance the problems in the future!

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I_Love_HanaE

6 лет назад, # |

Is there any nlog solution for F or it's just optimized nlog2?

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mkisic

6 лет назад, # |

+28

How we were supposed to locally test D?

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minibowser

6 лет назад, # ^ |

Write your own local judge, or don't write with mistakes.

Anyways, if you submitted and get a wrong answer on the first test, it will tell you what went wrong, for example "invalid move for king", which is still helpful.

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6 лет назад, # ^ |

+15

I scaled down the problem by a factor of 111, i.e. 9x9 board and 6 rooks in play.

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6 лет назад, # ^ |

That's a really clever line of thought.

I can see that 1 dimensional properties and 2 dimensional properties of the problem will remain the same. But maybe, just maybe, the properties of the problem that require an interaction of a 1D vs 2D would not hold.

I'm sorry that I'm very vague, but I'm not very clear about it myself. I also can't think of an example of such a property.

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golions

6 лет назад, # |

← Rev. 2 →

How to solve E? I thought of using finding a maximum spanning tree and every not used edge would be a flipped edge, but I didn't know how to implement a dsu for directed graphs.

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6 лет назад, # ^ |

← Rev. 2 →

+24

binary search the answer, then the problem becomes to check whether it has cycle.

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golions

6 лет назад, # ^ |

Thanks

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YuGo

6 лет назад, # ^ |

Hey, one quick question, would we reverse all the edges whose c is less than or equal to mid?

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6 лет назад, # ^ |

+15

No, find a valid topological ordering for the graph with > mid. Now, just orient the edges <= mid according to this order.

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lclpsoz

6 лет назад, # ^ |

I don't see why I can ignore edges to check of cycles just because I can choose the orientation of them on the directed graph.

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6 лет назад, # ^ |

Given that you ignore them, you have a topological ordering. Now, you can direct them using the topological order (from the vertex that comes first in the order to the one that comes second).

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lclpsoz

6 лет назад, # ^ |

Thank you!

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6 лет назад, # ^ |

can you explain what u mean by bs over the answer? what answer

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6 лет назад, # ^ |

← Rev. 2 →

When a problem is asking for the minimum k satisfies a condition T, and k satisfies T => any k' > k satisfies T, and for a number k' you can check whether k' satisfies T.

You can binary search k. If k satisfies T try k smaller, else try k larger.

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6 лет назад, # ^ |

ok skylinebaby i understand .

regarding this question everytime we should choose a k using binary search.

then we can build a new graph with edges > k , if this graph forms a cycle , then answer will obviously will > k , and if not then answer will be <=k.

this way we can get the best k .

now , how to proceed further . which edges to revert so that graph becomes acyclic .

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6 лет назад, # ^ |

← Rev. 3 →

You can topological sort the graph with edge > k, because the problem don't need to minimize the number of edge which should be reversed, you can simply reverse every edge which has the wrong direction against the topological sort.

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arjun_9426

6 лет назад, # ^ |

so is it reduced to can we make topological order of edges > mid, because if we can do so we can ALWAYS adjust remaining edges? . please help, i m not finding the way that how remaining edges can cause problem to topological order for why we need to orient remaining edges to topological order

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6 лет назад, # ^ |

← Rev. 3 →

Yes, if we can make the topological order of edges > k, we can always adjust remaining edges < = k, because we can just follow the topological order. On the other hand, if we can't make the topological order of edges > k, that means there is a cycle with edge all > k, and we can't satisfy the original condition by reversing those edge < = k.

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arjun_9426

6 лет назад, # ^ |

oh thanks bro, just got it accepted :D

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T1duS

6 лет назад, # |

Wasted the whole time on C. Dunno why got WA on D. Coded E just after the contest ended. Gonna kill myself if it passes.

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Caring_is_a_disadvantage

6 лет назад, # ^ |

+20

So, are you dead?

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MirrorCraze

6 лет назад, # |

← Rev. 3 →

+14

Honestly speaking, the gap between C and D is very large. C is the normal Div.2 B and B is the normal Div.2 A.

Surely you didn't send problems to the problem pool like what Arkady did on the problem B. I mean it was like, n=6, m = 6 and 112556 then the answer is 000001

Need solution on D too. It really looks like pigeonhole, but I don't know how to do it.

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T1duS

6 лет назад, # ^ |

-24

D and E were easier than C

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MirrorCraze

6 лет назад, # ^ |

+23

Well, depends on perspective IMO.

For me, I think C is easier than D and E. You just need to write some geometric formula and derive it.

Well, actually I can't tell much because I didn't finish D and E.

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M_H_H_7

6 лет назад, # |

+18

C solution

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BogdanRazvan

6 лет назад, # ^ |

+24

Geometry problem ---> T̶r̶y̶ ̶t̶o̶ ̶s̶o̶l̶v̶e̶ ̶i̶t̶ google it.

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M.A.H.M.O.O.D

6 лет назад, # |

← Rev. 2 →

What were the hacks for B and A? I managed to hack someone on B because his implementation had a time complexity of O((m — n) * n). Are there any other hacks for B?

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Harpae

6 лет назад, # ^ |

I passed pretests on A but failed system tests by thinking that you could opt to not remove any tabs. The case 3 2 1 1 1 will break this

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TheRedLegend

6 лет назад, # |

How to solve C guys? I could not come with a solution during the contest :(

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Adhami

6 лет назад, # ^ |

+22

Solve your homework on triangular formulas. If you did not take them yet, well you are rekt.

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6 лет назад, # ^ |

← Rev. 4 →

The centers of the n outer circles make a regular polygon with n sides. Now focus on the centers of 2 adjacent outer circles and the middle circle. You have a triangle with side lengths (r+R), (r+R) and (R+R). You can also find all the angles in the triangle. Finally, you can use cosine theorem to get a formula: (2*R)^2 = 2*(R+r)^2 — 2*(R+r)^2*cos(A) [A is the center angle, actually A = 360/n]. Then, you can simplify and use the quadratic formula, or use binary search.

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kushan02

6 лет назад, # ^ |

What I did was visualized that the centers of the outer circles when connected formed a regular polygon with n sides. That polygon has its center coinciding with the center of the inner circle. We can calculate the length from the vertex of the polygon to the center of the inner circle using trigonometry and hence there is only one variable i.e R (here side of the polygon will be 2R). I am almost sure of my solution, I didn't take time to prove what I conjectured, so will come to know exactly if my hypothesis is right when final results are out.

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sun_of_Ice

6 лет назад, # ^ |

This is my idea,may be inefficient.

draw some cases,we can find:

The center of inner circle can make a Isosceles triangle with two conterminous outer circles' center,and two angles will be $\text{[math]}$ ,difine this as θ

then we have (R + r)² = R² + (R·tan(θ))²

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Prakash_Gatiyala

6 лет назад, # ^ |

← Rev. 3 →

If you connect the centers of two adjacent little circles and the center of the big one, you'll get a triangle. The sides of this triangle have lengths r+R,r+R and 2r

. A little trigonometry will get you that the top angle is

θ=2arcsin(r/r+R).

Since you want the small circles to form a closed ring around the big circle, this angle should enter an integer amount of times in 360° (or 2π

if you work in radians). Thus,

θ=360°/n.

From this, you can compute that

r=Rsin(180°/n)/1−sin(180°/n).

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6 лет назад, # ^ |

All you gotta realize is that centers of outside circles must be 2R away from each other. How to get centers? Well just do x = (r+R)cos(t) and y = (r+R)sin(t), where t = (2pi/n)*i. If you want to visualize draw two circles tangent to each other with radius r and R. They will be r+R away from each other, and assuming one of them is at 0,0 the other's coordinates can be gotten from sin and cos.

Binary search: if distance < 2R the circles are too big otherwise they are too small.

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6 лет назад, # |

← Rev. 3 →

Please tell me how to debug intractive problem or what is pretets 2 for D. Idea for D: Place king in 500,500. Divide chessboard in 4 pieces. Each piece has one corner as king position and one corner as chessboard corner. Find which of this 4 pieces has smallest number of rooks. Go to opposite corner with king moving only diagonally. Convince yourself it always works.

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gen

6 лет назад, # ^ |

+19

This is correct and actually needs at most 1000, not 2000 steps! Really cool problem.

I debugged on 9 x 9 grid with 6 pieces.

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sniffleheim

6 лет назад, # ^ |

Unfortunately, one must check that the king doesn't move to a space already occupied by a rook (which I failed to do). This is probably where the factor of two comes from.

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gen

6 лет назад, # ^ |

If we try to move to a cell with a rook, it means we are trying to move diagonally! Otherwise the king was already hit. Therefore, we can just move either horizontally or vertically to a cell adjacent to this rook. And then we are hit and the game ends. So the number of moves will really be at most 1000.

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6 лет назад, # ^ |

I did it as well. Can anyone take a look and tell me what is wrong with this? Even passed pre1. 48356184

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6 лет назад, # ^ |

I did the same with WA6. I think what matters is how you get to the corner. You have to try to move diagonally as much as possible. After considering this, got AC!

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6 лет назад, # ^ |

Thanks for reply but, already found mistake it was checking if(x!=500 and y!=500) instead of (x!=500 or y!=500) ;)

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kayak

6 лет назад, # |

← Rev. 2 →

I'm surely misinterpreting the problem/making some really silly mistake, but I think the king can't guarantee having a check ?

Define S_i to be the configuration where the black rooks are at the cells {(1, 1), (1, 2), ..., (1, i - 1), (1, i + 334), (1, i + 335), ..., (1, 999)}. Initially start with S₁ and the white king at (1, 1). Now at some step, if the king's x coordinate changes from x to y (with |x - y| ≤ 1) then it's easy to see we can move some rook so that the rooks configuration become $\text{[math]}$ , so the king can never guarantee a check ?

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gen

6 лет назад, # ^ |

But in your example all the pieces share the row, so the king is already hit by all the rooks.

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kayak

6 лет назад, # ^ |

Ooops thanks, it was embrassing.

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vivek_shah

6 лет назад, # |

← Rev. 3 →

Just curious to know, why does this submission on problem B doesn't get runttime error when he declares array of size exactly 100000.
My hack test

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6 лет назад, # ^ |

Why would it get runtime error? n, m <= 100000 so it seems fine.

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vivek_shah

6 лет назад, # ^ |

← Rev. 2 →

array size has been declared as 100000. hence, array element range is between 0-99999. but he does arr[aa]++;
hence it should get runtime when aa = 100000. Sorry, correct me if I am wrong.

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6 лет назад, # ^ |

Ah, so he is accessing an element outside of the array. If he'd used std::vector instead of a plain array he would have gotten a Runtime Error. However, since he's using a plain array, his solution is doing undefined behaviour, which is bad, but most often it won't be a problem because he's accessing memory right next to the array.

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TheStory

6 лет назад, # ^ |

When you declare an array.Usually you will get an array a little bigger than you declare.That works in most computers.Include my computer,Codeforces computer and our school computer.And even a[-1] won't get Runtime Error in some computers.

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6 лет назад, # |

OMG! I had a clean solution for D, but didn't consider the case of eating a rook. Didn't look at the CF error message :(( It was saying king ate rook

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Kemal

6 лет назад, # |

← Rev. 2 →

+76

Trash contest...
1500 point problem solved by 3.5k and D problem solved by only 125 people. Also E problem is just classic know problem added binary search(here is the solution). I think F also range version of the XOR max problem(here is solution).
If you want to do such contest, better to not do. Educational Rounds much better than this kind of contests...

→ Ответить

6 лет назад, # ^ |

Problem D was not actually that difficult. The entire contest was consistently simple IMO, which is fair.

I do agree that this contest is more like an Educational round, tho. Because it's very easy to google solutions to some problems.

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Adhami

6 лет назад, # |

+10

I do not want to judge you strictly. But What the hell.

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mohamedeltair

6 лет назад, # |

+40

I understand that it is not easy to make a well balanced contest, but the big jumps in difficulty of problems are really obvious in most of recent codeforces contests.

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pantho

6 лет назад, # |

how to solve B?? Can anyone please explain??

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6 лет назад, # ^ |

I got pretest pass by maintaining a set {1...N} and erase element x when I see x. If I saw x after I erased it, store that information. When set is empty, re-construct the set, without the elements I stored information about before.

For example, 3 9 1 1 2 3 3 3 2 1 3

Set : 1 2 3 (initial)-> 2 3 -> 2 3, (1 seen once more) , 3 (1 seen once more) -> X (first Round), Reconstruct only 2, 3 -> 2 -> 2, (3 seen once more) .....

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Dontdoprepaitisatrap

6 лет назад, # ^ |

maybe you will get TLE set is too slow gratus907

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6 лет назад, # ^ |

Isn't set insert and erase both O(log n) operation?

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NothingToShow

6 лет назад, # ^ |

It was clear that whenever i get all type of elements atleast multiple of n times then i can create a contest. What I did was maintained a frequency array freq for each type of element. As soon as i get element of type X i increased it's frequency say new frequency is z. Also i maintained an array for keeping count of how many elements i have of particular frequency z. If i have n such elements that means i can create a contest so i incremented answer by one.

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6 лет назад, # |

← Rev. 2 →

Is it possible to hack someone's solution after the contest? Or how can I practice hacking (besides educational rounds)?

recurze, your B solution will probably run out of time on a test like:

100000 100000
1 2 3 4 ... 99999 1

I tried to hack it during the last 2 minutes of the contest but forgot to put an endl after the 1st line, so it wasn't accepted.

My the very first hack :D

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recurze

6 лет назад, # ^ |

I was more concerned about the inner loop but figured it'll only run (m/n) * n times so all good but didn't realize the if condition would run O(n). I tried to port my old solution using sets to using frequency array and didn't think much.

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6 лет назад, # ^ |

← Rev. 2 →

My own solution got TLE on test 41, lmao.

I guess, using std::set was the best thing to do.

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recurze

6 лет назад, # ^ |

Any structure with support for key-value pair (i, frequency[i]) would do the trick I suppose.

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saint_coder

6 лет назад, # |

+13

Codeforces turning into Codechef :P

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APwhitehat

6 лет назад, # |

How to solve E?

Here's my idea: Run a DFS from any node, let onstack[i] denote if the ith node is currently being processed, also intime[i] denote the scan time of ith node(similar to topological sort algo). Now maintain a segtree for onstack nodes only, and use RMQ query where index is the intime of nodes. Whenever we find a backedge we calculate the minimum weight edge in the cycle using the segtree. Is this idea correct? Is there an easier way?

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6 лет назад, # ^ |

+25

I'm not sure if this is easier, but this was my solution.

Task E Statement:

Given a weighted directed graph, convert it into a DAG by reversing edges. You have to minimize the maximum weight of the edges that you reverse.

If the given graph G = (V, E) is already a DAG, then the answer is 0.

Let us choose a value k arbitarily. Let us call an edge heavy if it's weight is greater than k, else it is light.

$\text{[math]}$ $\text{[math]}$

If G_heavy = (V, E_heavy) forms a DAG, then we can find a topological sorting of V in G_heavy. We can then arrange edges in E_light to be consistent with the topological sort.

We can find the minimum value of k so that G_heavy doesn't have cycles using binary search.

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decrypto96

6 лет назад, # ^ |

Are you reversing the direction of all the light edges? If so, then there might be a case where the original direction of a light edge might be a part of one cycle and after reversing it, it becomes part of other cycle.

→ Ответить

6 лет назад, # ^ |

← Rev. 2 →

No, not all the light edges. We find a topological sort of G_heavy. Then we can ensure that all the light edges are oriented so that they are consistent with that order.

We will only reverse those edges that are inconsistent with the order.

This is my submission, but I'm not sure how readable it is. I don't use the terms heavy and light in the code. It also hasn't passed system tests yet.

48354300

EDIT: Passed system tests.

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Killever

6 лет назад, # ^ |

← Rev. 3 →

what if the selected edges created 2 graphs, now we have 2 topological sorts what if we need to connect node from the first graph to another node from the second graph. example

graph 1 with the visiting order
A->B->C->D
4  3  2  1
graph 2 with the visiting order
E->F->G->H
8  7  6  5

now we need to connect nodes (D, E) but order[D]<order[E] so we reverse it and we add it to the answer, but if we didn't reverse it the graph is still DAG. how we handle this case, it seems I'm missing something.

→ Ответить

6 лет назад, # ^ |

what if the selected edges created 2 graphs

You mean a disjoint DAG? That's a DAG as well, and it has a topological sort as well.

In your example, there are many valid topological orders. Let us choose any one, say "EFGHABCD".

If we have edge between D to E, then we must have the direction as E--->D (because that is what the topological order tell us).

If the given edge is D--->E, then we will reverse it, else we won't.

Hope it is clear now.

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Killever

6 лет назад, # ^ |

Thanks for your respond, yea I understand that. but my question is, In the example, we can generate 2 topological sort

ABCD->EFGH and EFGH->ABCD

let's say we have edge E->D if we selected the first topological sort then we need to reverse the edge if we selected the second topological sort then we don't need to reverse the edge. so I see that selecting the topological sort actually affect the results, but it's not clear for me why selecting arbitrary topological sort is ok, did you got what I mean?! :D

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6 лет назад, # ^ |

I think I understand your question.

The observation is correct that choosing a specific topological sort affects the final resulting Graph (after reversing edges).

The claim is that we could use any of it. But once we choose a topological sort, we can't change that.

I think the best way for you to get this intuition but by generating examples.

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amr0

6 лет назад, # ^ |

Sorry if this is obvious but how is minimizing the maximum weight of reversed edges the same as minimizing the sum of the weights of the edges we reverse.

→ Ответить

6 лет назад, # ^ |

← Rev. 2 →

We aren't minimizing the sum.

the same as minimizing the sum of the weights of the edges we reverse.

I can't find this quote in my comment.

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overhyped

6 лет назад, # |

-20

Trash. Again.

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trickster78

6 лет назад, # ^ |

It's den2204's first time creating a contest. Cut him some slack!

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6 лет назад, # |

← Rev. 3 →

Pretests of E are weak. I did not get TLE for the following DFS routine:

void dfs(int u) {
  visited[u] = 1;
  for (int v : g[u]) {
    if (visited[v] == 1) { // cycle and hence do not continue }
    dfs(v); // missing visited[v] == 2 check
  }
  visited[u] = 2;
}

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sevlll777

6 лет назад, # |

Good round! Thanks for problem D!

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starboy_jb

6 лет назад, # |

+114

contestant.

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265918

6 лет назад, # ^ |

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6 лет назад, # |

If I submitted two solution for same problem and both are right then which will be judged first or second one?

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starboy_jb

6 лет назад, # ^ |

second. first will be skipped.

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6 лет назад, # ^ |

I think only the last solution you've submitted that passes the pretests will be taken into account.

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Mistra

6 лет назад, # ^ |

The last solution that passed pretests will be judged.

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HalfFragment

6 лет назад, # ^ |

second, you got skipped verdict on first

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6 лет назад, # |

+31

Yeah yeah contest was slightly unbalanced but I still think it was a good contest, especially if it's his first one. Tbh we've seen much worse here and I actually hope that den2204 will be the problemsetter for some later contests too. I mean, problemsetters also have to learn how to do it properly and I believe this contest was actually pretty good for his first one. Gratz man !

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FuyuMai12

6 лет назад, # |

I'm so regretful to spend 14 minutes to think for a solution in C.

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6 лет назад, # ^ |

Worse. I spend around 15 min think that error between 1.0000000 and 0.9999999 is greater than 1e-6. I have to actually put these values in formula to check the error.

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6 лет назад, # ^ |

You don't have to set constant value PI that much long

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FuyuMai12

6 лет назад, # ^ |

Well, I took it from my partner's template. Still I have a few things in my life that requires such a number.

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6 лет назад, # ^ |

hmmmm...

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berserker

6 лет назад, # ^ |

+21

If you don't know that already you can also use

const double PI = acos(-1);

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6 лет назад, # ^ |

That's amazing!!!

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Ahmed-

6 лет назад, # ^ |

Better than spending 19 minutes using tan instead of sin and don't know what's wrong :\

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6 лет назад, # ^ |

I somehow made the same mistake lol

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do_good_

6 лет назад, # ^ |

which ide u used

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FuyuMai12

6 лет назад, # ^ |

I work with Kate text editor and Terminal.

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antontrygubO_o

6 лет назад, # |

+113

Solution for D (if anyone is interested)

Move the king to cell (500, 500) in less than 1000 moves. Now board is divided into 4 squares by king's column and row. At least one of these squares contains no more than 166 cells. Then move king to the corner opposite to that square by diagonal. We will get to the corner in 499 moves, but the other player will have to move all >=(666-166)=500 rooks to avoid checks. So he will not have enough moves to avoid checks.

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6 лет назад, # ^ |

thought I was close to getting it during the contest but your solution seems much simpler than mine :()

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mouse_wireless

6 лет назад, # ^ |

Oh... you can move diagonally...

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linkret

6 лет назад, # ^ |

+31

I don't like these type of tasks, where every number needs to be so precise.

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nskybytskyi

6 лет назад, # ^ |

← Rev. 2 →

-7

In fact, it is possible to solve this problem in 998 steps with efficient implementation using the same logic, hence the number of moves is not as precise as you think.

Moreover, 166 is not just an arbitrary number, it is rather an integer part of 666/4. Therefore you could have write 666/4 everywhere instead of 166 and get the same result without having to care about all the precise numbers.

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6 лет назад, # ^ |

+10

If n = num_rooks and m = board_size (m is odd), then the condition for this to be a winning strategy was that

$\text{[math]}$

In the current example, it boiled down to $500 > 499$, which is just enough for this equality to hold. This is what he means by too precise.

But almost all pigeonhole problems seem to have this property of just managing to work. So, I really enjoying this problem despite not being able to solve this.

Open Question: Is there any winning strategy if this inequality doesn't hold?

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linkret

6 лет назад, # ^ |

← Rev. 2 →

Maybe you could generalize it, by checking the "quadrants" of every single position, not just the center. And then choose the closest one, instead of the center.

To me this looks like it's optimal, but still doesn't always win.

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Rinne

6 лет назад, # |

vjudge88's solution for problem F is a naive copy of https://blog.csdn.net/ShadyPi/article/details/79939990

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6 лет назад, # |

-11

Is there no hacking for this contest? I want to hack!!

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lunga

6 лет назад, # |

How much time after the contest can we submit solutions again?

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AmanGupta_

6 лет назад, # ^ |

After system testing completed

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Determination

6 лет назад, # |

Anybody came up with a good solution for E ?

→ Ответить

linkret

6 лет назад, # ^ |

Binary search the answer, then check if there exist a cycle with all costs > mid?

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Miraak

6 лет назад, # ^ |

← Rev. 2 →

Let's solve it with a binary search. We want to check that the answer is at least k. Let's remove all edges with weight less than k (we can choose any direction on this edges). Notice that if our graph contains any cycles then the answer is more than k, otherwise the answer isn't greater than k.

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raidencode

6 лет назад, # ^ |

and how restore answer? if we change all edges with cost lower or equal to k its not create another cycle?

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Miraak

6 лет назад, # ^ |

We know that there are no cycles, so we can find a topological ordering. After that we can direct our edges, which were removed, the right way

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Siriuslight

6 лет назад, # ^ |

Can you please explain the use of topological ordering in more detail.

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Miraak

6 лет назад, # ^ |

If we rearrange our vertexes in topological order then every edge will be directed from left to right. Some edges before removal were directed from left to right, we won't change their direction, but if an edge was directed from right to left we'll change its direction.

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6 лет назад, # |

Honestly, task C is nothing to do with coding competition. Very hard to understand problem statements.

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6 лет назад, # ^ |

-11

Devic, Can I know why?

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SIUUUUUUU

6 лет назад, # ^ |

Yes, you can.

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6 лет назад, # ^ |

-11

Oh, thank you

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6 лет назад, # ^ |

Not only because it can be googled. It's a pure geometric task which takes one line to solve in terms of code. There is no place for optimization, algorithmization, choosing the right data structure and so on.

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6 лет назад, # ^ |

I thought the same thing.

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6 лет назад, # ^ |

Basically it is more like D, E, F is too hard for most of Div.2 coders (including me. I ran into E for 80 min and I couldn't solve it). For them(from submission record, I think this includes both you and me), only meaningful problem was A, B, and C and one of them (C) were something like an basic geometric question for mathematics textbook or so. I agree with you — many people could write less than 5 lines to get it write after working on paper, especially when you write in python and don't really care about real number precision stuff. I think it would have been nicer if D was easier, so that more people could actually enjoy programming more.

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Odin-

6 лет назад, # |

-21

It was impossible to hack! pretests are stronger than they suppose to be!

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AliShahali1382

6 лет назад, # |

-6

system testing started yes!!!

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6 лет назад, # ^ |

Yes!!!

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6 лет назад, # |

Problem D can be solved without being able to move the king diagonally.

→ Ответить

6 лет назад, # ^ |

← Rev. 2 →

Works up until test #6 :)

Sorry this is not quite right. You have to move diagonally to the corner from the center. If you can't move diagonally, then you take one step either vertically or horizontally and terminate.

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6 лет назад, # |

I got TLE on problem B :https://mirror.codeforces.com/contest/1100/submission/48351000

How do I speed this up?

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antontrygubO_o

6 лет назад, # ^ |

Sorry, you just have wrong approach... Your works in O(nm) only

→ Ответить

6 лет назад, # ^ |

← Rev. 2 →

Thanks. I will try to get a better approach. Also, can you point out which part of the approach is wrong?

→ Ответить

6 лет назад, # ^ |

You seem checking every time if he can make contest. This is waste of information. Think about the case, when he knows that he needs 1, 2, 3, 4, 5 and he got 4. He don't have to re-check everything again. Right after seeing 4, it is already known that 1, 2, 3, 5 is missing without checking again.

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6 лет назад, # ^ |

Hi. I have modified the solution a little bit : https://mirror.codeforces.com/contest/1100/submission/48360651

Now getting a Runtime error. How do I test this?

→ Ответить

6 лет назад, # ^ |

Not quite sure because I'm not currently able to run or read carefully on your solution. Though I suppose the problem is that your vector val and group is both size m+1. Shouldn't one of them be size of n+1? Seems index out of range is being accessed.

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6 лет назад, # ^ |

← Rev. 2 →

Goddammit! Thanks. Got it. I guess this is why people allocate an array of size 10e5 and keep iterating over the fixed range. Miles to go.

→ Ответить

6 лет назад, # ^ |

Glad you like it :)

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Mohammad.H915

6 лет назад, # |

-8

→ Ответить

SamuraiBebop

6 лет назад, # |

+52

D is a very nice problem. Thanks.

→ Ответить

6 лет назад, # |

Pretest for E wast too weak

→ Ответить

6 лет назад, # |

-32

The worst contest in my life

→ Ответить

caa_tomsk

6 лет назад, # |

For problem D got WA on test 40. Is there any way to know what that test was? Here is my submission.

→ Ответить

6 лет назад, # ^ |

I have the same problem as you. Has there been some modifications on codeforces which don't let you see the test you got wrong answer on ?

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caa_tomsk

6 лет назад, # ^ |

One difficulty is that this problem is interactive, so the input kind of depends on what you program prints. So you would need initial positions + the algorithm that moves the rooks.

→ Ответить

caa_tomsk

6 лет назад, # ^ |

← Rev. 2 →

Never mind, I found my bug: I have

while (kx != 500 && ky != 500)

but it should be

while (kx != 500 || ky != 500)

Argh!

→ Ответить

6 лет назад, # |

-11

Even the system tests are weak for E. (Attaching previous comment below)

Pretests of E are weak. I did not get TLE for the following DFS routine:

void dfs(int u) {
  visited[u] = 1;
  for (int v : g[u]) {
    if (visited[v] == 1) { // cycle and hence do not continue }
    dfs(v); // missing visited[v] == 2 check
  }
  visited[u] = 2;
}

→ Ответить

r_pi

6 лет назад, # |

+10

Good contest , just +1 shift was needed i.e D = E , E = F, questions were very good ! den2204

→ Ответить

Adibov

6 лет назад, # |

Now i have to wait 9 days for comeback (-_-)

→ Ответить

6 лет назад, # |

+25

My friend GreenGrape should teach den2204 how to prepare the pretests. Dear den2204, can you imagine that someone can use map to store the edges and write in the statement, that there are can be absolutely same edges, or at least include it in the pretests.

→ Ответить

Koschei

6 лет назад, # ^ |

+17

I agree, my solution failed exactly because of this issue, which is a shame because the problem was pretty nice. :/

→ Ответить

im0qianqian

6 лет назад, # |

← Rev. 3 →

Excuse me, I used C++17 to submit my solution during the competition, but I got Wrong answer. After I tried to use C++11 and C++14, I got Runtime error, so I want to ask this. Why? (My solution involves dividing by zero error)

int vx = (tx - x) / abs(tx - x);
int vy = (ty - y) / abs(ty - y);

C++11 Runtime error on test 6: 48359273

C++14 Runtime error on test 6: 48359354

C++17 Wrong answer on test 6: 48360585

→ Ответить

juanigsrz

6 лет назад, # ^ |

+28

Division by zero is undefined behavior in C++

→ Ответить

im0qianqian

6 лет назад, # ^ |

← Rev. 2 →

That is to say, the error of dividing by zero in Online Judge may get uncertain results. Is that true? I always thought that if the pointer is out of bounds or divided by zero, it will return Runtime error, thank you.

→ Ответить

juanigsrz

6 лет назад, # ^ |

Indeed, undefined behavior means at that point the program could do anything completely unexpected (i.e., "undefined"), and it usually varies depending on the computer and the compiler it's being ran on. So it doesn't make much sense to analyze WA, RE, TLE or whatever veredict whenever you're doing undefined operations. Another common one: accessing an array over its defined bounds.

Sometimes it might be Runtime Error because that sector of memory doesn't belong to you, sometimes it actually does and it's initialized with zeros, some others it belongs to you as well but it hasn't been initialized at all (contains a random value). All of these three cases could happen running exactly the same code.

→ Ответить

im0qianqian

6 лет назад, # ^ |

Thank you for your help. I'll be more careful in programming and make fewer similar mistakes in the future. Another point is that I may not choose to use C++17 to submit the solution in the future. QAQ

→ Ответить

6 лет назад, # |

How to solve E if to reverse each edge different controllers are used i.e. minimize the sum of cost all edges we have to reverse?

→ Ответить

6 лет назад, # ^ |

If we can solve this then same procedure can be used to solve it for directed graph with weights 1. The latter problem is minimum feedback arc set problem. This problem is known to be NP-hard.

→ Ответить

tusg25

6 лет назад, # ^ |

I misread the problem during the contest but I think the problem might be reduced to minimum cost Hamilton path and thus NP complete(I don’t have a proof though).

→ Ответить

Siriuslight

6 лет назад, # ^ |

Same thing happened with me also. I thought different controllers are required for different edges. :(

→ Ответить

AwakeAnay

6 лет назад, # |

+36

Here's my idea with proof for D: Greedy: Take the king to the centre. Then consider the number of rooks in each of the 4 squares formed by lines passing through the king. Look at the square with least amount of rooks. Walk the opposite way.

Proof: From the center, you can reach each corner in less than 500 moves. The square with least rooks must have number of rooks <= 666/4 by pigeon hole principle. So, there are atleast 500 rooks in the other 3 squares, combined. Now, Dasha has to move >= 500 rooks within < 500 moves, which is not possible. Hence, we are done.

→ Ответить

6 лет назад, # ^ |

Ah, so that's why those numbers were chosen. Very nice.

→ Ответить

nskybytskyi

6 лет назад, # |

+12

Порадовала чисто математическая задача С. Порадовала интерактивная задача D. Расстроили относительно слабые претесты в D и E. Возможно, стоило бы поменять задачи E и D местами.

А так все очень даже хорошо как для первого раунда, так держать den2204!

→ Ответить

6 лет назад, # ^ |

Я бы сказал, что претесты по Е были крайне слабые

→ Ответить

doped.silicon

6 лет назад, # |

← Rev. 2 →

DELETED

→ Ответить

SAeed

6 лет назад, # |

I have a general question that is kind of related to this contest.

In today's contest there was a solution in my room that solved Problem B using brute force solution with complexity O(n²). The solution was supposed to give TLE on a test that has (n = m = 10⁵) and all the numbers appearing starting from 1 and ending with 10⁵.

I wasn't able to hack this solution due to test case size which was about 512 KB, while codeforces seems to have the upper limit for a test case on hacking set to about 265 KB. Is there a reason for this low limit? Shouldn't it be a little higher? Specially since lately most problems seem to have bounds about 2.10⁵ rather than 10⁵ like before, wouldn't this limit make it harder to hack solutions that fail due to TLE on large test cases?

MikeMirzayanov, perhaps you could kindly provide an answer?

→ Ответить

Datenshi

6 лет назад, # ^ |

For hacks with test case size greater than the limit you can send the generator for your hack.

→ Ответить

SAeed

6 лет назад, # ^ |

You mean sending the generator code rather than the test case itself?

→ Ответить

Datenshi

6 лет назад, # ^ |

+26

Yeah, you can send the generator code instead

→ Ответить

lohit_97

6 лет назад, # ^ |

Aha! I asked a similar question to the authors during the contest and they replied "No Comments" to it. :(

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6 лет назад, # |

+37

It's a shame C and E were well known. I liked those problems. (and I didn't know them, so I couldn't figure out E in time 0w0)

But nevertheless I think it is a good contest! Good job den2204

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6 лет назад, # ^ |

E well known.where? how to solve e

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6 лет назад, # ^ |

Binary search on the operators and then do this

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6 лет назад, # ^ |

← Rev. 2 →

but the above link describes the way in which we have to add edges , how can it be applied here

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6 лет назад, # ^ |

Imagine you want to check whether you can make the graph with C amount of operators.

Well notice that every edge > C is fixed and every edge <= C can be adjusted to fit the resulting graph.

If you can make a graph of edges with edge.c > C and find a topological sorting of the graph, then C works because you can adjust the other edges to fit the graph.

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6 лет назад, # ^ |

ok , so suppose we choose the best k .

then how to proceed further using topolozical sorting , how to revert edges so that graph becomes acyclic , since no. of reversions need not be minimum , so we can choose any edge .

but how ?

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6 лет назад, # ^ |

let us say we have the topological sort (I would suggest bfs based): {a,b,c,d,e}

And then we have this edge (b,d). We would not reverse that.

But we have another edge (c,a). It violates our toposort, so we would reverse that.

As for implementation you can store the positions of every element in the topological sort.

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SPatrik

6 лет назад, # |

← Rev. 3 →

-68

Problem D. Correct submission: 48360170 My contest submission: 48346037 This 1 character cost me 150 rating points (-65 instead of about +90). The worst part in it, that it passed pretests when it is wrong 1 out of 4 times. This should have failed on pretests. But now I am in the same position as people who couldn't solve D. I know that I made a terrible mistake with that x instead of y, but I still have the question: Why does Codfeforces judging system is so unfair? EVERYONE who gets wrong answer on pretest have the chance to correct it, while EVERYONE who passed pretests with a wrong submission will lose ALL of the points for that problem. This is just a game of luck. In My opinion there would be 2 fair way to judge with a similar system. A: There are no pretests, only maintests. That would be much harder to make a good solution, but it would be fair. B: pretests=maintests. This would make a little easier to make a good solution, and it would be fair as well.

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TadijaSebez

6 лет назад, # |

My wrong submission for D that I skipped during the competition passed system test. Paradoxically I would be first if I didn't come up with correct solution.

Wrong submission: https://mirror.codeforces.com/contest/1100/submission/48362464

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Maho

6 лет назад, # |

По идеи у дшки же ведь нет решения. Даша ведь всегда может сходить так чтобы он проирал. Я условие раз 5 перепрочитал. Не понимаю. Обьясните

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_bom_bom_

6 лет назад, # ^ |

У меня такой же вопрос был. Я задал во время конеста и предложил следующую страту: Где то находится король, возьмем вертикальную полосу ширины пять, две слева, две справа и вертикаль где стоит король, аналогично горизонтальную полосу. получился крест, расставим ладьи на главной диагонали чтоб они не попадали в крест. Если после хода короля, если ни одна ладья не попала в крест, то стоим на месте, если попала (а может и две сразу) то берем любую из двух или одной и перемещаем как можно дальше от креста на свободную позицию на диагонали, для второй ладьи(если она есть) остается еще один ход для перемещения из креста. Мне сказали что решение всегда есть

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balalaika

6 лет назад, # ^ |

Предложенная страта некорректна, ибо никак не доказывается момент, что нет такой страты, что при движении короля в крест никогда не попадет больше двух ладей. А это не так.

Корректное решение уже не раз описывалось, так что просто протранслирую на русский.

За <= 500 ходов можно перейти в клетку (500, 500). Тогда горизонталь 500 и вертикаль 500 разделят всю доску на четыре четверти. Т.к. ладей 666, то минимальное кол-во ладей в четверти не больше 666 / 4 = 166. Пусть мы нашли такую четверть. Тогда все, что нужно — это идти по диагонали в противоположную от этой четверти сторону.

До угла всего 500 ходов. За каждый ход уменьшаются размеры остальных четвертей. А в них как минимум 666 — 166 = 500 ладей. Итого за 500 ходов обязательно наткнешься на хоть какую-нибудь ладью.

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Farabi

6 лет назад, # |

-19

I can only handle A,B,C after contest.Bt during contest time I was a great looser

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meaningIess

6 лет назад, # |

There is a problem that harder than F and I revise it wrong :( Actually if I make the array in size 1e6 then I can get Happy New Year :(

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WA_A_PLAYER

6 лет назад, # |

I wanna ask sth about E, i use the dfn[], get the time when i into a node, but others get the time they leave the node, as a result i get WA, i hope someone can explain it. Thank you so much.

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WA_A_PLAYER

6 лет назад, # ^ |

i use dfs to solve,bfs get in or out can solve it,so i wanna konw why use dfs to get the into time cannot solve it

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evildanlapokin

6 лет назад, # |

Спасибо) Хороший раунд, мне все очень понравилось)

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djq_cpp

6 лет назад, # |

I know it's an accident but problem F really coincides with a problem in a training contest proposed by whzzt called "异或"(XOR, in English).

In that problem, you will be given a sequence of n integers a[] and m queries {li, ri, di}. For each query, you need to choose a subset of a[li, ri], compute the XOR-sum of all numbers in this subset and XOR the result by di so that the final result is maximized. You need to output the final result. Constraints: n, m <= 10^6 and 0 <= ai, di < 2^30.

Anyway, D is awesome.

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MagicSpark

6 лет назад, # ^ |

https://blog.csdn.net/ShadyPi/article/details/79939990 This is the article

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lychao

6 лет назад, # |

In problem E, is there someone like me that easily get a binary search and Topological sorting solution to the answer of worker but failed on the construction of answer of edge?

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Prakash11

6 лет назад, # |

← Rev. 3 →

Problem A

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6 лет назад, # |

How to solve e. my idea was do a dfs from a certain node. keep trace of the minimum weight . if a cycle is formed. we will reverse the edge with minimum weight and then again start dfs from the reversed node.

will it work?

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berserker

6 лет назад, # ^ |

No because reversing the minimum weighted edge doesn't work because reversing the wrong edge might create a new cycle for which reversing cost can be much higher.

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kissu_pari_na

6 лет назад, # |

← Rev. 3 →

+14

I think there is a critical test case which was not given in the dataset of problem C. I found a user’s two accepted solution with different output. If I am not wrong then some user may not be able to pass the dataset if this test case is valid. I am giving the test case and the two solutions link below-
Test case: 3 100
First accepted solution
Output of first solution: 646.4101614984
Second accepted solution
Output for second solution: 499.9999999418
Please check den2204

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MikeMirzayanov

6 лет назад, # ^ |

+17

Thanks, added the test into testset for practice. Also I wrote to the writer about it.

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1910903

6 лет назад, # |

Our coach 2778 (rank 448) Rating +24 1876->1900 became Candidate Master ???

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Zyrax

6 лет назад, # |

"Тут мог быть ваш смешной комментарий"

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6 лет назад, # |

-31

Regarding tasks A-D:

A — easy, but you have to state in bold, that i can be any number. Because usually, it is 0/positive.

B — ok, but usually the number of problems would be described as n and would go first.

C — this task is the worst. As I stated in my older comment, it's pure geometry task which takes one line to resolve. There is no algorithm, no space for optimization, just geometry. I believe, that some people just googled it and I'm mad at myself, that I haven't done it also and spent the time to solve this task.

D — OK

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6 лет назад, # ^ |

← Rev. 2 →

+21

I think you did the right thing.

I could have searched up both C and E and instantly solved them. But I didn't. I don't think you should be mad for putting an earnest effort in a contest; looking for solutions on google will never help you develop problem-solving skills. The people who just googled it are doing themselves a disservice, and I question why they even do codeforces.

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6 лет назад, # ^ |

-17

Yeah. I wouldn't google E or any coding task during the contest. But solving C didn't help me in terms of my coding problem-solving skills. The problem is interesting because you need to see, that if you connect centers of external circles, you've got regular polygon. But it's nothing to do with programming(((. Maybe I'm more disappointed, that I've spent time on this task instead of solving D faster.

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minibowser

6 лет назад, # ^ |

+30

Many hard CP problems are based entirely on observations instead of hard algorithms. Programming is only one part of CP.

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6 лет назад, # |

← Rev. 2 →

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Nightmare05

6 лет назад, # |

Can someone please explain me how is Idleness limit calculated?

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Siriuslight

6 лет назад, # |

+17

den2204 will there be an editorial?

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Kyochi

6 лет назад, # |

← Rev. 2 →

Someone can explain me how to solve B with a time complexity of O(m) and not O(m*n) ?

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https://mirror.codeforces.com/contest/1100/submission/48354732

6 лет назад, # ^ |

If you loop the O(m) elements and keep track of a frequency array you can realize at each point you can either

1) hit a contest — O(n) update

2) not hit a contest — O(1) update

Seems like O(n*m) right? Nope!

Observation: you can hit a new contest up to O(m/n) times. In total hitting a contest can only contribute (m/n)*(n) = O(m) to your complexity. So actually the overall complexity is O(m), regardless of n. This is the basic idea behind amortized analysis: looking at worst case of how much each operation contributes instead of the whole algorithm. It is very likely you already thought of this solution in contest!

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kushan02

6 лет назад, # ^ |

Can you please explain more, how exactly can I identify these types of problems?

And if you don't mind can you explain why my solution gives a TLE, I thought of a very similar approach.

Thanks a lot for asking, I am just getting started with CP!

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6 лет назад, # ^ |

Your code is O(nm) because your loop O(n) everytime you check a new element. You want to only loop O(n) when you hit a contest, which would make overall code O(m).

for (int i = 1; i <= n; ++i) {
    if (ar[i] == 0)
        return false;
}

Instead of this you want to keep track of "how many cnt[i] > 0" which can be done by updating every time you add or subtract.

My code

How to identify these problems?. Intuitively, you know you should analyze complexity more carefully when the algorithm does a heavy operation rarely. Here is works out nicely, because the O(n) update is infrequent enough so that it is linear overall.

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6 лет назад, # |

-30

i am sorry if this comes across hateful, but this isn't an interdisciplinary contest. why did we get maths in this contest? i think there was a misunderstanding somewhere. can we get a rating compensation for this? doesn't seem fair ..

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6 лет назад, # ^ |

-6

Geometry problems are not endemic to codeforces.

As a side note: The idea of certain math concepts as "unrelated" is actually harmful to your learning. I learned this the hard way.

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6 лет назад, # ^ |

+21

Some math and geometry questions fall under the umbrella of competitive programming.

Notable math topics: Number Theory, Combinatorics, Probability.

Notable geometry algorithms: Convex Hull, Line Sweep.

These topics are well known, and a competitor shouldn't be surprised if they appear in a contest.

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6 лет назад, # ^ |

-32

sorry but i can't understand this. what has math to do with programming? they are two completely different things. why mix them together?

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farmersrice

6 лет назад, # ^ |

+10

CS is literally Math 2.0

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