Codeforces Round #671 Editorial

4 года назад, # ^ |

Um, for your information, there is a button called "Click here for more information" at the bottom of the page of every submission. Click on it and you can see all tests and know what kind of test data you fail on.

BTW it's really important to make good index calculation :) And not don't blame hacks, though this solution is hackable.

→ Ответить

Lithosphere

4 года назад, # |

I'm still struggling with B...

→ Ответить

sirearsh

4 года назад, # ^ |

try making out patterns on paper and see how the pattern goes, they gave un hint in sample that answer was going to be in 1 to 30 only. so you could simply precompute these 30 values.

→ Ответить

tadek

4 года назад, # ^ |

← Rev. 2 →

You can solve it through recurrence. The first stair case (a single green square in the picture in the problem statement) with will have $$$1$$$ cell. Lets denote the number of cells in the first stair case with $$$a_1$$$ so $$$a_1 = 1$$$ cell. The next possible stair case (the second possible stair case) will have $$$a_2=6$$$ cells. It's made out of two green cells (two $$$a_1$$$s) and one yellow square which has $$$2^2$$$ cells (as in the picture in the problem statement) so $$$a_2=2*a_1+2^2$$$. The third possible stair case will have $$$a_3 = 2 * a_2 + 2^4$$$ cells. I hope that by this point you can see that $$$2^4$$$ corresponds to the red square from the picture in the problem statement? So now the general formula for the number of cells in a nice stair case is $$$a_i = 2 * a_{i-1} + 2^{2*(i-1)}$$$. Now all you need to do is count $$$a_i$$$ up to some value so that you will meet the problem constraints. Then count a prefix sum array of the $$$a_i$$$s and then you can greedily iterate over it up to a given $$$x$$$. Here's my code 93245756.

→ Ответить

Lithosphere

4 года назад, # ^ |

Thanks, I get your explanation, not sure if its really a proof but close enough. I wasnt confident enough in the contest to procede with this idea

→ Ответить

4 года назад, # ^ |

Don't forget to use 1ll if you want to make some power of two ( use 1ll<<x instead of 1<<x), that cost me about 40 min to debug during the contest...

→ Ответить

Madara__Uchiha__

4 года назад, # ^ |

Bro what does this 1LL does and what do you mean by 1ll<<x and 1<<x

→ Ответить

4 года назад, # ^ |

1LL means 1 but in type long long. Just the same as (long long)1

1ll<<x means shift the bits right x times in binary. This is how I do for getting powers of two in C++. Bit operations are useful to me :v

→ Ответить

Madara__Uchiha__

4 года назад, # ^ |

Author's solution is confusing me can you help me bro

→ Ответить

4 года назад, # ^ |

yeah sure what do you want to know? I will do my best to answer.

maybe you can check this first: https://codeforces.me/blog/entry/82817?#comment-698848

hope it helps

→ Ответить

arpitP

4 года назад, # ^ |

hi pls can you tell me why do we need to use 1LL and not just 1?

→ Ответить

kiviad

4 года назад, # ^ |

So that there is no overflow of number in int

→ Ответить

arpitP

4 года назад, # ^ |

← Rev. 2 →

just out of curiosity i have written the code without using 1LL but it still giving the desired outputsolution without using 1LL pls can you tell why this is showing accepted then

thanks

→ Ответить

alg0man

4 года назад, # ^ |

← Rev. 2 →

Because you have i already ll long long int type One side of expression is long long int type it is enough

→ Ответить

rum3r

4 года назад, # ^ |

Why are we doing prefix sums?

→ Ответить

4 года назад, # ^ |

You don't necessarily need it. We are searching for the smallest n such that a1 + a2 + ... + an < x(the input val), so if you do prefix sum, you can search more conveniently. Again, this is not necessary.

Pls upvote if this helps :)

→ Ответить

A_Gentle_Name

4 года назад, # |

Jury's solution for D2 can't opened..

→ Ответить

DumbCarlsen69

4 года назад, # |

+23

Please make strong pretests from next time. Nothing is more painful than getting your problem A wrong after the system testing :/

→ Ответить

shishyando

4 года назад, # ^ |

+53

First time creating a contest is hard, some mistakes may be made.... We will be even more careful next time.

→ Ответить

0xF5

4 года назад, # ^ |

← Rev. 2 →

ㅤ

→ Ответить

NBonaparte

4 года назад, # ^ |

same here

→ Ответить

GuptaJi

4 года назад, # ^ |

how u come up to 2^k-1 diectly..in div2b can u explain a bit. Thnx

→ Ответить

yesbutno1685

4 года назад, # ^ |

+27

There's a really useful tool called messing around with the numbers

→ Ответить

decoder__

4 года назад, # ^ |

Can you share the link for the same?

→ Ответить

4 года назад, # ^ |

← Rev. 2 →

+12

First, one can easily see the basic solution is one stair (after understanding the problem, which took me something like 10 min).

Then, the sample input explanation tells you that another nice stair is which have 3 stairs.

Lastly, (maybe) you would notice that the image in the description has 7 stairs, and there are only squares with a side length of powers of two.

So I assumed that nice stairs must consist only squares that have side length of powers of 2.

Next is to prove it (you can choose to prove it by AC, but that's not what I'm gonna do here) using mathematical induction (or something similar): 1. assume that a solution with n stairs ( a_n ) exists. then you can see that by duplicating this solution and attach them to a bigger square in the middle (like the yellow square in the description), you can make solution ( a_{n+1} ), as you just created a new nice stair contains (2n+1) squares and (2n+1) stairs.

Since the base case is 1 stair with 1 cell, the only solutions must be something in this series.

I'm lazy so I'm not gonna write down the formulas. Hope this make you understand better.

And, make sure to use 1ll instead of 1 when using 1ll<<x for powers of two. This cost me 40 min approximately during contest, which results in not having enough time for implementing pE and broke my dream to 1900+.

→ Ответить

justanotherlad

4 года назад, # ^ |

This proof is not satisfactory enough. At best, it says that it works for 2*a_i +1 if it works for a_i. How do you still know there aren't any other numbers in the middle for which it works?

→ Ответить

4 года назад, # ^ |

Oops my bad. I forgot to mention that this is what i went through during the contest. As you say, this do not prove that there is no other solution, but I thought this was correct and gave it a try. If anyone knows the full proof pls let me know ><

Thanks for pointing that out :)

→ Ответить

Pradyumna_

4 года назад, # |

Please make tight pretests I realized my mistake after the contest in Problem A. Please try to set some pretests because passing two pretests I can then console the answer as well

→ Ответить

ashanksingh1997

4 года назад, # |

+10

My contest was ruined because of B. I will look for pattern and think less next time .

→ Ответить

HumblX

4 года назад, # ^ |

I found the pattern in the "sum of first n numbers formula" during the contest. I just figured that we have to increase n by a power of 2 in this formula (n*(n+1))/2 and subtract it from x until you run out of cells. I just made this conclusion from the first two good staircases and tried my luck. here's my submission: https://codeforces.me/contest/1419/submission/93215283

→ Ответить

believebelieve

4 года назад, # |

-22

"If the single last digit is odd, then Raze wins, else Breach wins." It is not mentioned that If the single last digit is even then Breach wins. How can you assume that ?

→ Ответить

Norrius

4 года назад, # ^ |

+17

If the single last digit is odd, then Raze wins, else Breach wins

If the last remaining digit is not odd, it is even. That should be obvious.

→ Ответить

memoryallocator

4 года назад, # ^ |

← Rev. 2 →

I didn't understand your question. Any digit is either odd or even. Hence, if the single last digit is not odd, then it is even.

→ Ответить

believebelieve

4 года назад, # ^ |

My bad, I misinterpreted it -_-.

→ Ответить

parad0x_

4 года назад, # |

Great Contest !! Good work Guys

→ Ответить

Hooded_Hood

4 года назад, # |

long statements were obvious and low pretests are obvious too because of this much of WAs after the contest

but out of those, problems were good

→ Ответить

JMan_Zx

4 года назад, # |

On question e: So, I don't know why, but I misunderstood the meaning of coprime (yeah I know) to be divisible, so two numbers that are coprime means (what I thought) a number is divisible by the other.

So I was solving a more restrictive case of the question. Well, not that it matters. I wasn't able to get it in time anyway.

Thanks for the fast tutorial!

→ Ответить

__invariant__

4 года назад, # |

I am unable to get the explanation for B. I know the pattern , but can someone prove it please ?

→ Ответить

shinigami_

4 года назад, # ^ |

See, if you just try it out by yourself, you'll notice that stairs are only possible for n values: 1,3,7,15, 31....., i.e 2i+1. And also the cells needed for a particular value of n, is actually n(n+1)/2 . So one could just loop while sum is less than x and updating n with 2n+1.

→ Ответить

justanotherlad

4 года назад, # ^ |

It's 2^n - 1, not 2i+1.

Also, I wonder too why it only works for Mersenne numbers. Wish it had a more mathematical/some number-theoretic argument.

→ Ответить

shinigami_

4 года назад, # ^ |

Yes yes, actually the i in 2i+1 is the prev value in the sequence, hence the same thing. Sorry for the confusion.

→ Ответить

https://codeforces.me/blog/entry/82817?#comment-698848

4 года назад, # ^ |

See this:

hope its helpful ;)

→ Ответить

tenth

4 года назад, # |

← Rev. 3 →

+13

https://artofproblemsolving.com/community/c6h34314p213007

A was too hard though :)

→ Ответить

meet2mky

4 года назад, # ^ |

Please teach me some googling tricks.

→ Ответить

Norrius

4 года назад, # |

← Rev. 2 →

What's the implementation idea for E? Even with the editorial, I still don't see how to efficiently sort all divisors into these "buckets" based on what prime divisor they have. Or is O(number of divisors * number of prime divisors) good enough?

Edit: thanks to everyone who responded. I forgot just how sparse primes are. Turns out, you don't even need to precompute primes, just checking for primality with a simple $$$O(\sqrt{n})$$$ function is enough: 93278287

→ Ответить

Lusterdawn

4 года назад, # ^ |

The maximum "number of divisors" for $$$n \leq 10^9$$$ should be around 1000 so it is enough to pass

→ Ответить

cstuart

4 года назад, # ^ |

← Rev. 2 →

Before solving any queries, precompute all "small" primes up to $$$\sqrt{10^9}$$$. Then any number up to $$$10^9$$$ can have at most one "large" prime factor.

Now to solve each query:

For each $$$n$$$, iterate through the "small" primes to find out which "small" primes are prime factors of $$$n$$$. You will not TLE because there are only a few thousand "small" primes.

Using this you can find the "large" prime factor if it has one. There will be at most $$$9$$$ prime factors in total because the product of the first $$$10$$$ primes is more than $$$10^9$$$.

Then, iterate up to $$$\sqrt{n}$$$ to find all factors of $$$n$$$. For each factor, you can check all the prime factors. Again, you will not TLE because of the limit on the total number of factors.

→ Ответить

skittles1412

4 года назад, # ^ |

The problem guarantees that

the total number of divisors of n for all test cases does not exceed $$$2⋅10^5$$$.

And number of prime divisors is a small number, as it is at most 10 (2*3*5*7*11*13*17*19*23*29 = 6e9)

→ Ответить

pandastic

4 года назад, # ^ |

Number of prime divisors is less than 10. And number of divisors is bound roughly by square root of n (it is only a loose bound, but already enough).

→ Ответить

vok8

4 года назад, # ^ |

← Rev. 5 →

More Simple (Easy to Understand) Solution for E:

Insert all the factors of n in a set.

Set the first value as s.begin()

After that for each index i, find the first number from current set (using just simple iteration) whose gcd with ans[i-1] is not 1. And put that particular value at the ith position.

That's it. You will be able to fill the whole array this way. No matter what is the value of n.

It can be proved intuitively, mathematically, and with examples too. (Left to readers)

For the number of insertions,

The whole array won't need any insertions. Because we filled it in a way, that needs 0 insertions, for such arrangement.

So, just check gcd(ans[0],ans[no. of factors-1])

If it's 1, then insertions req = 1

Else, insertions req = 0

I solved it, this way.

→ Ответить

Norrius

4 года назад, # ^ |

← Rev. 2 →

It can be proved intuitively, mathematically, and with examples too. (Left to readers)

I'm really curios about this, can you at least provide some intuition?

This is definitely much easier to implement, but I don't see why this process will never get itself in a corner. You essentially have a greedy algorithm, and it's a bit surprising to me that it works.

I think there are two big questions: (1) why is there always a non relatively prime number available, and (2) why will the last one never be relatively prime with the first number (the first one must be prime, so basically why is the last number a multiple of the first one?)

Edit: Ok, I see why (2) is correct: the last number you pick will always be n, so it's of course a multiple of the first divisor.

→ Ответить

Marshmello

4 года назад, # ^ |

← Rev. 2 →

Because n always come at the last position as you said and it is not coprime to any of the earlier numbers.

→ Ответить

vok8

4 года назад, # ^ |

← Rev. 2 →

I got a simple intuition while solving,

It was like:

If I take 1st element as smallest number (It will always be prime), and 2nd element as the smallest available non co-prime number with the 1st element, then, there will be a number in the factor list which will be non-prime with the 2nd element, always.

Let's take an example:

n = 6 Factors = 2,3,6

ans[0] = 2 (We will set this)

ans[1] = 6 (gcd(2,6) != 1)

Now, if we have 2 & 6 in the factor list, one can easily say that 3 and maybe some other multiples of 3 too, will be there in the factor list, which will be non co-prime with 6.

So, this will be valid for filling out the whole list.

This was my intuition and an informal proof, for this solution.

→ Ответить

yogi23

4 года назад, # ^ |

can you please explain, how will there always be a non co-prime number in the divisor set available? can this be proved? because this method is really great!Thanks!

→ Ответить

ss_loves_AC

4 года назад, # ^ |

I don't know why this simple solution works 93280623. Is test cases are Weak?

→ Ответить

deependra227

4 года назад, # |

I have used the same approach as mentioned in the editorial for Problem A & C but I am getting wrong answer.
Can anyone please help me out?
Problem A my submission.
Problem C my submission

→ Ответить

cstuart

4 года назад, # ^ |

Problem A: "num" can be many digits long so it cannot fit in the int data type. You should store "num" as a string instead.

Problem C: You missed out one case. From the editorial: "If at least one account is already infected we can infect all other accounts in a single contest."

→ Ответить

codeTheworld3101

4 года назад, # |

← Rev. 2 →

In question A it is not given that positive integer is string(it could be taken as int type too).

→ Ответить

WitchOfTruth

4 года назад, # ^ |

Which type you use is your personal affair However, it is given that it has up to a thousand digits, which does not leave much choice

→ Ответить

awasthishubh

4 года назад, # ^ |

It's just stdin. You can read it as int or string but since n <=10^6, that integer value as a whole will be huge!

→ Ответить

pussydestroyer1337

4 года назад, # |

+19

nice div2E

→ Ответить

yesbutno1685

4 года назад, # |

+20

I think you overcomplicated D2... It's literally D1, just a tiny bit different in calculating the number of cheap ones.

→ Ответить

pucuk16

4 года назад, # |

+12

why i can't see jury solution..?

→ Ответить

robotolev

4 года назад, # ^ |

Yeah, it's a real problem. Fix jury's solutions, please!

→ Ответить

150iq

4 года назад, # |

← Rev. 2 →

In Problem C, I thought rating can't go beyond [-4000,4000] so messed up :(

I am curious though, how would we solve this problem if rating can't go beyond this range.

Any ideas?

→ Ответить

kdjonty31

4 года назад, # |

-9

After seeing the number of successful submissions in D2 and E, I really wish I hadn't tested this round and would've given it live. Would've been a really great opportunity to go back to purple. ;__;

→ Ответить

PalForce

4 года назад, # ^ |

https://medium.com/@harryjobz/sages-birthday-1c68ffc70987 Simply pick all the even positions first in the sorted array. I have explained how here

→ Ответить

kdjonty31

4 года назад, # ^ |

Umm, okay.
But, why did you reply this to my comment? I had solved till E while testing. o_O

→ Ответить

4 года назад, # |

https://codeforces.me/contest/1419/submission/93253990 Can someone please tell why this code is failing at test case 2. It is giving the right answer when i put in those individual test cases, but for some reason when i am putting them collectively, its failing to give the right ans.

→ Ответить

wolfie.

4 года назад, # |

-8

one of the best codeforces round i have taken part in .. great problems, enjoyed a lot :) :)

→ Ответить

snorkel

4 года назад, # |

There was not a necessity of including the problem D1 in this contest.

→ Ответить

coolboy7

4 года назад, # |

Problem D is solved more than Problem A.. weird..

→ Ответить

snorkel

4 года назад, # ^ |

It's not weird XD this problem indeed has the same difficulty as A.

→ Ответить

vedhant

4 года назад, # |

← Rev. 3 →

+20

D2 can be solved in a very simple greedy fashion. I simply sorted a[] and first filled all even positions for the final order and then filled all odd positions in the increasing order of a[i]. See my submission for better understanding — https://codeforces.me/contest/1419/submission/93275649

→ Ответить

jack07

4 года назад, # ^ |

I think sorting takes O(nlogn). So solution can also be O(nlogn). Right?

→ Ответить

vedhant

4 года назад, # ^ |

Yup, but i found this to be very straightforward and elegent.

→ Ответить

rgnerdplayer

4 года назад, # ^ |

I am practicing after contest and I also came up with this solution intuitively but I wonder how to prove it, do you also have a proof?

→ Ответить

rgnerdplayer

4 года назад, # ^ |

Sorry never mind, I saw the proof on somewhere below.

→ Ответить

mike51

4 года назад, # ^ |

Yes, seeing many of submissions use greedy. Why this greedy approach can work? Any proof or idea can help me better understanding on this? Thanks!

→ Ответить

0xF5

4 года назад, # ^ |

← Rev. 2 →

ㅤ

→ Ответить

ashutosh_aks

4 года назад, # |

← Rev. 2 →

On clicking the link for jury solution of E, it shows "You are not allowed to view the requested page." P.S. It is fixed now.

→ Ответить

trail99

4 года назад, # |

When I click on Jury Solution I get an error "You are not allowed to view the requested page".

→ Ответить

Amel__

4 года назад, # |

+20

I have a question about D2.

My solution is to just order n/2 biggest elements at odd position in increasing order, and n/2 smallest elements at even position in increasing order.

After I solved E, just about 10 minutes were left so I submitted this solution without any hope. But it passed system test .. Can someone prove my solution?

→ Ответить

Svlad_Cjelli

4 года назад, # ^ |

+18

Let's call a number "big" if it's bigger than the median, "medium" if it's equal to the median, and "small" if it's smaller.

You can easily see that an optimal solution can be constructed that consists only of small and medium numbers. In your solution, all the small numbers are automatically cheap. And the number of cheap medium numbers is bounded by the number of large numbers (minus 1), which your solution also gets.

So it's impossible to do better.

→ Ответить

hvk3

4 года назад, # ^ |

Assuming that your proposed solution is 1-indexed, consider the following. On placing the smallest (n / 2) elements in ascending order in the array at even positions, it's ensured that the two smallest elements in the larger (n / 2) elements will be matched against the smallest element in the entire array, In case of any other ordering of the two smaller elements in the array, against which an element from the smaller n / 2 elements may be compared, there is always an alternate ordering where the comparison is never worse.

For instance, if the two smallest elements in the larger (n / 2) array are x and x + y, and they are currently compared with some s, where s is not the minimum of the array, replacing s with s — t (t > 0) never yields a worse comparison result.

→ Ответить

WitchOfTruth

4 года назад, # ^ |

← Rev. 9 →

I also solved this way

My thoughts

The array will look like this

\/\/\/\/\/\/

So obviously we will not get more than n // 2 items and we should consider n // 2 smallest

Also, for n < 3 the answer is 0

Now, let's look at elements [n // 2], [n — 1] and [n]

If [n - 1] > [n // 2],

then everything is fine, we captured the most expensive item we could hope for, there is no better use for [n — 1] and [n]. We can add one to the answer and forget [n] and [n // 2], effectively reducing the problem to size (n — 2). We don't forget about [n — 1], but on the next step we will have to place it at exactly the place where it already is.

If [n - 1] == [n // 2]

Let's consider [k] such that [k] is the rightmost element less than [n // 2]. Then all elements in range k + 1... n // 2 are absolutely unreachable no matter what we do as they're equal to [n — 1]. And all elements in range 0... k are reachable by elements n // 2... n because these elements are certainly bigger

→ Ответить

dhruv7888

4 года назад, # |

Though I liked the questions , and also the editorials were fast , I would like to request you to make better pretests , as it hurts man when one of your solution which you assume has been passed gets failed, it really hurts man. Today I have seen many of them for question 1st and 5th , hope this doesn't happen again , as it really really hurts

→ Ответить

WitchOfTruth

4 года назад, # |

There is no access to jury's solutions

→ Ответить

robotolev

4 года назад, # |

← Rev. 2 →

Can someone help me with the search for errors in my solution of E? I can't understand tester's comment on WA2.

wrong answer Integer parameter [name=divisors[1252]] equals to 0, violates the range [2, 735134400] (test case 1)

Where is the problem in my answer for this test? I can't see full correct answer, I can't see jury solution right now, I don't know where is error in my program :(

→ Ответить

albjeno

4 года назад, # ^ |

← Rev. 2 →

It means you have not output all divisors of 735134400. There are 1344 in total. The judge has read the zero in the second line as a divisor not as the number of moves. That is why you get this error message.

Edit: Since 1 is coprime to all numbers, you only have to print 1343 divisors.

→ Ответить

robotolev

4 года назад, # ^ |

Thanks!

→ Ответить

decrypto96

4 года назад, # |

Its nice to have these kind of "adhoc only" round sometimes. Even the idea for F problem is very easy to come up with (implementation is definitely tedious). Nice round!

→ Ответить

dblark

4 года назад, # |

Thanks for soooooo fast tutorial! And also thanks for the great contest, it improved my coding skills greatly lol.

→ Ответить

FairyWinx

4 года назад, # |

← Rev. 2 →

The author's solution E is not correct, in particular, for example 12. It will make 2, 12, 6, 3. And the answer will be 1, although it is clear that it is 0 (that is, for 2 prime and at least 4 divisors)

→ Ответить

atom0410

4 года назад, # ^ |

← Rev. 2 →

They haven't explained how the following case will be handled exactly:

$$$ {p_1}^{q1}.{p_2}^{q2} $$$ where ($$$ q1 > 1 $$$ or $$$ q2 > 1 $$$). 12 falls in this category, although they have handled this case in their jury solution.

→ Ответить

atom0410

4 года назад, # ^ |

This can be easily handled if after making the structure mentioned in the editorial, one starts filling up divisors from the last prime $$$ (p_{n}) $$$, this way we can ensure that there will be always something in between $$$ p_{n} $$$ and $$$ p_{1} $$$.

→ Ответить

LHQing

4 года назад, # |

I fst *2...

→ Ответить

pulkit14

4 года назад, # |

not a good contest

→ Ответить

clyring

4 года назад, # |

← Rev. 2 →

It's also possible to solve problem F with the observation that, given any minimum spanning tree of a graph, that graph is connected by edges of length at most $$$t$$$ if and only if that minimum spanning tree has no edges longer than $$$t$$$. (This is equivalent to saying that Kruskal's algorithm works.) Adding one vertex and at most four edges to a graph doesn't change its minimum spanning tree very much, only removing at most three of the edges in the original minimum spanning tree, so it's possible to calculate the minimum value of $$$t$$$ for any plausible relevant detachment location in $$$O(5^2)$$$ with a modified Prim's algorithm on an 5-vertex graph associated with that location. (See 93293641.)

→ Ответить

abhiaps

4 года назад, # |

-6

Framing of B question was very confusing. It took me more than 1 hr to just know what the question is saying.

→ Ответить

girishgarg

4 года назад, # |

In B, the problem is basically choose the maximum count of numbers from (1, 3, 7, 15 ....) and form any number less than equal to x. Why can we solve it greedily? Shouldn't we be applying dp?

→ Ответить

LucisHeirusArdor

4 года назад, # ^ |

Because we need to find the maximum different nice staircases using 'at most' n cells, and if we use from small nice staircases to big nice staircases It always have the correct answer. Using dp, if he ask 'how many ways' using exaclty n cells.

→ Ответить

girishgarg

4 года назад, # ^ |

I still don't understand. Isn't the concept similar to coin change dp problem?

→ Ответить

TheScrasse

4 года назад, # ^ |

← Rev. 2 →

In this specific problem, you need dp if you have to use exactly $$$n$$$ cells.

Greedy is enough if you have to use at most $$$n$$$ cells.

→ Ответить

TheScrasse

4 года назад, # ^ |

Wait, why are you downvoting?

→ Ответить

girishgarg

4 года назад, # ^ |

Please don't downvote him

→ Ответить

machinepainter

4 года назад, # |

Am I the only one who thinks that C could have been phrased a little better ?

→ Ответить

ayush3

4 года назад, # ^ |

Not just C

→ Ответить

puranjaymishra

4 года назад, # |

Loved the contest! I think D questions were a bit too easy, but hey I'm a newbie so I shouldn't complain!Also a quesion on Omen could've been more interesting ;)

→ Ответить

Coolcodings

4 года назад, # |

In problem stairs:

If we take staircase of height 6, then it can be an answer, with 1 square of size 3, 2 squares of size 2 , and rest will be squares of size 1, they will be disjoint , only corners of same size square will touch.....check if this case is right or wrong.

→ Ответить

ayush3

4 года назад, # ^ |

← Rev. 2 →

I have the same doubt

→ Ответить

partheshsoni

4 года назад, # ^ |

+12

Here, the total number of unique boxes are 7, it should be exactly 6.

→ Ответить

BaOH2

4 года назад, # |

← Rev. 2 →

who can tell me the train of thought ( is it right? ) of F? ( or tell me where it is thanks )

i can't understand the solution

thank you so much

→ Ответить

partheshsoni

4 года назад, # |

shishyando: Although the contest was good, the problem statement of C was not really clear. After reading the tutorial only, I figured out that if I have one infected account, I can infect all others by making all others equal to infected one, then change the infected one. Reading the problem statement, I didn't find any clue of that, what I interpreted while reading was, that newly infected accounts and the previously infected account (from which new infection occurs) should have same ratings even after the contest. I think mentioned word "instantly" didn't really justified the phenomenon you were trying to convey. It may be possible that only I found this thing to be difficult to understand. I hope you take this as a constructive criticism.

→ Ответить

GeoMetrix123

4 года назад, # |

Ok but E was just USAMO 2005 right?

→ Ответить

4 года назад, # |

https://codeforces.me/contest/1419/submission/93303906 Please tell what is wrong here.

→ Ответить

partheshsoni

4 года назад, # ^ |

← Rev. 2 →

Why are you not getting Segmentation fault in these lines?

for (ll i=0;i<n;++i){
          if(a1[i]>a1[i-1]&&a1[i]>a1[i+1]){
              ++cnt;
             
          }
      }

→ Ответить

4 года назад, # ^ |

can you please explain further

→ Ответить

thesudhir

4 года назад, # ^ |

← Rev. 2 →

 for (ll i=0;i<n;++i){
          if(a1[i]>a1[i-1]&&a1[i]>a1[i+1]){

Your if condition is wrong. Count will increase if a1[i] is smaller than a1[i-1] && a1[i+1].
for i = 0, a1[i-1] is not defined, and for i = n-1 , a1[i+1] is not defined.

Correction :

for (ll i=1;i<n-1;++i){
          if(a1[i]<a1[i-1]&&a1[i]<a1[i+1]){
              ++cnt;

→ Ответить

https://codeforces.me/contest/1419/submission/93326069

4 года назад, # ^ |

i did as you said but now it is failing at the very first test.

→ Ответить

Santosh_Dragon

4 года назад, # |

← Rev. 2 →

I have an alternate solution for D2. Sort the array, place the first half on odd indices and second half on even indices. For example if the sorted array is 2,4,6,8,10; then make the array as 6,2,8,4,10. After this, count the number of elements which are less than their neighbours. If there are repeated elements, it would still be a valid arrangement to find maximal no. of cheap spheres.

This is in a way similar to your solution, where I am using the same concept that if we can find x cheap, we can definitely find x-1 cheap spheres too. By this method, you can directly get the maximum amount of cheap spheres possible.

→ Ответить

nagaraj41

4 года назад, # ^ |

Yes, most people who did greedy used this. It's easier to see.

→ Ответить

Rahat_Khan_Pathan

4 года назад, # |

Still can't understand problem B. Can anyone help me?

→ Ответить

-grace-

4 года назад, # ^ |

If you managed to find the pattern, then the problem becomes pretty easy. Here's my submission- 93223145

→ Ответить

arpitP

4 года назад, # |

what does this line mean ? A staircase with n stairs is called nice, if it may be covered by n disjoint squares made of cells.

→ Ответить

rickbaishuak

4 года назад, # ^ |

It means if a staircase can be completely covered with n non-overlapping squares that don't go out of the staircase, it's called nice. (look at example)

→ Ответить

rickbaishuak

4 года назад, # |

In E, what if k=2 and q1+q2>2, n=p1^q1+p2^q2. Then p1*p2=pk*p1, so it doesn't work. For example: n=24, 2 4 8 24 6 3 6

→ Ответить

streaker_rules

4 года назад, # |

Why simple sorting and greedy placing did worked for me in problem D2 submission

→ Ответить

__Apocalypse__

4 года назад, # |

Could someone please provide a formal proof of why the greedy solution in D2 works. I understand till the idea that picking the smallest (n-1)/2 elements is the most beneficial way as it's always better to keep the possible candidates for the center as low as possible. But I'm not able to proceed beyond that in the proof. Thanks in advance.

→ Ответить

Dontgiveup20

4 года назад, # |

https://codeforces.me/contest/1419/submission/93311817 can anyone please tell me why I am getting wrong Answer

→ Ответить

UNOwen98

4 года назад, # |

Can someone tell me what is the complexity of this solution for problem E? 93314845

During the contest I was sure this would TLE, but voila!!!

→ Ответить

2018LZY

4 года назад, # |

I think the statement of the ahead two problems is a little bit long.....

→ Ответить

Shantanu

4 года назад, # |

Why to make D2 complicated (by using binary search) if this simple approach is enough.
Solution

→ Ответить

ctyx

4 года назад, # |

Please somebody tell me what am I missing in C: https://codeforces.me/contest/1419/submission/93322918

→ Ответить

BetoSCL

4 года назад, # ^ |

In the case in which all the elements are equal in your code you have 2 options, if they are equal to x the answer is 0 but in the opposite case the answer must be 2 because there is no way that the sum of the differences is 0, so you have to infect n-1 and leave one uninfected to make up the sum for what you need 2

→ Ответить

jalotra

4 года назад, # |

Thanks for nice problem set and quick editorial.

→ Ответить

jimm89

4 года назад, # |

← Rev. 2 →

I am failing test set 10 on problem E because of incorrect number of transformations on case 25, yet when I compare it with correct solutions, it gives the same number of transformations. Is there a bug? Or am I missing something?

Thanks in advance

→ Ответить

Chao5

4 года назад, # |

https://youtu.be/fHyJObr8EdI My solutions ( A-D )

→ Ответить

hmantv_83

4 года назад, # |

← Rev. 2 →

Hi everyone! I am not able to understand the logic of the part ~~~~~ else if (cnt > 0) { cout << 1 << '\n'; } ~~~~~ in the jury solution of the problem Killjoy at line number 21. If cnt represents the number of ratings equal to x initially, how does it being greater than 0 helps in determining the contests to be 1?

E.g. If we have x = 70 and the ratings as [70,72,74], the answer should be 2 contests but the solution should show it as 1.

→ Ответить

Farrius

4 года назад, # ^ |

← Rev. 3 →

The answer is 1. Imagine we sum the difference to 70 for all numbers, we call this number k. k = (70 — 70) + (72 — 70) + (74 — 70) so k = 6. It may seem that you need another contest but because 70 = 70 we can infect this account before the contest start. And now we can sum or substract to this k. The process will be: 1.Infect account with rating 70. 2.After the first contest accounts will be [76, 70, 70].

If (cnt > 0) it means that we can inffect one account before the contest. So we will need just one contest to infect the rest.

→ Ответить

hmantv_83

4 года назад, # ^ |

Hi, thanks for the explanation. Also did you mean that in step 2 of the process, after the first contest, the ratings will be [76,70,70] i.e. we will compensate k=6 by adding the already affected account by 6 and subtracting it from the disinfected accounts?

→ Ответить

Farrius

4 года назад, # ^ |

Ups you are right, edited

→ Ответить

sayedramishali

4 года назад, # |

-9

CAN ANY BODY TELL THAT IN PROB C OUR NEW RATINGS CAN GO BEYOND -4000 OR +4000 ???

PLEASE tell

→ Ответить

ApraCadabra

4 года назад, # ^ |

New ratings can be any integer. so the answer is YES , it can go beyond -4000 and +4000

→ Ответить

sare

4 года назад, # |

My code for problem D accepted D2(93375350) but failed in D1(93375360)

→ Ответить

paulzrm

4 года назад, # |

I'm afraid the problem E is a bit easier than common...

→ Ответить

paulzrm

4 года назад, # ^ |

And many of my friends said that they have difficulty understanding the problems...

→ Ответить

nutboy13

4 года назад, # |

neal actually make D2 a greedy problem, how to prove it, can someone help me ?

→ Ответить

mango_lassi

4 года назад, # |

Why are the limits in F so tight? Even after 4 TLEs and a bunch of constant optimisation, my $$$O(n^2 \log 10^9)$$$ (93436282) barely passed in 1996ms.

→ Ответить

shishyando

4 года назад, # ^ |

Jury's solution passed in 46ms, so we think the limits are ok.

→ Ответить

Broken_dreams_

4 года назад, # |

Thx

→ Ответить

YLWang

4 года назад, # |

← Rev. 2 →

There exist a solution of F with time complexity of $$$O(2^m n\log^2 n\log T)$$$ where $$$m$$$ is the number of components that is $$$\leqslant 4$$$ .

It uses scanline and Segment_tree Divide (sorry for cannot explain it in English excatly) to count if there is a point coverd by all the components.

Upd : I've improved my algorithm to $$$O(2^m n\log n\log T)$$$ which allows n up to $$$10^5$$$. We can just turn range modify to range query so that we don't need segment_tree Divide.

code

→ Ответить

4 года назад, # |

← Rev. 3 →

Hello!

I have solved F in the following way and am getting WA6.

Algorithm: The difference with the solution in the editorial lies in checking whether certain value is valid. The way I check it is that after performing a DFS on the graph with a limit $$$T$$$ on the distance, we get a list of components. I leverage the fact that the point to be added must share either $$$X$$$ or $$$Y$$$ coordinate with each of the components. So I pick the smallest (in number of vertices) component and iterate through the $$$X$$$ and $$$Y$$$ coordinates of its points. For each, x0, I compute the possible intervals for $$$Y$$$ coordinates. To do that I compute such intervals for each component one by one and intersect these sets of intervals. Before computing the intervals, I've first sorted the points in the components based on its $$$X$$$ or $$$Y$$$ coordinate depending on for which coordinate I'm computing the feasible intervals. This results in the intervals being sorted in ascending order. In turn, this makes it easier to intersect the intervals in linear time. In the end if the intersection of these intervals is non-empty, then we can find such point, otherwise we can't.

Can anyone see any flaws in the approach or is it rather implementation error?

Any feedback highly appreciated!

Thank you!

→ Ответить

4 года назад, # ^ |

YLWang does this sound similar to your approach?

→ Ответить

4 года назад, # ^ |

shishyando Have you considered or come across a solution similar to the above? Thanks

→ Ответить

4 года назад, # ^ |

Found my mistake, it was extremely subtle typo, copy-paste error. In one place, instance of somePoint.x, I had somePoint.y. Took me hours of debugging...

→ Ответить

YLWang

4 года назад, # ^ |

I think the main idea of your approach is right. Maybe details got wrong.

You can use generator and brute-force to check your solution.

→ Ответить