Ивановские разборки...
Все задачи были подготовлены на Polygon мной, поэтому если вы столкнулись с какими-либо проблемами (мы знаем о решениях по задаче A, которые не прошли системное тестирование, но прошли претесты), то я приношу вам свои извинения.
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Идея: shishyando
Авторское решение: pastebin
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Идея: Artyom123
Авторское решение: pastebin
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Идея: shishyando
Авторское решение: pastebin
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Идея: Artyom123
Авторское решение: pastebin
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Идея: shishyando
Авторское решение: pastebin
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Авторское решение: pastebin
Good job!
Fast blazingly fast
Thanks for quick editorial.
I see people down-voting, I just liked the editorial and i was thankful for the contest and the quick editorial, now ,whom did I hurt or what was wrong in this comment for these downvotes?
There are some trolls who use fake accounts for mass downvoting.
it doesn't and shouldn't matter
True.
Thanks for the speedy Tutorial man!
Thanks!
For D2 My solution got hacked for testcase 67 Can anybody help me? This is my submission link: https://codeforces.com/contest/1419/submission/93235884
That was a system test, not a hack test
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.
I'm still struggling with B...
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.
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.
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
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...
GL
Bro what does this 1LL does and what do you mean by 1ll<<x and 1<<x
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
Author's solution is confusing me can you help me bro
yeah sure what do you want to know? I will do my best to answer.
maybe you can check this first: https://codeforces.com/blog/entry/82817?#comment-698848
hope it helps
hi pls can you tell me why do we need to use 1LL and not just 1?
So that there is no overflow of number in int
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
Because you have i already ll long long int type One side of expression is long long int type it is enough
Why are we doing prefix sums?
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 :)
Jury's solution for D2 can't opened..
Please make strong pretests from next time. Nothing is more painful than getting your problem A wrong after the system testing :/
First time creating a contest is hard, some mistakes may be made.... We will be even more careful next time.
ㅤ
same here
how u come up to 2^k-1 diectly..in div2b can u explain a bit. Thnx
There's a really useful tool called messing around with the numbers
Can you share the link for the same?
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+.
This proof is not satisfactory enough. At best, it says that it works for
2*a_i +1if it works fora_i. How do you still know there aren't any other numbers in the middle for which it works?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 :)
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
My contest was ruined because of B. I will look for pattern and think less next time .
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.com/contest/1419/submission/93215283
"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 ?
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.
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.
My bad, I misinterpreted it -_-.
Great Contest !! Good work Guys
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
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!
I am unable to get the explanation for B. I know the pattern , but can someone prove it please ?
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 ofn, is actuallyn(n+1)/2. So one could just loop while sum is less thanxand updatingnwith2n+1.It's
2^n - 1, not2i+1.Also, I wonder too why it only works for Mersenne numbers. Wish it had a more mathematical/some number-theoretic argument.
Yes yes, actually the
iin2i+1is the prev value in the sequence, hence the same thing. Sorry for the confusion.See this:
https://codeforces.com/blog/entry/82817?#comment-698848
hope its helpful ;)
https://artofproblemsolving.com/community/c6h34314p213007
A was too hard though :)
Please teach me some googling tricks.
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
The maximum "number of divisors" for $$$n \leq 10^9$$$ should be around 1000 so it is enough to pass
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.
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)
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).
More Simple (Easy to Understand) Solution for E:
Insert all the factors of
nin 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) whosegcd with ans[i-1]isnot 1. And put that particular value at theith 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, theninsertions req = 1Else,
insertions req = 0I solved it, this way.
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.
Because n always come at the last position as you said and it is not coprime to any of the earlier numbers.
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 = 6Factors = 2,3,6ans[0] = 2(We will set this)ans[1] = 6(gcd(2,6) != 1)Now, if we have
2 & 6in the factor list, one can easily say that3and maybesome other multiples of 3too, 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.
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!
I don't know why this simple solution works 93280623. Is test cases are Weak?
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
Problem A: "num" can be many digits long so it cannot fit in the
intdata 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."
In question A it is not given that positive integer is string(it could be taken as int type too).
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
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!
nice div2E
I think you overcomplicated D2... It's literally D1, just a tiny bit different in calculating the number of cheap ones.
why i can't see jury solution..?
Yeah, it's a real problem. Fix jury's solutions, please!
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?
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. ;__;
https://medium.com/@harryjobz/sages-birthday-1c68ffc70987 Simply pick all the even positions first in the sorted array. I have explained how here
Umm, okay.
But, why did you reply this to my comment? I had solved till E while testing. o_O
https://codeforces.com/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.
one of the best codeforces round i have taken part in .. great problems, enjoyed a lot :) :)
There was not a necessity of including the problem D1 in this contest.
Problem D is solved more than Problem A.. weird..
It's not weird XD this problem indeed has the same difficulty as A.
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.com/contest/1419/submission/93275649
I think sorting takes O(nlogn). So solution can also be O(nlogn). Right?
Yup, but i found this to be very straightforward and elegent.
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?
Sorry never mind, I saw the proof on somewhere below.
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!
ㅤ
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.
When I click on Jury Solution I get an error "You are not allowed to view the requested page".
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?
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.
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.
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 < 3the answer is 0Now, 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 // 2are absolutely unreachable no matter what we do as they're equal to [n — 1]. And all elements in range0... kare reachable by elementsn // 2... nbecause these elements are certainly biggerThough 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
There is no access to jury's solutions
Can someone help me with the search for errors in my solution of E? I can't understand tester's comment on WA2.
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 :(
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.
Thanks!
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!
Thanks for soooooo fast tutorial! And also thanks for the great contest, it improved my coding skills greatly lol.
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)
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.
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 $$$ p_{1} $$$.
I fst *2...
not a good contest
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.)
Framing of B question was very confusing. It took me more than 1 hr to just know what the question is saying.
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?
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.
I still don't understand. Isn't the concept similar to coin change dp problem?
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.
Wait, why are you downvoting?
Please don't downvote him
Am I the only one who thinks that C could have been phrased a little better ?
Not just C
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 ;)
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.
I have the same doubt
Here, the total number of unique boxes are 7, it should be exactly 6.
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
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.
Ok but E was just USAMO 2005 right?
https://codeforces.com/contest/1419/submission/93303906 Please tell what is wrong here.
Why are you not getting Segmentation fault in these lines?
can you please explain further
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 :
i did as you said but now it is failing at the very first test.
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.
Yes, most people who did greedy used this. It's easier to see.
Still can't understand problem B. Can anyone help me?
If you managed to find the pattern, then the problem becomes pretty easy. Here's my submission- 93223145
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.
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)
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
Why simple sorting and greedy placing did worked for me in problem D2 submission
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.
https://codeforces.com/contest/1419/submission/93311817 can anyone please tell me why I am getting wrong Answer
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!!!
I think the statement of the ahead two problems is a little bit long.....
Why to make D2 complicated (by using binary search) if this simple approach is enough.
Solution
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
Thanks for nice problem set and quick editorial.
https://codeforces.com/contest/1419/submission/93326069
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
https://youtu.be/fHyJObr8EdI My solutions ( A-D )
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
cntrepresents the number of ratings equal toxinitially, 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.
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.
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?
Ups you are right, edited
CAN ANY BODY TELL THAT IN PROB C OUR NEW RATINGS CAN GO BEYOND -4000 OR +4000 ???
PLEASE tell
New ratings can be any integer.so the answer is YES , it can go beyond -4000 and +4000My code for problem D accepted D2(93375350) but failed in D1(93375360)
I'm afraid the problem E is a bit easier than common...
And many of my friends said that they have difficulty understanding the problems...
neal actually make D2 a greedy problem, how to prove it, can someone help me ?
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.
Jury's solution passed in 46ms, so we think the limits are ok.
Thx
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
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!
YLWang does this sound similar to your approach?
shishyando Have you considered or come across a solution similar to the above? Thanks
Found my mistake, it was extremely subtle typo, copy-paste error. In one place, instance of
somePoint.x, I hadsomePoint.y. Took me hours of debugging...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.
Thanks!
I sorted it out as mentioned in my last comment. The bug was that when sorting points based on $$$Y$$$ coordinates, when y's are equal, I should compare x coordinates, but I was comparing x coordinate with the other point's y coordinate. Silly typo. Once that's fixed, passed all tests in 249ms in Java 11. Here's the accepted submission
In terms of finding the error, I was expecting it to be a genuine error so was reading through more complex parts of the solution :) The way I found it was to take fully visible tests from CF tests. I had to run on 30 tests before finding a failing example. For this problem, to generate valid test cases where the answer isn't -1 requires a bit more careful impl than just generating random points. I was reluctant to do that.
The harder version of Sage's Birthday got solved by greedy approach.
The problem D2 is rated 1500 when I've seen the same problem in Div2A at 800-1000, is there a reason why, Its definitely not a 1500 level problem if I was able to solve it in less than 10 mins (on the time of writing this I am rated 1331).
Why do we have to use binary search in D1 (easy version) or even D2? Can't we just sort the elements and reorder them in an optimal way?
залупа полная я не понимаю