Hori's blog

Codeforces Global Round 27 Editorial
By Hori, 4 weeks ago, In English
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author: Hori

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author: willy108

Code: 288397529

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author: Hori

Code: 288397556

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author: ChatGPT and oursaco

Code: 288397595

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author: lunchbox

Code: 288397611

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author: turtletortles

Code: 288397640

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author: sum

Code: 288397661

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author: Benq

Code: 288397685

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author: Hori

Code: 288397713

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4 weeks ago, # |
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Okay so D is done by ChatGPT but also oursaco. oursaco, if you don't mind, can you please tell a little bit about how you contributed to the creation of that problem? Like did you just write the prompt or did the chatGPT stuff have to be revised?

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    3 weeks ago, # ^ |
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    Asked chatgpt to make a problem which was initially solving the problem for a single array with no i < j constraint (this was an existing div2a). I thought the i < j constraint would make it more interesting and solved for it. Then the solve for every prefix part was added. Also made sure to plug it back into o1-preview to make sure it couldnt solve XD

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4 weeks ago, # |
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Code is not accessible.

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    3 weeks ago, # ^ |
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    Fixed, thanks for letting me know

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3 weeks ago, # |
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cant see tutorial of D, but is it possible to solve D using knapsack?

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    3 weeks ago, # ^ |
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    I didn't solve it, but I think D is to be solved using the observation: if $$$a_i=2^{p_i} \cdot {q_i}$$$, there should be $$$p_i$$$ operations that greedily decrease $$$A_i$$$ and greedily increase the member of the prefix with greater or equal index and max $$$q$$$.

    With this, we can compute the optimal sum for each prefix separately in quadratic time. Or, we can compute for each prefix using the the previous prefix in linear time (somewhat like DP).

    Essentially, we can process the $$$i^{\text{it}}$$$ prefix as adding $$$A_i$$$ to an array $$$p$$$. By adding $$$A_i$$$, we offer a new $$$q$$$ to all of the elements that precede it. We obviously want all of $$$A_1, \cdots, A_{i-1}$$$ that are using their operations to increase an element with a lesser $$$q$$$ to instead increase $$$A_i$$$. All of the elements that precede the nearest element with a greater $$$q$$$ will not change, but those that follow will now use their operations to increase $$$A_i$$$. We can find the nearest greater element with a greater $$$q$$$ using monotonic stacks, which are well explained by the USACO Guide or CPH. With prefix sums, we can find the the factor by which $$$A_i$$$ increases, and monotonic stacks have a linear time complexity, so the solution is done in $$$O(n)$$$.

    I don't see the intuition for Knapsack DP as the problem isn't looking for a subset, but maybe you're on to something.

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3 weeks ago, # |
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C is the hardest div2C in history. Even harder than D2 from the last div2 round (rated 2187 on CList).

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3 weeks ago, # |
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Yo, i've just completed tutorial for problem D in Hackmd : link

Sorry for not directly publishing tutorial in codeforces, i prefer using hackmd :)

First time getting upvoted, thanks a lot !

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3 weeks ago, # |
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C odd case: since n%2==1 then l will always be 1 so it is easier to just set the last 4 to 1 3 n-1,n

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    13 days ago, # ^ |
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    Please tell me , in A why they are not using the formulae which is given . And also why moves of first element is counted as (m+1) but when i find distance using formulae i am getting (m).

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3 weeks ago, # |
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Can someone share their thought process of D like how and why it works

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    3 weeks ago, # ^ |
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    you have to collect all the 2's in terms of powers and give them to the best successor possible, then calculate the sum of the remaining odds which are not considered

    Ex1:1 6 5
    for this, you have to give all the 2 powers to 5 which is better over 3

    Ex2:4 1 3
    same as above but the format is different, first you will accumulate at 1 then accumulate at 3

    Ex3:4 7 2 2 2
    Here for the first 2 elements, 4 will be accumulated into 7
    For 3 elements, since there is no benefit of sending 4 to 2 over 7, we leave it unchanged
    For 4 elements, first, we will accumulate the previous 2 and then compare 4 with 7, since 7 is bigger, the powers remain unchanged
    For 5 elements, we will accumulate past 2 twos to get 8, which is in turn greater than 7, so we will send the 4 to 8 making it 32

    thus we will be storing all the accumulated indices and powers in a stack, all the odds remaining after removing powers of 2 in other variables unaccumulated_odds, whenever we get a new element we try to pop elements in the stack, and try to accumulate values at new element same as shown in example 3

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    3 weeks ago, # ^ |
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    think like this . from a given i to previous 31 even numbers you will have a number with or without shifting 2's greater than 1e9 . since every ai is bounded by 1e9 . all other even numbers before that will lose thier 2's to largest element

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3 weeks ago, # |
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Can E be solved by simulated annealing?I submitted serveral times, but no luck.288417811.This is the first time I wrote simulated annealing, so it may come with some mistakes. So I wonder whether it can be solved by such a approach or not.

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    3 weeks ago, # ^ |
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    I also tried but failed ;( multiple-test aren't so easy for SA to pass.

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3 weeks ago, # |
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No way in hell I can ever "notice" what the editorial tells us to notice in problem D.

Solved it by thinking about where to utilize the 2s in between the last number we have given 2s to and the current number i. If we can use these 2s to make a[i] greater than the last number we gave all the 2s, then we should give all the twos we gave the previous number to this a[i]. Else we re-use the answer from the last number we gave all the twos and give all the twos in between to a[i]. To maintain the last number we gave all the 2s we can use monotonic stack. Submission: https://codeforces.net/contest/2035/submission/288418798

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    3 weeks ago, # ^ |
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    Actually, I did notice it after thinking about it for a while. Not bad. Wouldn't have ever come up with it myself though. Thanks for the editorial.

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3 weeks ago, # |
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you can solve E using ternary search too.

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    3 weeks ago, # ^ |
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    talk is cheap, share code.

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      3 weeks ago, # ^ |
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      288768348

      runs in Clogz.

      C = 100 works.

      Idea: if you replace ceiling division in the formula with normal division, then the function can be simplified down to pA + t/A, where p and t are arbitrary constants. This is a valley array, and the valley can be found with ternary search. Then just search a bit to the side of that valley to account for any inaccuracies because of the division change.

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        3 weeks ago, # ^ |
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        I can't really prove the horizontal search, if anyone could share a counter case that would be cool.

        Also side note: I actually never had all the details ironed out before I started coding it, so ig papa-ka-para was right. upvoted :)

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          3 weeks ago, # ^ |
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          nice one. although, I tried a lot during contest to pass ternary search. but didn't work.

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    3 weeks ago, # ^ |
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    Lol no. You can maybe pass the current testset if you are lucky but it's not "solving".

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3 weeks ago, # |
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Bro C was just too difficult for me.

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3 weeks ago, # |
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for C, we can also approach it greedily, observe if n is odd then last operation is and so we can get to less than or equal to last operation, so we choose last to be n and second last to be n-1 as (n &(n-1)) = (n-1), but what if we try to make the first bit of (n-1) on? then we can make it (n), but how to do it? observe if we place 3, 1 before it we would get the first bit ON which would be carried forward to get the maximum value as n. Also in the case of n if it is not a power of two, n would have a bit which isn't present in the sequence so we can greedily place it at the end, and have a permutation such as 1, 2, 3..., n. Rest is nicely explained in the editorial, thanks for it!

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3 weeks ago, # |
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Am I the only one who found B to be harder than C and D? (and D easier than C)

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3 weeks ago, # |
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In D, does the solution described for the entire array ("We iterate backwards on the array [...] is the largest, we do nothing") not work on

11 1 6 9 4 7 4 4 10 3 2 3

in sample? I get 1313 when it should be 1568

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    3 weeks ago, # ^ |
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    When checking if an element a is the largest, you need to check it after removing its powers of 2. In the example, when it gets to the number 10, the largest is already 2*3=6, so it turns the largest into 10, even though if you were to add the 2 from 10 to 6, you would get 12, aka a bigger result. So what it should be doing instead is checking if 5 is larger than 6, and since its not, it will then multiply 6 by 2.

    If its still not clear, you can consider the times a number can be divided by 2 as its potential. This potential can go to the base of the current number (the current number/its potential), or to the largest number calculated before. So, returning to the example, the 8th element has potential 2, and base 5. If I give the current number the potential, I end up with 10, but if I give it to the largest number (which equals 6), I will end up with 12.

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3 weeks ago, # |
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One more way to solve F is (same observation(and logic) but different implementation) :

Do a normal binary search on answer , but this time instead of checking for only mid , check from mid-2*n+1 , mid , even if one turns out to be true(say for num) , hi = num-1 else lo = mid+1;

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3 weeks ago, # |
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For D solution, why 31? 2 divisors

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3 weeks ago, # |
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can someone help me in proving my solution for C? I brute forced some cases of $$$ 2, 1, 3, 4, .... , n $$$ and noted that we get the required $$$ k $$$ somewhere always. based on this, my solution is just $$$ x + 1, x + 2, x + 3, ......, n, 2, 1, 3, ....., x $$$ for some x such that the resultant $$$ k $$$ for $$$ 2, 1, 3, ... x $$$ is maximum and $$$ n - x $$$ is even. I cannot formally prove why the sequence $$$ 2, 1, 3, 4, ... x $$$ gives the best k always.

Submission

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3 weeks ago, # |
← Rev. 2   Vote: I like it +8 Vote: I do not like it

I just don't know why $$$O(n^2\log{a_i})$$$ does not fit in 4 seconds in problem F.

UPD: It seems to be my problem. I was submitting in G++ 17 instead of G++ 20, which is based on 32-bit architecture.

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    3 weeks ago, # ^ |
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    We knew the TL was tight for some people but were also aware of a very fast N^3 solution and wanted to cut it.

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      3 weeks ago, # ^ |
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      In such cases, it might be better to modify the constraints a bit.

      For example, you could use 0 <= a_i <= 10^5 instead of 0 <= a_i <= 10^9.

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        3 weeks ago, # ^ |
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        I guess that would have made sense. However, there were a lot of optimizations that where any of them could make it pass (although unoptimized passed comfortably for many people).

        • Adjust the bounds of binary search to look for only a better answer
        • Skip half of the binary searches by only checking the correct parity of the answer
        • Unrolling the dfs order and iterating instead of doing recursion
        • Optimizing to $$$\mathcal{O}(n^2 \log n + n \log a)$$$
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3 weeks ago, # |
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I am very bad at solving problem like B. Can anyone tell me what topic do i have to learn to solve this kind of problem. Thanks in advance.

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3 weeks ago, # |
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I am confused, in problem D , we can have $$$a_i$$$ > $$$a_j$$$ but it is more optimal to divide $$$a_i$$$ by 2 and multiply $$$a_j$$$ by 2

example: 20 and 9

20 + 9 = 29

but we can do: 20 / 2 / 2 = 5

and 9 × 2 × 2 = 36

and the sum becomes 41 > 29

what am I missing here ?

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    3 weeks ago, # ^ |
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    Consider this case in a "distribute the $$$2$$$ factors" manner, now we have $$$20=5\times 2^2$$$, and it's pretty clear that giving the $$$2$$$ factors to $$$9$$$ is better than to $$$5$$$.

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      3 weeks ago, # ^ |
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      what do you mean by "distribute the 2 factors" ? do you mean reducing all numbers to their odd form (devide them all by 2 untill they become odd) Edit: yeah I got it now. it wasnt mentioned in the editorial though. Thanks!

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3 weeks ago, # |
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However, E can be solve in $$$O(z^{3/4})$$$ , only by using some optimizations in brute force. My Code

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    3 weeks ago, # ^ |
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    Please explain the logic you used for optimization

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      3 weeks ago, # ^ |
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      Every time increase $$$d$$$ by $$$k$$$ and use a second operation, it will be end in at most $$$O(\sqrt{\dfrac zk})$$$ times.

      If you choose to increase $$$d$$$ by $$$i$$$ and always choose the second operation from then on, the additional cost is $$$ix+\lceil\dfrac z{d+i}\rceil y$$$ . You can find that there are at most $$$O(\sqrt z)$$$ different $$$\lceil\dfrac z{d+i}\rceil$$$ and always use the smallest $$$i$$$ with the same $$$\lceil\dfrac z{d+i}\rceil$$$ will be better. So now is $$$O(\sqrt{\dfrac zk}\min(\sqrt z,k))$$$ .

      If $$$\sqrt z\le k$$$ , then $$$O(\sqrt{\dfrac zk}\cdot\sqrt z)=O(\dfrac z{\sqrt k})\le O(z^{\frac 34})$$$ ;

      If $$$\sqrt z>k$$$ ,then $$$O(\sqrt{\dfrac zk}\cdot k)=O(\sqrt{zk})<O(z^{\frac 34})$$$ .

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3 weeks ago, # |
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Hori

In the editorial of E problem, you have mentioned this statement.

a * b >= 2 * z , this implies that min(a,b) <= sqrt(2*z) ? Why is that ? How did we reach there ? 

===> a * b >= 2 * z

===> min(a,b) * max(a,b) >= 2 * z

===> sqrt ( min (a,b) * max(a,b) ) >= 2 * z

Just to replace few values, lets take a = 5 , b = 5, and z = 5, then a * b >= 2 * z holds true, but min(a,b) <= sqrt(2 * z) doesn't hold true.

Please can you elaborate a little here ?

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    2 weeks ago, # ^ |
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    When z=5 and a=5, b=2 satisfies the inequality. Therefore, min(a, b) <= sqrt(2*z) is still satisfied. While there are definitely, other solutions, b=5 is one as you correctly point out, they are simply worse than the solution we just found as they cost more.

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3 weeks ago, # |
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is system testing done till this round !!?

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2 weeks ago, # |
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For solution E, I don't understand how we got $$$\dfrac{a\cdot b}{2} \ge z$$$

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2 weeks ago, # |
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I think Problem B very Medium hmmm

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13 days ago, # |
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In G1 editorial, while counting cnt(i,j), it should the case that j < m < i and not the other way right?

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12 days ago, # |
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Could you please help me, what do I do if it is showing "N/A" in the solution Code page? Thank You

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12 days ago, # |
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Can we solve D using priority queue. We store pair of {lowest value a[prev] can get, a[prev]} and check if a[i]>pair.first then we can perform the operation and pop the pair. We will then maintain answer vector where ans[i] += a[i-1] — pair.second + pair.first + a[i] * (pair.second/pair.first)

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9 days ago, # |
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I can't understand C's solution,could someone help explain the principles clearly,it's too abstract for me.

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3 days ago, # |
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Hori, the code is not available

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