We hope you enjoyed the contest as much as we enjoyed writing it! Thank you especially to PCTprobability for your very entertaining submissions to problem E.
2246A - farmpiggie and Subset Sum
Try to eliminate a large class of possible sums.
Think about parity. Is it possible to eliminate all odd numbers?
Try to construct the solution iteratively, by adding each new element.
2246C - 0mar and Alternating Sums
For a subsequence to have an alternating sum of zero, it must contain an even number of elements from every group of equal numbers.
The number of ways to pick an even size subsequence is equal to the number of ways to pick an odd size subsequence.
If the array is of size $$$n$$$, there are $$$2^{n-1}$$$ ways to pick an odd size subsequence.
Try to find a simple formula for the case where there are only positive elements.
The formula is $$$2^{n-d}$$$ where $$$d$$$ is the number of distinct elements. This is because you must pick an even size subset from every group of equal numbers, and multiplying all the choices together gives you $$$2^{n-d}$$$.
Do casework on whether or not you select an odd number of $$$-1$$$ elements.
2246C - 0mar and Alternating Sums
First, let's solve the problem for only positive integers (i.e. $$$1 \leq a_i \leq 10^9$$$). For a subsequence to have an alternating sum of zero, it must contain an even number of elements from every group of equal numbers.
The alternating sum of the subsequence $$$b_1, \ldots, b_m$$$ is the sum $$$b_1 - b_2 + b_3 + \ldots + (-1)^{m+1}b_m.$$$ If $$$m$$$ is even, we can pair the terms into the form
Since each of these terms is non-positive, the sum is zero if and only if all the terms are zero, which happens iff the subsequence contains an even number of elements from each group of equal numbers.
If $$$m$$$ is odd, we can group the terms as follows:
Since $$$b_1 \gt 0$$$ and all the other terms are non-negative the alternating sum cannot be zero.
Since the number of odd and even length subsequences of a group are equal, the number of subsequences of the array which contain an even number of elements from each group is the product of $$$2^{(\text{group size} -1)}$$$ over all numbers, which simplifies to $$$2^{(n - d)}$$$, where $$$d$$$ is the number of distinct elements in the array.
Let $$$S = {1,2,\ldots,n}.$$$ Our claim is equivalent to the statement that $$$S$$$ has $$$2^{n-1}$$$ subsets of even size. Let $$$T$$$ be a subset of $$$S$$$ which does not contain $$$1$$$. Then $$$|T \cup {1}| = |T| + 1,$$$ so its size has opposite parity. Furthermore this set is unique for $$$T$$$. Thus there are exactly $$$2^{n-1}$$$ pairs $$$(T, T \cup {1}),$$$ and each pair contains exactly one even size subset, so there are $$$2^{n-1}$$$ subsets of even size.
Now we consider the full problem, with $$$a_i$$$ possibly equal to $$$-1.$$$ In this case there exist subsequences with alternating sum zero which do not contain an even number of elements from each group.
Note that the number of subsequences containing an even number of $$$-1$$$ elements is $$$2^{(n - d)}$$$, since they will cancel each other, hence this reduces to the situation discussed above.
It remains to count the number of subsequences which contain an odd number of $$$-1$$$ elements. Let the total number of $$$-1$$$ elements in the array be $$$c$$$. There are $$$2^{c-1}$$$ ways to pick an odd number of them. Given that our subsequence contains an odd number of $$$-1$$$ elements, the positive elements of the subsequence must have an alternating sum of $$$-1.$$$ This occurs precisely when there exists a pair $$$(v,v+1)$$$ such that both $$$v$$$ and $$$v+1$$$ occur an odd number of times, and the remaining elements all occur an even number of times in the subsequence.
Using the same notation as in the previous proof, we see that $$$b_1, \ldots, b_m$$$ must again be even length since otherwise the alternating sum will be positive.
In the case where $$$b_1, \ldots, b_m$$$ has even length, we need exactly one of the terms $$$(b_i - b_{i+1})$$$ as grouped previously to be equal to $$$-1.$$$ This is precisely the characterization given.
For any value $$$v$$$ such that both $$$v$$$ and $$$v+1$$$ occur, we can choose the pair $$$(v,v+1)$$$ to be the pair with odd counts. This "flip" does not change the number of choices, since in each group the number of odd and even length subsequences are equal.
Therefore, the answer is $$$2^{(n - d)}$$$ if $$$-1$$$ isn't present; otherwise it is $$$2^{(n - d)} \cdot (\ell + 1)$$$, where $$$\ell$$$ is the number of values $$$v$$$ such that both $$$v$$$ and $$$v+1$$$ occur in the array.
2246D - diss_quack and Array Game
Try to solve the problem with no increments.
Do casework on whether or not there is an odd element in the array.
If there is an odd element, the answer is
Otherwise you must first remove the minimum LSB from all the elements and then this answer is forced.
This is true because if there is an odd element Bob can always put it in position 2, and force Alice to operate on each element individually.
Try fixing the minimum LSB over all $$$a_i$$$.
2246E - lce4113 and Security Game
You only need to determine one bit of $$$v$$$ in order to determine $$$b.$$$
The case $$$o = x$$$ needs to be handled carefully.
Try randomization.
2246F - Whoname and Unsorted Array
Consider the inversion count to check whether or not it is possible.
It is not possible if $$$n$$$ is even and the initial inversion count is odd.
Try sorting the elements in the suffix.
There is a clever $$$3$$$-move operation when you can't add the next element to the suffix.
There is a solution that works in $$$\frac{5n}{3}$$$ operations.
Brute force suggests that it is always possible in at most $$$n$$$ moves.
Auto comment: topic has been updated by 0mar (previous revision, new revision, compare).
B was so good, I like it
likewise.
I don't like it
Now that the contest is over, can I know? (Also I love how it's in red)
If there's 3.5 million total tests (if t <= 10^5 and 40 pretests) then there's a 0.65% failure chance. Maybe I'll see if I can submit my lucky number and make it fail lmao
What is your failure probability? There are 40 pretests.
Same as intended, 1 in 2^-29. (Some of my friends / me initially thought of the 2^-15 way, I presume that is supposed to instantly fail) I plugged this into python
Assuming independence. If not then maybe I need to startfopen("/dev/urandom")ingyou joke but there's kinda a reason hacks are disabled; lack of independence would do things to the problem
If you randomize properly (like the classic mt19937 with chrono steady clock), I don't think hacks (assuming they would be in the format like, give fixed v and which operation and b) should affect you.
You would think this is true, except as it turns out if the only layer of defense you use is mt19937 + chrono steady clock, you actually can get hacked! mt19937 is actually very well known for being hackable if you learn enough information about the state, so at the very least a naive implementation that would work as it is in a hackless environment shouldn't work. There is a stronger random you can use to defend against this, but we're getting to the point where the crux of the problem becomes a cryptography check and not an algorithms check.
Oh that's interesting. What part of it is weak? I know that mt19937 is not crypto-secure but it should be good enough for this case right?
Here are all the attacks I am aware of:
It's actually well known that mt19937 only needs 624 integers (or more precisely, 19937 bits) that have their state leaked in order to figure out the future state.
Guess what, your outputs to $$$m0/m1$$$ (or at least one of them) necessitate being random (randomizing everything else but leaving both $$$m0$$$ and $$$m1$$$ deterministic makes it so the grader can effectively be fully adaptive; I leave the explanation for why this bricks your solution to when they later edit the editorial).
Even if you ration your bits used calling rng(), you will realistically not be able to save all your allowed bits before they run out at around ~1000-1500 queries, which is really far away from the 100000 you need for AC.
By now, your state has been exposed; the hacker can now do the math to figure out the state, take it for themselves, run your code to figure out the next rng() calls, and become fully adaptive. Thus, this rough sketch of the protocol bricks any solution that just calls rng() directly.
Note that solutions that waste random rng() calls exist, but 1) they're very much not natural, and 2) it only makes the math more annoying to code in many cases (depends on how many bits you waste, I'm not 100% sure how to attack this version), though this is still probably enough since the round doesn't have an open hacking round. Either way, they rely on you knowing how an adversary hacks such PRNGs, which again is not the point of the problem.
I don't think you can do that. I would agree with you iff you could see my first 600ish outputs and then construct your failing test case afterward, but in the hack format, you must specify the complete multitest ahead of time.
If the attack model is one where I can generate the next test case based on the previous, then all bets are off, you have to use a CSPRNG that's seeded with good entropy (like from the OS, in this case steady clock will not save you because the attacker can brute force over a few hundred million possible seeds to find which one you used).
Well, I think it depends on whether or not generator is allowed I believe? I think there have been some cases where generators been allowed, and some cases where hack format is "To hack, manual and write your u/v", but please correct me if this is false. At the very least, any sufficiently evil author could've done this against a weak enough combination (e.g. just a single PRNG like mt19937 or splitmix64 that gets the next seed)
But yes, I agree that if the hack format was manual then you would have no way of doing this because you cannot first decrypt the state. I would be included to agree that such a hack rule would've also be fine, if your goal is instead to allow only the hacking of deterministic solutions. So in short, yes I would agree it simply depends on which model the hacks are in.
As for the seed, I'm pretty sure that you do have some options to chain together, except your seed gets fed into some ChaCha20 implementation or similar CSPRNG that can cleanly fit on CF. I believe this blog has a strong enough combination that considers start time, process time, and a random pointer value as your initial entropy.
Editorial has been updated to include a proof that a deterministic solution doesn't exist.
I think it worked out fine because it didn't seem like anyone with a 2^(-29) sol didn't pass, but we probably should have made it 60 bits instead of 30 to make the failure probability much lower than 0.65%.
I think 60 bits might be too much because then splitting in half would've worked. I think 0.65% is probably fine although I would definitely crash out if I WA23'd or something like that after waiting like 2 mins to judge lol. Maybe 40 bits could've been better but I also would have to admit that it does kinda risk exposing the answer (since 30 bits seems like your typical constraint that doesn't mean anything obvious). Overall E is super absolute cinema I liked it a lot
I asked whether adaptive value of v depends on ty, and recieved the same answer lol
Good contest and fast editorial. Upvoted.
Although it's sad to see how far off the ratings are gonna be from those predictions, due to all the cheaters and such (at least that's what it seems like after checking the ratings on clist). Well, it does make me feel better about my performance ngl.
From no where C is going to be 1600-1700 . It would be around 1300-1400 .
Either author misjudged the difficulty of the problem or mass usage of AI by participant affected its rating a lot .
The same apply for B.
C felt 1800 to me. It's quite hard, it's not your standard DP, you have to make some extra observations (like how if you have an odd number of elements you have to have an odd number of -1s and everything must be paired up except for one pairwise difference which must be +1) and then do 2 cases.
there is no dp in C? you just need a basic understanding of combo to know that sum of even binominal coefficients is the same as sum of odd binomial coefficients.
Oh idk. I did it super scuffed DP way (you can check my submission lol). I thought of calculating directly with math but it seemed way too complicated when you have a case like
-1 [2 2] [4 4] [5 6] [9 9]You can first just calculate the total for cases where each element appears even no. of times by just multiplication of even possibility for each element which is just (nC0 + nC1 + nC2 + ... ) which is just 2^(n — 1). For the case where -1 appears once and some consecutive elements appear odd times, just add the even possibility value / even possibility of -1 * frq of -1. Maybe I over complicated it
C felt like 1600ish rating to me when I did with math. CLIST predicts it will be 1300 though TwT
The last several contests have all felt heavily underrated on CLIST in my opinion
Difficulty of this problem depends a lot on "boldness" of font of this text: "array is non-decreasing" :)
i figured out B somehow, but what was that question C? I feel like authors are trying to put harder problems at C nowadays to counter cheaters but legitimate users are being screwed this way.
it seemed fine to me after 1-2 observations
I think a lot of people used AI because the problem looks like it should be rated around 1800. After the contest, I asked ChatGPT how to solve it and to rate that problem and it said it should be around 1800-2000.
i like d(even tho i couldnt solve it in contest)
how fast
c get wrong just becasue not using mod in final ans (crying emoji crying emoji)
ngl idgaf fr
If hacks were enabled, how would you theoretically hack most solutions? At least a simple explanation of what features of the problem itself you can exploit is sufficient for this.
Is there a way to defend against it? Give one such valid way that is "reliable enough", i.e. a strong enough adversary could still possibly get in but with sufficiently low probability that it does not matter.
In fact, the first question is already answered in the editorial, because you will know how the solution determine m_0 and m_1. Thus, the second question has no solution.
C is straightup from some math olympiad.
Btw D can be solved in $$$O(n\log A)$$$ using prefix minimum
Can you explain? And are you counting the cost calculation (with bitcount and popcount) as O(log A)? Because if you count it as O(1) instead then the solution described in the editorial is already O(n log A).
(And if we count the cost calculation as O(log A) then the whole analysis becomes more complex because technically addition is logarithmic in the length of the integers involved too.)
This requires one more observation. First we fix the LSB to be at least $$$x$$$ for all numbers. It is always optimal to pick the minimal number at least $$$a[i]$$$ that has that LSB. You can precompute all of these values and their cost for every element.
This is true to get a number with the same LSB that’s different, you must first go through a number with a greater LSB which is smaller.
OK but how does this reduce the time complexity?
What is the complexity you claim for that algorithm, assuming arithmetic operations including bitcount and popcount are $$$\mathcal{O}(1)$$$?
The editorial's solution is $$$O(n\log^2 A)$$$ with $$$O(1)$$$ cost calculation; it does a search over both $$$k$$$ and $$$b$$$, both of which are $$$O(\log A)$$$.
No, because it only considers the values of $$$b$$$ between $$$0$$$ and $$$\log A$$$ that are divisible by $$$2^k$$$.
Take a look at this submission, which uses no precomputation: 382321482. I claim it performs only $$$\mathcal{O}(n\log A)$$$ calls to
Cost()and other trivial arithmetic operations per test case. (using $$$A = \max a_i$$$).This follows from the fact it has three nested loops:
The first loop is clearly $$$\mathcal{O}(\log A)$$$.
The second is clearly $$$\mathcal{O}(n)$$$.
For the third loop, consider that min_cost = $$$\mathcal{O}(\log A)$$$, but the right shift means it is halved each time $$$k$$$ is incremented, so the total time complexity is: $$$\mathcal{O}(n×\log A + n×(\log A)/2 + n×(\log A)/4 + n×(\log A)/8 + ... + n×1) = \mathcal{O}(n×2×\log A) = \mathcal{O}(n \log A)$$$.
You are right. If carefully implemented and analyzed the editorial solution is also $$$O(nlogA)$$$. We were describing a different way to implement it.
My post contest discussion stream for ABCD can be found here
Upsolving E, I saw that failing submission feedback has a "model solution interaction log". Is this new? I have solved interactive problems in the past and never seen something like this. IMHO this is too big of a hint and spoiled this solution. Is there a way to disable this (but keep my own interaction log)?
In B solution, why does the alternating sum need to be -1 for the positive elements? should not need to be +1?
It's technically correct, but I can see why it is confusing. If you take an odd number from the $$$-1$$$ group, then the alternating sum after that would start with a negative. So that negative alternating sum would need to be $$$+1$$$ overall. However, the way we defined alternating sum is that it starts with a positive number, so we can just negate the value we need to reach and use the same term.
Does that make sense?
Thanks that clears it up
I really liked C! I didn't realise that #ways to pick even number = #ways to pick odd number and that made my solution a lot more complex :P
It isn't... But they r related.
what
For E, can the interactor determine $$$v$$$ based on one of the previous $$$ \lt t$$$ test cases?
If not, could save a bit of coding time by just hardcoding a random value
You can consider each test case to be independent.
B is really hard
wo cha dian xin le bro
Same I easily solved C (Because I know combinatorics due to JEE prep) also the D in an hour. But B killed me!
C was so good
B is nice problem.
Anyone got a small hint for Problem F?? I'm implementing the greedy approach in Python, but it keeps giving WA on test 2. The checker says a valid solution exists, while my code prints
-1. The issue seems to be around this part:I've been debugging for quite a while but still can't spot the mistake. Am I missing an edge case, or is my greedy approach fundamentally flawed? A small hint would be really appreciated!
I have a much more complicated solution for B. consider this array $$$a$$$:
$$$\frac{1}{1}$$$ , $$$\frac{1}{1} - \frac{1}{2}$$$ , $$$\frac{1}{2}-\frac{1}{3}$$$ , $$$\frac{1}{3}-\frac{1}{4}$$$ , ... $$$\frac{1}{n-1} - \frac{1}{n}$$$ , $$$\frac{1}{n}$$$
The sum of this array is $$$2$$$. Since $$$\frac{1}{i-1} - \frac{1}{i} = \frac{1}{i \times (i-1)}$$$ , we can find out $$$ 2 \div \frac{1}{i \times (i-1)} = 2 \times i \times (i-1)$$$ is an integer. That satisfy division constraint.
So it seems that we can multiply each $$$a_i$$$ by $$$lcm(1,2,...,n)$$$ to get an integer array. But for n up to 50, $$$lcm(1,2,...,50) = 3099044504245996706400 \gt 10^{17}$$$ , which is too large.
To decrese the lcm , we can delete some big primes $$$p$$$ (like 31,37,...). To do this, we can replace ($$$\frac{1}{p-1} - \frac{1}{p}$$$ , $$$\frac{1}{p} - \frac{1}{p+1}$$$) with ($$$\frac{1}{p-1} - \frac{1}{p+1}$$$) so $$$p$$$ wouldn't appear in lcm and we would't violate division constraint.
Another issue is sometimes $$$\frac{1}{n}$$$ would equal to $$$\frac{1}{i}-\frac{1}{i+1}$$$ (like $$$i=2,n=6$$$) , but we can handle it by change $$$n$$$ to $$$n-1$$$ or $$$n+1$$$.
implement here (with many hardcode number):382360757
This make me wonder what the minumum value of the maximum $$$a_i$$$ among all valid $$$a$$$ with length $$$n$$$. maybe a bonus? Is it doable?
I think I have construction for $$$a_i \leq 840$$$. I'll edit when I confirm it works.
Edit: see the following submission: 382388477
Can i know the logic behind your hardcoded value or its just satisfying the condition somehow ???
Maybe it's dp approach like this
Not completely sure; the construction is mostly found by optimized brute force. That being said, I came across many recurring values, so it seems that there are some desirable properties that lend themselves to small constructions. I guess the most obvious property is that numbers with a lot of divisors tend to work well as the sum, but there are probably some other ones that help.
hell yeah
i have a doubt in D, say i have fixed X zeros in the suffix of the binary rep. of each number.(now Alice can divide the whole array for atleast X times before any odd number appears in the array)
say, the transformed array is B ,WHY do i have to check for some more numbers in the range [B[i],B[i]+32] , like what advantage does it give? can't i just check for B[i] and move on? (any examples where it helps?)
My A/C submission with this extra check
My submission Without this extra check,(WA on tc3)
Thanks!!
try
I struggled with this as well. Here is my reasoning:
For a fixed $$$k$$$ and $$$a[i]$$$ we want to zero the first $$$k$$$ bits of $$$a[i]$$$. Define $$$p = 2^k$$$ as the least power of 2 greater than or equal to $$$a[i]$$$.
Alice is left to choose some $$$b$$$ such that $$$b \geq a[i]$$$ and $$$b \equiv 0 $$$ (mod $$$p$$$).
The cost after such incrementing to b is $$$(b - a[i]) + opCount(b / p)$$$. We wish to minimize this function.
The important observation is that opCount is NOT a smooth function.
It is often smooth: for example opCount(6) = 4 and opCount(7) = 5.
However, consider that opCount(127) = 13 yet opCount(128) = 8.
Thus simply choosing the least multiple of $$$2^k$$$ may or may not be optimal.
I'd be curious to see if there's a less brute force way of finding the optimal choice of $$$b$$$ besides just trying $$$a[i]$$$ + $$$d \in [0, 32]$$$.
you should provide the source code at the end always
It's a little surprising that the difficulty of problem E is probably above 2600 which may(I think) contradict with what authors and testers had predicted.
If the final ratings are 800 — 800 — 1300 — 1800 — 2600 — 2800 as clist shows,It will give out
1.The biggest D~E difficulty gap so far in 2026.
2.The most complex counting problem on C which is just 1300.
The fact that C is 1300 is crazy to me. We weren’t very worried about the D -> E gap and were very worried about the B -> C gap.
I feel that a difficult problem showing up early in a contest pushes people to cheat. If somebody gets stuck on E or F, it's an "oh well I will place fine". When people are at risk of going 2/6 on a contest, they panic and turn to AI.
C really needs observation for at least experts, and maybe a bit math (or you'll enumerate c(cnt,an even number) to calculate 2^{cnt-1} and write a code which is hard to debug like me) so it's crazy for it to be 1300.
I think there are many participants using ai so that the difficulty is much lower than we think.
no way, I almost got the idea on E on a virtual contest, and spent the current last 3 hours understanding D
In F the second case can also be solved with 4 operations:
So I still don't understand how to solve C.
What is confusing you?
D > E
In case someone finds this useful, I had a different approach to problem D.
First, we try to solve without Alice's initial moves.
For each $$$a_i$$$, it will be decremented a certain number of times (say $$$u(i)$$$) and divided by 2 a certain number of times (say $$$v(i)$$$).
There's no way for Alice to decrement more than one number at a time, so she will definitely have to do at least $$$\sum_{i=1}^{n} v(i)$$$ operations, and at most $$$\sum_{i=1}^{n} u(i) + v(i)$$$. Next, if Bob can always ensure that an odd number is second in the array, we can see that this maximum value will be reached. As soon as there is any odd element in the array, we can see that Bob will always be able to make swaps so that it is second.
Therefore, the game after the initial part of the array consists of two parts.
PART 1: all $$$a_i$$$ are even, there is nothing Bob can do to prevent Alice from getting $$$n$$$ divisions by 2 for the cost of only one operation.
PART 2: there is at least one odd element of $$$a$$$. On his turn, Bob can always swap this element into the second position, meaning that Alice gets to do only one division.
Now we can find the answer for a fixed array (with no initial operations).
First, let's say $$$S = \sum_{i=1}^{n} u(i) + v(i)$$$ (note since the $$$a_i$$$ aren't big we can precompute $$$u(x) + v(x)$$$ for all possible $$$x$$$). We find the biggest power of 2 that divides all $$$a_i$$$ (say $$$p$$$) and we know that Alice will get to do $$$n log(p)$$$ operations in only $$$log(p)$$$ steps, and will have to do all the other operations separately. Therefore, Alice will be able reach an empty array in $$$S - (n - 1) log(p)$$$ operations.
Now, we have to figure out how Alice can do better by using her initial moves to increment the numbers. We'll consider 17 different possible values of $$$p$$$ (from $$$1$$$ to $$$2^{16}$$$, since $$$2^{17}$$$ is too far above $$$10^5$$$). For each possible value of $$$p$$$, we have to go through $$$a_i$$$ and consider the cost of incrementing $$$a_i$$$ to a nearby value of $$$p$$$.
Why "nearby" and not just the next multiple of $$$p$$$? Consider $$$p=1, a_i = 15$$$. Even though $$$a_i$$$ is already divisible by $$$p$$$, $$$u(16) + v(16) + 1 \lt u(15) + v(15)$$$, so its worth increasing $$$a_i$$$ to 16. Checking with a bruteforce, I found that for $$$a_i \lt 10^5$$$, the maximum $$$k$$$ such that there exists $$$x \le 10^5 : u(x) + v(x) \lt u(x + k) + v(x + k) + k$$$ is $$$k = 11$$$, so for each $$$a_i$$$, we only have to check multiples of $$$p$$$ within 11 of the very next multiple of $$$p$$$.
To summarise, we just have to go through the 17 values of p and for each one find the total cost of increasing all the $$$a_i$$$ to the optimal value near to the next multiple of $$$p$$$, and take the minimum of these 17 values.
Submission: 385742595
P.S.: thanks to authors for a really great contest!
Is it just me or just B is just hard for its rating? Or just contests getting harder over time? or maybe i am just trash