HHSgddgd, Codeforces!!
I am thrilled to invite you to participate in Codeforces Round 1132 (Div. 2), which will take place on Nov/06/2026 17:35 (Moscow time) and probably won't contain a single $$$\text{XOR}$$$ problem. You will be given 6 problems, two of which will be divided into subtasks, and 2 hours and 15 minutes to solve them.
All problems were authored and prepared by ot_dre, d0esn, amof and me.
ALERT ALERT The round may contain interactive and communication problems. We recommend familiarizing yourself with these formats beforehand.
I would also like to express my heartfelt gratitude to everyone who made this round possible:
- nifeshe for making the coordination process so enjoyable and discussing the problems
- Vladosiya for discussing the problems at an early stage
- all our testers:
- all our VIP testers: suprend, ColobocCodeforces, employed, Rvess, __baozii__, temporary1, ne_justlm, linearspace, robert9524, Nyemot, A_G, AksLolCoding, nik_exists, m3tr0, amsraman, Proof_by_QED, Stathis, sammyuri, KWAMP, Pie_TealCompressorDragon, Dragos, flammifer, h4cker3, jeroenodb, samsoom, aryanc403, ItzsleepyXD, catgirl
- MikeMirzayanov and KAN for the Codeforces and Polygon platforms
- You for participating in the round!
The scoring distribution is: $$$(500 + 500) - 1000 - (1000 + 1250) - 2000 - 2500 - 2750$$$.
Congratulations to the winners!!
Div. 1
Div. 2
First AC
The editorial will be published soon.
First time to see Problem A divided into subtasks:)
And a round with both interactive and communication problem
In fact we have it before in Round 1071, since its F is a communication problem and its G is an interactive one.
https://codeforces.me/contest/1775
A1 and A2
what it means by problem divided into subtasks??? i am newbie:)
It means the problem is intentionally split into variants of itself, each with smaller (easier) constraints than the original. This allows for partial scoring , a contestant who can't solve the full problem can still earn points by solving the easier variants. It bridges the difficulty gap and makes scoring farier.
what is this score dist
No xor? :(
He said probably. So maybe he's joking. I need to prepare for my first interactive and communication tasks (which i will never solve cuz of skill issue :skull:)!
As a tester, ubfeiaeiuabfeiabfieauifenhliuofajs
As a participant, i think that our dear tester wanted to say, that he forced to test this round, otherwise he will never see pancakes again (:sob:).
maybe this is a secret massage that we need to find out what it is :D
As a tester, I lowkey tested two weeks after getting sent the invitation :)
i guess we are in NXN(No XOR November) now
Problem C in the last Educational Round. Streak broken on 4th day itself :D
Interesting scoring distribution:)
As a tester, I ❤️ interactive and communication
As a
VIP
tester, I'm very important.
Sooooo 2 xor problems?
ouuu shiii actually none
This is my first time seeing A1 and A2
me too
Imagine A is an interactive problem ::skull
As a tester,
I will not spoil anything
a tester is not supposed to spoil anything! Why did u spoil that!
I was expecting a rickroll lol
That is not the most spoiled spoiler I have ever seen.
lol!
Whats an communication problem ?
Hmm maybe click the link
Bro made up some friends just to set up a cf round
Why does clicking on problem authors' profiles in this post lead to an anime video.Or is it just me?
NO XOR PROBLEMS YAYYYYYYYYYYYY!
Hope to escape 1000 rating trap.
it will take u atleast 3 contest to do that, i mean u cant get 112+ from this round unless u did till d, and im assuming that at best u can do till C
anyway, best of luck!
Thank you!
good luck to everyone :) thanks to the authors for preparing this round :D
What's the GIF blud :sob:
A1: n<= 5 A2: n<= 50
As a Temu tester I'm sure that I'll get my rating to 0 now(yesterday I got -76).
wth is A subtask bro T_T
how can you POSSIBLY split div2 A into subtasks?!?
also i have a feeling every question is going to be revolving around XOR in some way xd
ohh no..... ohhhhhhh noooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo
if i didnt solved till problem C then im gay, if i didnt solved till D then im lesbian
....... so thats what my fate has decided...... yes i accept my fate.
Will I be surrounded by interactive problems, communication problems and XOR problems
As an author, why am I not on the writer list in the contests tab :(
Contest without XOR problem is the GOAT contest.
Time Changed of this Contest?? It is showing it delayed to 15Min
great contest! eugenechka.boyko.2_0-0 orz, but :sob could only solve A1, but appreciate the problems!
WHY NO XOR PROBLEMS Still contest with communication and interactive is awesome btw
$$$B$$$ is stupid
How does it work?
I mean, how to handle numbers that occur multiple times?
I noticed one thing, when the array is sorted, all elements will be >= their previous average. When the array is reverse sorted, none of them will be good.
So then like, choose the $$$k$$$ last of them in a sorted array to be the good ones, and put the first $$$n - k$$$ of them at the back.
But this WA'd, so I reversed the first $$$n - k$$$ of them and then put them at the back and it passed.
But after doing this you still have to iterate and check if theres exactly $$$k$$$ good ones.
It took me too long to get this + no idea about the proof, hence I really didn't like this problem.
Ah, ok.
So, if same values are there multiple times, they make a certain number of k-values impossible. :/
Anybody explain why is that?
if value repeats multiple times repats k time and if k == n then there would be answer otherwise not .
like basically if pick value one since all value are same the nect number we pick would be same as one we average sum right , so is values repats n times then it make sense to use them
is what i thought.
don't know how would it work when there is mix of multiple same values and not same values
Like I think this can be a proof
The largest k numbers shall always be greater than or equal to the average when they are being considered, no matter what. So if the (k+1)th largest value is equal to the maximum value, then there are atleast k+1 values which will be greater than average, so this is the NO case. Just checking with maximum value is enough for this because, we need atleast two distinct values in the first (k+1) largest values.
For every other arrangement, we can consider the first k largest values in increasing order, then consider the rest of the values in decreasing order (why? because if we take the increasing order for the rest of the elements, a smaller number goes in before a larger number, creating a possibility that the average will be equal to the larger number in the future).
This construction guarantees the k largest values are all greater than equal to average, and the rest are lesser.
I think this code will make it clearer: 397601716
make sense!!
Same.. it just passed.. I observed the pattern but could not proof it for a larger cohort
no lol
NOOOOOOOOO I didnt have time to submit D ahhhhhhhhhhhhhhhhhhhhhh
In D; graph can be completely different from the start. how can we really put any data in it?
Yeah it is really hard to find an invariant in that one
I Was thinking if we can limit the operations which can be done, because k>=2 so we can not touch paths of length 3 or less... but I couldn't solve it.
invariant: 2 color the nodes, 2 coloring never changes over operations. for node u>=2 if u has same color as node 1 determines u'th char in s
woww, that's clever! thank you
but in case of a star, we cannot determine if the string is 11...1 or 00...0.
I hope you to help me.
as far as i understand its 000...0 because nodes has different color from 1. we set node 1 as a reference in the solution cuz string is n-1 length
I've just solved this tricky D. absolutely tricky problem ever.
I treated 11...1 as a special case, and gave a graph containing a triangle (a cycle of length 3) for that case. This works because a cycle of odd length will persist after the operations (the length of the cycle may change, but there'll always be an odd-length cycle).
Nice problemset
D is so peak
Almost solved D but was 5 minutes too slow:(
Great problems!
How did you solve it?
Best D div 2 and best 2-steps problem ever seen. F was so good too but I spend too much time on C so that couldn't solve F in time -_-
wonderful contest
In C1, do the n + 10 operations really check? I didn't even care about it
yeah so C2 writes the checker for it... like they can run the C2 code solution to find minimum number of moves in C1
problem C2 counts minimum moves to change
atob.. so same function can count how many operations we can use to convert given array to answer in C1so the author's solution for C2 is the checker for C1, I guess.
ahhh, yes, ok, thanks
I think, it's just a buffer. The tightest constraint is obviously n
How does C1 work?
Replace the number with the very next prime, all the elements will have gcd 1 or the number itself, and there is always a prime between x and 2*x
we can print next bigger prime for each number.. there is a theorem which says there is always a prime between
xand2xso next prime will have increment less thanx[i]for everyx[i], so total increment will be less than sum of all array valuessee, anyways, we are performing one operation on each ai, so total operation is n<=n+10. (if u have tried to convert ai to its just next prime)
In fact, this restriction blocks a method: changing a_i to the nearest 2^k. This method requires up to 2n operations.
solved till C2 but I was only sure for C1 that my solution is correct, others were like based on feeling that I am correct.
I was sure that ans <= 3 for diff < 25 but was brute forcing for only primes in the range [1, 30] which I only realised after the contest |-|
whoa.. don't tell me it has something to do with goldbach conjecture that every even number can be written as sum of two primes
although I did something like dp for smaller gaps, for larger gaps there should be a prime in between where we can jump
I dunno about the conjecture or whatever, but just the fact that even if n grows to 1e9, the proximity of consecutive primes is ln(n) which comes out to around ~ 23, so if the diff >= 25, you can easily make a jump to the largest prime <= b, I just brute forced that the gcd of any number x and the difference between any prime number greater than x and x is = 1, so just jump to that largest prime <= b, and then the diff becomes < a, and since a is a prime, the gcd of the diff and a = 1. For diff < 25, I kinda half guessed that it's always possible in 3 moves, and brute forcing only takes 25^3, which when multiplied with the number of test cases, should be well under the TL.
ok, thanks for sharing this, hope to get clarity with editorial on this.
Same look at mine, I have no idea why it works :397655764
ok, that means it is always possible to do it in 3 moves
can you please tell me, why you used the number
100and not some other number ?Not sure why it works, but my thinking was that there are at list 25 primes (1 to 100), and multiplying all of them would exceed 1e9, so there should be 2 or 3 prime jump (+x) I can use without any problem.
I just tried it by taking only two extra primes: 397658647
Or C2 might also have poor test cases like C1.
ok, I will read editorial to understand what all is happening in this question
In C1, why finding the nearest prime for each number does not work ?
Wdym it doesn't? I did exactly that and it got accepted
I hope you mean nearest GREATER prime, because that is what I did and passed
It does, check 397601571
It works, if b > a, gcd(a,b) = gcd(a,b-a).
the nearest greatest prime number will work because Bertrands Postulate guarantee's a prime number between n and 2n for even postive n, so the next largest prime number for a given a_i will need less than a_i steps to reach it. Thus the condition of total sum of x's is satisfied
fym bru, it always works
How does A2 work?
for a1 i applied n*n/4 formula making sure to make right angle
for a2 i considered n as sum of power of 2^(ai) and applying same formula but didnt work?
i did the same thing for A2, but it didn't work, for A1 this was my code:
Can anyone share code for A2?
my solution
tysm bro! (its also looking somewhat similar to A1)
because you should apply each operation to exactly one point. for example; in "1, 2" (so "2, 4"), you say that making 3*3 is optimal. But you can't divide 4 to 1+3 or etc.
But the logic is similar to A1, you should make the points as nearest as possible. So you can divide array to 2 subsets with greedy.
You can't split a single move into two moves.
Bertrand's postulate ftw
hi,can someone explain why this solution doesn't work 397603019
1 1 2 2 as a counter example I think
oh yes its thank you
nice contests thx :)
that was my bad contest
Commenting after a long time. Hope to do better next time. :)
For C2, I know if A and B are far enough apart, you can either jump directly from A -> B or jump to A -> (largest prime < B) -> B.
But how do we know that for the case where A and B are close, you can always do it in <= 3 moves.
I did DP for smaller gaps, but didn't prove that it has to be 3 steps.
but later ( after contest ) one idea came to me.... that if gap is even it can be written as sum of two primes ( goldbach's conjecture is true for all numbers below 1e9 ) and if it is odd we can add 1 and then gap become even.. so max 3 moves.
Good point, but it's not a complete proof because even if $$$b - a = p_1 + p_2$$$ that doesn't guarantee $$$gcd(a, p_1) = 1$$$ and $$$gcd(a + p_1, p_2) = 1$$$; it's possible that $$$a = p_1 × p_2$$$ for example, or that $$$a + p_1$$$ is a multiple of $$$p_2$$$.
Yea I'm interested to see the actual proof for this. I didn't get it in contest, but I just guessed that you probably don't need that many.
Also, it's kinda dumb but I didn't think finding the greatest prime <= B would work since there's like (1e4 cases) * (~100 numbers you have to check) * (sqrt(n) = ~1e4) to see if something's prime. I didn't think that would've passed but it did comfortably.
there is a better check and bound
it works in <= (t = 1e4) * (lpg — 1 = 281) * (log (A = 1e9) = 30) * 2 = 1.686e8, which is better (and works in c++) than your 1e10 bound, where lpg is largest prime gap
397676508
oh wow !!.. thanks for mentioning it .. now I am glad I did DP
You can also jump A -> smallest prime that doesnt divide B -> B, and you dont need that observation for <=3 moves you can just bruteforce because they are pretty close to each other when there is no prime like that between them
How do you brute force it? It looks like it can be around a 100 difference, so unless you know that the number of moves needed is <= 3, you can't just try all combinations.
i tried for <=15 because it says so in task, you can just track all number from a to b and then have 2D dp dp[i][number of moves], and then one loop goes theough number of moves and then you go through all the numbers between a and b and then try to go from every less number(you check if you can get to that number in one less move becuse dp already covered that and then if you can then check if you can go to i and update dp, check my submission it will be easier to understand.
Oh that makes sense, thanks!
You can do BFS
More guesswork than patrick jane
c2 is absolute goysloppery wtf
d was so cool but i have skill issue so no solve
I think C1 is not a good problem because it relies on the conclusion that there's a prime between [x,2x], but it is really hard to prove rigorously. It punishes people who try to work out the problem step by step and rewards people who like guessing.
agreed
Felt the same while solving it
same 2 u,I think C1 & C2 actually test whether contestants have the guts to just go for it,not their ability to solve the problems independently.
After all , as a contestant not a mathematician , if you don't know Bertrand-Chebyshev Theorem and Jacobsthal's Function , it's hard to justify your code even if it's correct .
facts dude
for problem C2, let vector a = {i such that gcd(i,a)=1 and a <= i <= 2a} vector b = {i such that gcd(i,b)=1 and b-a <= i <= a}
should size(a X b) < 10^8?
my solution's time complexity is O(a X b) and it passed
I was seconds away from having the first solve on A2, would've been cool
I wanna discuss how to solve E with someone.
I had divided the given graph into connected components. Then, I found that we can shrink the searching area into 2 or 1 components within 2*ceil(log_2(k)) queries. So, the remaining is how to find exact edge among 1- or 2-component bipartite graph using at most 33 — 2 * ceil(log_2(k)) queries. where, 33 — 2 * ceil(log_2(k)) >= 11.
please publish editorial before next round starts.
now the editorial of the next round is already there but for this contest "The editorial will be published soon."
he he, I touched Expert rank and came back , but still no editorial for me :( :(
please authors, I want to know how to prove the correctness of the solutions in this round, aaaaahhhh!!!
https://codeforces.me/blog/entry/158544
oh, thanks
Can you publish editorial? I need to know how to solve C2 and D
Editorial?