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By awoo, history, 6 months ago, translation, In English

Hello Codeforces!

On Jan/18/2024 17:35 (Moscow time) Educational Codeforces Round 161 (Rated for Div. 2) will start.

Series of Educational Rounds continue being held as Harbour.Space University initiative! You can read the details about the cooperation between Harbour.Space University and Codeforces in the blog post.

This round will be rated for the participants with rating lower than 2100. It will be held on extended ICPC rules. The penalty for each incorrect submission until the submission with a full solution is 10 minutes. After the end of the contest, you will have 12 hours to hack any solution you want. You will have access to copy any solution and test it locally.

You will be given 6 or 7 problems and 2 hours to solve them.

The problems were invented and prepared by Adilbek adedalic Dalabaev, Ivan BledDest Androsov, Maksim Neon Mescheryakov, Roman Roms Glazov and me. Also, huge thanks to Mike MikeMirzayanov Mirzayanov for great systems Polygon and Codeforces.

Big thanks to the testers shnirelman and Minder for their valuable advice and suggestions!

Good luck to all the participants!

UPD: Editorial is out

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6 months ago, # |
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Hopefully I reached Expert :)

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6 months ago, # |
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I wish everyone a good contest!

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6 months ago, # |
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Good luck to every Contestant!

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6 months ago, # |
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My 3rd rated contest. Wonder if I could reach True Rating 1700(displayed 1400)

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6 months ago, # |
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Good Luck and Have Fun everyone!

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6 months ago, # |
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Good luck to everyone

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6 months ago, # |
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Hope this round aint Mathforces af. Usually, Edu rounds = Mathforces

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    6 months ago, # ^ |
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    ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤ

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6 months ago, # |
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What's the difference between an educational round and normal div 2 round?

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6 months ago, # |
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Such a short and clear announcement I hope problems statements be like this too!!!

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6 months ago, # |
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Hopefully I can get Pupil(not Newbie) :/

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6 months ago, # |
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BledDest Round

Spoiler
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6 months ago, # |
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Hope I can get Expert badge.

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6 months ago, # |
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Hopefully I reached Specialist :)

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6 months ago, # |
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Hope for a positive delta.

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6 months ago, # |
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How many problems I need to solve to become a Pupil in this contest?

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6 months ago, # |
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Educational round is overrated

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6 months ago, # |
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Educational round means you have to study gcd well :)

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6 months ago, # |
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l have nothing big,Pupil hop to be reached!

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6 months ago, # |
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I hope to bring the pupil back again. Good luck for everyone!

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6 months ago, # |
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As I screwed up in the last div3, I hope to solve 3-4 problems in this round :|

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6 months ago, # |
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anxiety

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6 months ago, # |
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bad explanation of first question

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6 months ago, # |
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[deleted]

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    6 months ago, # ^ |
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    and you are also using an alt, so shut up, do not complain of stuff you are personally doing

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    6 months ago, # ^ |
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    What happened to "romanian diversion strat"?

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6 months ago, # |
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This First question made me skip contest man :(

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6 months ago, # |
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Didn't realize E was easier ,should've tried that first.

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    6 months ago, # ^ |
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    thats apio2022 perm

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      6 months ago, # ^ |
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      Okay I start to implement it

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        6 months ago, # ^ |
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        I got 98.8 pts and stuck on it :(

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    6 months ago, # ^ |
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    D is a lot easier than E, E has just appeared before

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      6 months ago, # ^ |
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      Yeah and I thought D<F<E... E needs constructive skills...

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6 months ago, # |
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B was a good problem for such a simple concept but i made 2 wrong submissions.

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    6 months ago, # ^ |
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    Yes! Problem statements were clear and precise. But I must be blind... :(

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6 months ago, # |
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Was C that easy ? is it based on some standard algorithm/ some standard problem ? i was thinking in direction of Floyd Warshall algorithm. but could not come up with solution. I'm seeing around 7000 successful submissions. or it was based on something else like Dijkstra/DFS/BFS ??

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    6 months ago, # ^ |
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    i pre-calculate cost of go forward and backward

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    6 months ago, # ^ |
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    This problem maybe doesn't require the shortest path algorithm

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    6 months ago, # ^ |
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    C was a modified version of prefix sums

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    6 months ago, # ^ |
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    It can be shown that in order to go from point x to a point y, going back is never a good choice.

    proof
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6 months ago, # |
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Damn, the first question reduced my life by a few years. :')

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    6 months ago, # ^ |
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    I mean yeah. Like I did too many wrong subs in A :(

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6 months ago, # |
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I have skill issue

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6 months ago, # |
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I think E was way easier than it ought to be. Good contest nevertheless.

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6 months ago, # |
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Wasted too much time in problem A :/

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6 months ago, # |
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can someone please explain problem A? Thanks in advance.

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    6 months ago, # ^ |
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    this took me 90 minuts to figure out

    loop from 1 to n if(a[i]==b[i]&&c[i]!=a[i]) then yes and break if(c[i]!=a[i] && c[i]!=b[i]) yes and break

    print no if loop finished without any yes

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      6 months ago, # ^ |
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      but the question said if there is one wrong letter then its a NO??

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      6 months ago, # ^ |
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      Same lol!. Usually the first problem is too stressful than other problems

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      6 months ago, # ^ |
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      only condition 2 is enough

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      6 months ago, # ^ |
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      if a="ab", b="cd", c="ef" then what will be the template string t? suppose t= "EF", but it matches with string c also, right? Sorry if my question sounds dumb.

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        6 months ago, # ^ |
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        In order to print "yes," it should not match the condition with the third string. So, "EF" matches with the first and second strings, because none of them is equal to "ef,",the third string should not match the condition. This is what is happening in our case because "EF" means that the string should be different from "ef," but it is "ef," so it doesn't match the required condition. Therefore, we print "yes."

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    6 months ago, # ^ |
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    Count the number of positions with a!=b but( b==c or a==c) and also if all are equal then if the count is ==n it's no otherwise it's yes

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6 months ago, # |
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awoo Loved the D and people think LinkedList is overrated.

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    6 months ago, # ^ |
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    I actually used TreeSets/Ordered Set to track the previous and next poitners. Unfortunately was not able to solve. If someone could help me identify my mistake in the solve function (Java Code):

        static void solve(IO io) throws IOException {
            int[] params = io.readIntArray();
            int n = params[0];
            // int k = params[1];
            long[] attack = io.readLongArray();
            long[] defence = io.readLongArray();
            TreeSet<Integer> alive = new TreeSet<>();
            for (int i = 0; i < n; i++) {
                alive.add(i);
            }
            HashSet<Integer> active = new HashSet<>();
            active.addAll(alive);
            for (int i = 0; i < n; i++) {
                HashSet<Integer> dead = new HashSet<>();
                HashMap<Integer, Long> damage = new HashMap<>();
                for (Integer j : active) {
                    Integer prev = alive.lower(j);
                    Integer next = alive.higher(j);
                    if(prev == null && next == null)
                        continue;
                    else if(prev == null){
                        damage.put(next, damage.getOrDefault(next, 0l) + attack[j]);
                    } 
                    else if(next == null){
                        damage.put(prev, damage.getOrDefault(prev, 0l) + attack[j]);
                    }
                    else{
                        if(j - prev == next - j){
                            damage.put(next, damage.getOrDefault(next, 0l) + attack[j]);
                            damage.put(prev, damage.getOrDefault(prev, 0l) + attack[j]);
                        }
                        else if(j - prev > next - j){
                            damage.put(next, damage.getOrDefault(next, 0l) + attack[j]);
                        }
                        else 
                            damage.put(prev, damage.getOrDefault(prev, 0l) + attack[j]);
                    }
                }
                for (Integer j : damage.keySet()) {
                    if(damage.get(j) > defence[j]){
                        dead.add(j);
                    }
                }
                for(Integer j: dead){
                    alive.remove(j);
                }
                active.clear();
                for (Integer j : dead) {
                    if(alive.lower(j) != null){
                        active.add(alive.lower(j));
                    }
                    if(alive.higher(j) != null){
                        active.add(alive.higher(j));
                    }
                }
                io.append(dead.size());
            }
            io.newLine();
        }
    
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    6 months ago, # ^ |
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    bro u fell of

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    6 months ago, # ^ |
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    Still do think its overrated after writing this piece of art which even i can't read now Submission

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    6 months ago, # ^ |
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    Bro just use set

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6 months ago, # |
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I hate segment trees

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6 months ago, # |
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I got drown in caseworking C. Any ideas on avoiding, like, a dozen of caseworks?

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    6 months ago, # ^ |
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    you can create a Generic function that decrease the number of those caseworking.

    242254308

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      6 months ago, # ^ |
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      Wow! That was neat!

      I tried to partition the array into several groups, and each group contains two "center" cities where everyone leads to, and then hop between the groups before caseworking the best answer out when they are in the same group.

      That was painful. Those with a daring soul may see it here --> 242305848

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Stucked in Question B till last not able to optimise.

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my first E yaaaaaaaaaaay

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6 months ago, # |
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Anyone feel like the wording on A was weird?

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    6 months ago, # ^ |
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    Took me 20 minutes and out of the contest to understand the concept of "uppercase is lowercase this time".

    I do not understand how problemsetters cannot see that such a wording is bad.

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6 months ago, # |
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what is the approach for B?

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    6 months ago, # ^ |
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    Note that the sides of the triangles are powers of two. So, at least two of the sides have to be equal. This observation makes it a lot easier.

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      6 months ago, # ^ |
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      how to count the valid triplets?

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        6 months ago, # ^ |
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        Use combinatorics, create a map to keep track of frequency, for each frequency in map, you can choose 3 sides of same length NC3 (N = frequency of element and you can choose 2 same sides and 1 from smaller sides, NC2 * CF (cumulative frequency before the current element). Intuition is 2^k + 2^(k + 1) is less than 2^(k + 2) using GP formula you can prove, therefore 3 sides cannot be unique, isosceles and equilateral triangles can be formed

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E was basically APIO 2022 P3 with nerfed limits?

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How to solve F?

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6 months ago, # |
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Good contest

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6 months ago, # |
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I submitted B 14 times ;-;

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Did the authors intend to cut off solutions to D that uses two std::sets? I got TLE on my first attempt before finally switching to segment trees.

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    6 months ago, # ^ |
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    No, I used two std::sets as well and it passed in 700ms

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      6 months ago, # ^ |
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      same, my passed in 389ms

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      6 months ago, # ^ |
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      Same. It was meant for Set. At any point if you remove X elements, you will iterate for 2*X elements and out of which if you remove Y elements, next time you will iterate for 2Y elements. So overall 3*N operations will be carried out.

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      6 months ago, # ^ |
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      I see. I guess it was my implementation that was inefficient, then.

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    6 months ago, # ^ |
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    How did you solve it by segment trees? Like what was each Node storing, can you please share.

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      6 months ago, # ^ |
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      The segment tree was only used to get the global minimum of the remaining health points of each monster (i.e. $$$d_i$$$ minus the total attacks dealt to it). Each node stored the minimum of that value among all indices in its range, as well as the index that caused that minimum value. This allows us to simulate each round and find out which monster died during the round.

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    6 months ago, # ^ |
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    I used linked list — 265 ms.

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    6 months ago, # ^ |
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    I used 2 vectors and a static linked list. It's supposed to be O(n)...

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6 months ago, # |
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C was intuitively not too hard.. just too painful to implement.

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How to guess the solution of problem E?

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    6 months ago, # ^ |
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    Notice that this problem appeared in APIO and copy paste

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    6 months ago, # ^ |
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    maybe 10^18 gave me a hint towards set bits....

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    6 months ago, # ^ |
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    10^18 says it should be done in O(logn).

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      6 months ago, # ^ |
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      I'm pretty sure it can be done in O(n), using the fact that increasing array of length n contains exactly 2^n increasing subsequences. Oh, I meant n being the length of resulting array.

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    6 months ago, # ^ |
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    I think considering case of $$$1, 2, 3, 4, 5, 6 ...$$$ is very natural.

    Then understanding $$$1, 2, 3, ... , n$$$ gives $$$2^n$$$, should be enough to be able to construct solution

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      6 months ago, # ^ |
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      I tried this idea. I noticed, that if I have blocks, which look like: $$$|10 \; 11 \; 12|7 \; 8 \; 9|3 \; 4 \; 5 \; 6|1 \; 2|$$$, and their lengths are $$$a_1, a_2, \dots, a_k$$$, then number of subsequences is $$$(2^{a_1} - 1) + \dots + (2^{a_k} - 1) + 1$$$. Then I can factorize $$$x$$$ in sum of degrees of $$$2$$$. For example, $$$37 = 2^0 + 2^2 + 2^5 = (2^5 - 1) + (2^2 - 1) + 3$$$, so I use $$$a = {5, 2}$$$ and append two decreasing numbers at the end.

      This is $$$O(log^2{X})$$$ solution, which is incorrect.

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        6 months ago, # ^ |
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        Nice idea. I think after that you should have realized, that building solution using independent blocks would result in very lengthy sequence. And after that realization should have abandoned this approach and returned one step back to $$$1, 2, 3, 4, ...$$$

        If we add $$$3$$$ to the end of $$$1, 2, 3, 4$$$ we would get additionally as much sequences as there were sequences in $$$1, 2, 3$$$

        So in general adding $$$m$$$ to the $$$1, 2, 3, ..., n$$$ would increase number of sequences by $$$2^m$$$. That is I believe — final idea of solution.

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          6 months ago, # ^ |
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          I figured this out but was unable to implement within time :(

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6 months ago, # |
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I didn't like the round (yes, I am going back to expert again)

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    6 months ago, # ^ |
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    When will I be expert like you?

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6 months ago, # |
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is it div3?

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    6 months ago, # ^ |
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    A div four F had comparably equal ACs to today's D and lower ACs than E.

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      6 months ago, # ^ |
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      maybe because majority of high rated users don't write div4?

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      6 months ago, # ^ |
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      You're comparing an outlier with another outlier

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6 months ago, # |
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Horrible statement for D!

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6 months ago, # |
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Am I the only one who was struggling with problem B ;-;

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    6 months ago, # ^ |
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    no bro

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    6 months ago, # ^ |
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    Struggled with both A and B :)

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    6 months ago, # ^ |
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    Not sure why but I found A relatively harder than B. B was pretty much straightforward as triangle is only possible if sum of two sides is greater than third. Note that 2^0 + 2^1 + .... + 2^n < 2 ^ (n + 1). So you always have to take two same numbers. Reminder can be any number which is less or equal to this.

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      6 months ago, # ^ |
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      Yes agree B was trivial once u realise that fact

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        6 months ago, # ^ |
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        A was relatively hard, I was confused for like 6 minutes, then I just guessed based on testcases.

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      6 months ago, # ^ |
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      F missed the first line sides are 2^a[i] not a[i]

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      6 months ago, # ^ |
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      Great solution bro, I have solved A and C and couldn't solve B. LOL

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    6 months ago, # ^ |
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    i was struggling with B for 1 hour cause of overflow case. t-t

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6 months ago, # |
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E looks like easyer than D because you can find the solution on the Internet :)

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    6 months ago, # ^ |
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    Cheaters cheat what can we do :( I spent all my time on D

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6 months ago, # |
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OMG! Stuck on B for more than 30 minutes, not until the contest is finished did I notice that the length is $$$2^{a_i}$$$ not $$$a_i$$$ .

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    6 months ago, # ^ |
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    OH NO!

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    6 months ago, # ^ |
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    Just noticed it by reading your comment :/

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    6 months ago, # ^ |
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    Its 2^ai

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    6 months ago, # ^ |
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    the length is $$$2^{a_i}$$$. Stuck on A also because if this The string doesn't match the template if the given condition doesn't hold for at least one i.

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    6 months ago, # ^ |
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    Same here, i too thought the same and trying to figure out the sliding window solution!

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6 months ago, # |
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Problem E coincides with APIO2022 Problem C completely.

In APIO2022 Problem C — Permutation, It requires an array of length less than 90. In today's contest, it just has a limit that the array of length at most 200. Besides, In APIO, it also requires that elements should be distinct. It has been weakened.

I think that E<D.

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    6 months ago, # ^ |
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    noo it was the first E i did :(

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    6 months ago, # ^ |
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    I feel the same , I saw some submissions to E with 9 lines of code. Even if it didn't appeared in some contest , it would still be an easy problem and clearly not E or (may be even D) level.

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    6 months ago, # ^ |
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    I totally agree with you

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    6 months ago, # ^ |
      Vote: I like it -9 Vote: I do not like it

    how did you find that? Had you solved this problem earlier? I searched problem E for almost 20-30 minutes but was unable to find any similar problem.

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    6 months ago, # ^ |
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    how do u solve for < 90? I can understand upto 118

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6 months ago, # |
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Can someone tell me why my code for D get runtime error? Thank you in advance

My Code

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6 months ago, # |
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Wasted 10 minutes in Problem B before I noticed the power of 2.

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6 months ago, # |
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Me thinking that D was about decreasing health instead of just checking if damage is enough. :[

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    6 months ago, # ^ |
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    ohh yes I was thinking that too, I realised after a while..

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6 months ago, # |
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I submitted B 5 times and got WA.Because I thinked pow(2,0)=0.:)

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6 months ago, # |
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Can some high rated coder tell me how to become an expert asap?

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    6 months ago, # ^ |
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    You just need to work on yourself. Try solving previous contests in virtual mode or just try solving problems in archive. There you can choose the types of tasks in which you are bad. If you can't solve something, don't look at the solution right away, try to solve by yourself. But if you really can't, check solution and always write your code by yourself. Also there are some useful courses in edu, you can solve them too

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6 months ago, # |
  Vote: I like it +4 Vote: I do not like it

Div 2. where E solved by more than 1800 people,

bye-bye my rating 😂

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6 months ago, # |
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it's okay ill try again next round. btw can i participate div1 round?

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    6 months ago, # ^ |
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    yeah but div1 is harder. Div2 and div1 happens at the same time you can participate in div2 if that is better for your level.

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6 months ago, # |
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Can someone tell me does KasymK cheating in this round?

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6 months ago, # |
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I don't understand why my hack didn't work on problem B Test: 1 4 0 0 0 3 This is withing the constraints of the input given but it shows unsuccessful hacking attempt because expected answer is 1. Shouldn't expected answer be 0 since no triangles can be formed obviously.

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    6 months ago, # ^ |
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    can be formed from 0 0 0

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      6 months ago, # ^ |
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      definition of degenerate triangle is given in the problem(area must be greater than 0) do you really think a combination of 3 sticks of length 0 is a triangle?)

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        6 months ago, # ^ |
        Rev. 2   Vote: I like it +3 Vote: I do not like it

        the length of each side is 2^Ai not Ai

        so in case Ai == 0 the length is 1

        its shown in first line of statement

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          6 months ago, # ^ |
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          Yeah you are right I made a similar mistake during contest, my brain us melting.

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        6 months ago, # ^ |
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        bro the length is 2^ai not a[i] GG

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6 months ago, # |
  Vote: I like it -21 Vote: I do not like it

Decent contest. A and B is okay. C is somewhat fun but probably should only be limited to a<b (failed rhe caseworking like an idiot, but even then turning from the left-right variation to the other way around doesn't require much effort) Who tf allowed E to exist? Even I, a literal green, knows that it's APIO22's perm at first glance. Stealing problems from chinese OJs or some random unpopular place, while a scummy thing to do, would still only allow a small set of participant to get a major advantage. But stealing from fricking APIO??? What kind of coordinator even allowed that to exist lmao. What's next? Stealing literal IOI problems? Come on man what the hell

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    6 months ago, # ^ |
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    ......and the fact that anyone who takes CP seriously should know about the existence of oj.uz, which...allows for free copy pasting, is, uhhhh, funny. Stealing from a judge that allows you to see submissions post-AC is already bad, stealing from somewhere that you could ctrl+c ctrl+v for free..... Should've copied instead of debugging C smh

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    6 months ago, # ^ |
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    then why you didn't solve E?

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      6 months ago, # ^ |
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      Don't want to copy code + messed up horribly (unironically know that adding a numbern that is larger than all previous corresponds to adding a 1 at the end of current bit sequence but not knowing how to deal with x2 case). What do you expect from a Green again

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6 months ago, # |
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Anyone plz tell me how can you solved the Problem C or D, I was solved only A,B. Thankyou

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    6 months ago, # ^ |
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    Problem C was only the cost of going to X -> X+1 -> X+2 -> ... -> Y if $$$X < Y$$$, or X -> X-1 -> X-2 -> ... -> Y if $$$X > Y$$$. The reason is that if you go back to a closest city you have to retreat anyway, so we only need to calculate the cost of going to the left, or to the right. To do this efficiently we use prefix and suffix sums.

    Problem D for me was only implementation, I simulated the steps, but instead of iterating all the monster and removing them for $$$N$$$ steps I only save the monsters that are affected by removing the dead monsters. I think this is $$$O(n \log n)$$$?, because in each iteration the monsters affected are reduced. I really don't know and possibly get hacked. For removing monsters I use Linked List.

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      6 months ago, # ^ |
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      D is $$$O(n)$$$ if implemented properly because each monster is removed once = n in total, and each removed monster affects two in the next round = 2n. +n rounds = still $$$O(n)$$$.

      Something like this. My initial implementation was $$$O(N log N)$$$ improved by reading other solutions.

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      6 months ago, # ^ |
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      Thanks for sharing your idea.

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6 months ago, # |
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242304887

what is wrong with my solution!! this problem suck!!

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    6 months ago, # ^ |
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    In the combinations function you cast res to int, which can overflow in some testcases

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    6 months ago, # ^ |
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    your ll C(ll n, ll k) actually return a int instead of a long long.

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6 months ago, # |
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Going back to newbie :(

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6 months ago, # |
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Could someone explain why and tell me the number of increasing subsequences?

UPD: I found out why, thanks

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6 months ago, # |
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I actually liked E, It took me an hour to come up with a pattern, the pattern was related to bitmask of X. It could have been better if length of array was atmost 120.

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6 months ago, # |
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Can anyone tell me why my submission on D gets WA?

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    6 months ago, # ^ |
      Vote: I like it +3 Vote: I do not like it

    The problem is the first occurrence of this line:

    if (i==0||been[i]) continue;
    

    If the monster on the left is missing/dead, you still need to check the monster on the right. You should combine that if-statement with the next one, instead of skipping the rest of the loop body.

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    6 months ago, # ^ |
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    Take a look at Ticket 17262 from CF Stress for a counter example.

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6 months ago, # |
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What does this mean "The penalty for each incorrect submission until the submission with a full solution is 10 minutes"

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    6 months ago, # ^ |
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    When you commit the correct solution, you get points like you would have commited it 10 minutes * wrong times later.

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    6 months ago, # ^ |
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    assume that you sent problem B after 10min and get wrong answer on test bigger than 1

    this 10min will be added to your penalty

    your handle A: +(0:03) B: -1(0:10)

    your penalty will be 3+10

    sorry for being complicated

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    6 months ago, # ^ |
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    even if you send 100 wrong submissions, it doesnt count to your penalty until you submit a correct one. Each wrong submission costs 10 penalty points. A submission is not wrong if it is accepted or compilation error or your code fails in the very first test.

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    6 months ago, # ^ |
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    On the scoreboard, people are ranked by number of problems solved first (higher is better), and by time second (lower is better).

    The time is the submission time of all the solutions added together. For example, if you solve problem A, B and C exactly 10, 20 and 30 minutes after the start of the contest, your total time is 60 minutes. For every incorrect submission, 10 additional minutes are added to your time.

    However, and I think this is where the confusion comes from, the penalty time for incorrect submissions only applies to problems that you have solved. If you do not solve the problem, that problem adds 0 minutes to your time, no matter how many failed attempts you made.

    This prevents situations like where Contestant 1 solves problem A in 15 minutes. Contestant 2 solves problem A in 10 minutes, and then makes a wrong attempt to solve problem B. Giving contestant 2 a 10 minute penalty would make him rank lower than 1, which seems unfair, since both participants solved the same set of problems (only A) and participant 2 was strictly faster.

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6 months ago, # |
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Can Anyone please explain the logic behind Problem D ...

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    6 months ago, # ^ |
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    Consider the monsters before first day. Each monster gets some damage, foreach monster you can tell how many days it lives.

    Output answers until the first day monsters die, then remove all monsters that die that day. Repeat.

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    6 months ago, # ^ |
      Vote: I like it +10 Vote: I do not like it

    If you use the linked list approach, then the intuition is this: you can easily compute whether a monster dies in the current round, by adding the attack scores of its neighbors and comparing it to the monster's defense score. If the monster survives this round, it will also survive subsequent rounds, as long as its neighbors don't change.

    So the only monsters that you need to check on round X are the neighbors of the monsters that died on round (X — 1). (Round 1 is a special case: here you have to check all monsters.) If you store the monsters in a double-linked list, then you can efficiently remove dead monsters and calculate a list of their neighbors, which you use in the next round, etc.

    This way, you do at most O(N) work across all N rounds.

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    6 months ago, # ^ |
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    Video editorials for D & E here: https://www.youtube.com/@rishabhxchoudhary/videos

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6 months ago, # |
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Anyone knows why my submission for problem D gets wrong answer verdict? I am trying to maintain the previous and next monsters in prv[i] and nxt[i], code here: https://mirror.codeforces.com/contest/1922/submission/242332530

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6 months ago, # |
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How to solve F? Does it use some specific optimisation?

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    6 months ago, # ^ |
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    Naive O(N^3K) dp was just fine.

    Spoiler
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      6 months ago, # ^ |
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      Can't you just bruteforce every possible x in $$$O(N x)$$$?

      Oh, I see now. That different x's are optimal, in 3rd test case replace first all 2s with 3 and then entire array to 2.

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6 months ago, # |
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Can someone explain B?

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    6 months ago, # ^ |
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    we have to pick three powers a,b,c lets assume a<=b<=c. if a=b means 2^a+2^b=2^(a+1) => 100 + 100 ==> 1000 (2^3 addition in binary) the sum has to be less than longest side to have a valid triangle. which means a+1>c but c>=a so we have to take c such that it is equal to a

    if a!=b a=0100 b=1000 sum=1100 this sum is less than 10000 i.e we need c to be b<=c<b+1

    so we need to take all the pairs in which either 3 or 2 numbers are equal

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6 months ago, # |
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242343507 whats up with testcase 3 in B? how can be factorial of any number be zero?

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6 months ago, # |
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Can anyone explain how to approach problem D

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6 months ago, # |
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Video editorials for D & E here: https://www.youtube.com/@rishabhxchoudhary/videos

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6 months ago, # |
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can anyone tell me where i'm wrong in my solution for problem D? thank you.

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6 months ago, # |
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Can anyone help me point out what is wrong in my 242347538 for problem D?

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6 months ago, # |
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still trying to solve A

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6 months ago, # |
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Can someone explain why this solution 242350744 for problem C is giving runtime error on testcase 3.

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    6 months ago, # ^ |
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    Try long long type instead of int.

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      6 months ago, # ^ |
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      Done that already,seems like the problem was out of bound vector. My bad :(

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6 months ago, # |
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respectfully, write better statements please.

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6 months ago, # |
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E was surely much easier than D.

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6 months ago, # |
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This submission seems suspicious, the participant has included code from some other problem to avoid plag checks. His exact same code was available in a youtube video during the contest. MikeMirzayanov please look into this.

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6 months ago, # |
Rev. 4   Vote: I like it +5 Vote: I do not like it

Problem E, Possible Issue with Judge,

Verdict: Wrong answer on test 2, test case 29

Test case 29,

30

My output for test case 29,

7

1 100000003 2 100000002 3 100000001 4

Judge: Wrong answer The number of increasing subsequences in the printed array doesn't equal to x (test case 29)

I have myself counted 30 increasing subsequences in my output. Please look into this.

242373040

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    6 months ago, # ^ |
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    Ok this is really weird actually, I ran your code locally and on an online compiler for X = 30 and got the output of:

    8 100000004 1 100000003 2 100000002 3 100000001 4

    But when I ran your code on Custom Invocation on CF, I got the output that you described. Really strange.

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      6 months ago, # ^ |
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      Actually follow up on this, it prints out: 8 100000004 1 100000003 2 100000002 3 100000001 4

      When compiled with G++17 but it outputs:

      7 1 100000003 2 100000002 3 100000001 4

      When compiled with G++20

      This might be your problem...

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    6 months ago, # ^ |
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    The judge is correct. However, your code seems to be printing

    8
    100000004 1 100000003 2 100000002 3 100000001 4 
    

    instead of the output you mentioned for the 30 case when I run it on C++17. Here the answer is clearly > 30. But the output you mentioned in your comment will have 30 favourable subsequences.

    Curiously, I got the output you mentioned when I ran it in C++20, so you may have some kind of compiler specific error in your code.

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    6 months ago, # ^ |
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    You access outside the limits of your pos array (I think you meant to put j < pos.size() in the if?).

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6 months ago, # |
Rev. 4   Vote: I like it +7 Vote: I do not like it
A hack for C
How it works
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    6 months ago, # ^ |
    Rev. 3   Vote: I like it +17 Vote: I do not like it

    Actually there's another one. A lot of contestants don't use the prefix sum to solve C. They just calculated the distance one by one. So, the following can hack them.

    1
    100000
    ......
    100000
    1 100000
    1 100000
    ......
    1 100000
    
    The generator

    And, for problem B, there's so many people use a "cnt" array to count the appearance of a number. They just memset the cnt array in the beginning of each test case. So, just construct a test with 10000 test cases and make sure that the sum of n is 300000.

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6 months ago, # |
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Guys pehle ques me toh same same but diffelent...ho gya..:)

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6 months ago, # |
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I think A problem is bad,because its topic is vague

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6 months ago, # |
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I feel like this contest has tested the contestant's ability to judge time complexity 感觉这场就考验了选手对时间复杂度的判断能力

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6 months ago, # |
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Wow, 3 years since I last did competitive programming and I still choke under pressure of a CF contest XD

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6 months ago, # |
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Can anyone tell me a case where it is failing in B





int n; cin>>n; vi v(n); for(auto &it : v)cin>>it; map<int, int>mp; sort(al(v)); map<int, int>smaller; int cnt = 0, prev = -1; set<int>st; for(auto &it : v) { st.insert(it); mp[it]++; if(prev != it) { prev = it; smaller[it] = cnt; } cnt++; } debug(mp); debug(smaller); int ans = 0; for(auto &it : st) { int f = mp[it]; int cur = 0; if(f >= 3) { int NCR = ncr(f, 3); debug(NCR); cur += NCR; } if(f >= 2) cur += smaller[it]; ans += cur; } cout<<ans<<endl;
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    6 months ago, # ^ |
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    either 3 sides have to be same. (or) 2 sides have to be same and 1 side have to be lessthan other equal sides. ~~~~~ reason: In a triangle, sum of 2 sides have to be more than 3rd side. ~~~~~ As we are given in powers of 2, if 3 sides are different, then count of triangles are 0. if 2 sides are same, count = choose 2 of same size and other with size less than choosen size. if 3 sides are same, count = choose 3 from same size.

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6 months ago, # |
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this is code, i have written for E (Increasing Subsequences)

vector<int> solve(ll n){
    vector<int> v{1}; vector<ll> dp{2};
    while(dp.back()*2<=n){
        v.push_back(v.back()+1); 
        dp.push_back(dp.back()*2);
    }
    ll cs = dp.back(); int l = v.back();
    vector<pair<int, int>> ins; int s = v.size();
    while(cs<n){
        if(cs+1==n){
            l++;
            ins.push_back({l, 0});
            cs++; break;
        }
        for(auto it = dp.rbegin();it!=dp.rend(); it++){
            if(cs+*it <= n){
                cs+=*it;l++;
                ins.push_back({l, s-(distance(dp.rbegin(), it))});
                break;
            }
        }
    }
    for(auto &i: ins){
        int val = i.first; int p = i.second;
        v.insert(v.begin()+p, val);
    }
    return v;
}

Its running perfectly fine in my local device.

when submitted, i came to see that I got wrong answer when n=pow(10, 18). i checked my solution with the code:

ll count_seq(vector<int> v){
    int n = v.size();
    vector<ll> dp(n, 1);
    for(int i = 0;i<n;i++){
        for(int j = 0;j<i;j++){
            if(v[j]>=v[i]) continue;
            dp[i]+=dp[j];
        }
    }
    return accumulate(dp.begin(), dp.end(), 1ll);
}

I sent output of solve function as input to count_seq function. I am getting pow(10, 18) implying its correct answer.

I cant figure out whats wrong. please help me.

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    6 months ago, # ^ |
      Vote: I like it +8 Vote: I do not like it

    Can you tell me the submission id? So I can find why you're wrong more quickly.

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      6 months ago, # ^ |
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        6 months ago, # ^ |
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        In your main function, you should use ll n instead of int n to read the input data correctly.

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          6 months ago, # ^ |
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          Bro, this is error!!!. Thanks for correcting me. from now i will completey use ll instead of int. I have been poking my head for 3 hours. I dont know what can i do other than saying thankyou. I am just a newbie, Lets be friends. I need some guidance from pro's like you.

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6 months ago, # |
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For D when i run my code in VScode for the example testcase i am getting correct answer but when i submit or use the custom invocation i am getting wrong answer for the same testcase. Does anyone know the reason?. My Code.

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    6 months ago, # ^ |
    Rev. 3   Vote: I like it +8 Vote: I do not like it

    Your code contains an UB. In the following code block:

    for(auto itd = defence.begin(), ita = attack.begin(); itd != defence.end(); itd++, ita++){
        if(*itd < 0){
            died++;
            defence.erase(itd);
            attack.erase(ita);
            itd--;
            ita--;
        }
    }
    

    You tried to erase 2 iterators called itd and ita, but you immediately use them after the erasing. It is now allowed and that's why there are different output of your code.

    To correct, you should create a temporary backup for prev(itd), prev(ita) and then erase itd, ita. Finally, replace itd, ita with the previous backup.

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      6 months ago, # ^ |
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      Thanks for the reply. i found one more problem which was the reason for different output was in the first if block

      for(int i = 0; i < k; i++){
          if(i == 0 && (defence[0] - attack[1] < 0))
              defence[0] -= attack[1];
                      
          else if(i != k-1 && (defence[i] - (attack[i-1] + attack[i+1]) < 0))
              defence[i] -= (attack[i-1] + attack[i+1]);
      

      even when i = 0, the if statement is not executing because of the '&&' statement and the else if statement is running for 0 index and subtracting '-1 th' index which gives it a garbage value.

      what you suggested is another problem which i need to correct so thanks for that. so will this work??--

      for(auto itd = defence.begin(), ita = attack.begin(); itd != defence.end(); itd++, ita++){
          if(*itd < 0){
              died++;
              auto prev_itd = itd - 1;
              auto prev_ita = ita - 1;
              defence.erase(itd);
              attack.erase(ita);
              itd = prev_itd;
              ita = prev_ita;
          }
      }
      
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        6 months ago, # ^ |
        Rev. 2   Vote: I like it +8 Vote: I do not like it

        I think the following will be better:

        for(auto itd = defence.begin(), ita = attack.begin(); itd != defence.end(); ){
            if(*itd < 0){
                died++;
                auto tmp_itd = itd;
                auto tmp_ita = ita;
                defence.erase(itd);
                attack.erase(ita);
                itd = tmp_itd;
                ita = tmp_ita;
            } else {
                itd++;
                ita++;
            }
        }
        

        Your updated code still have some problem. If the itd points the first element of defence, you cannot to set the prev_itd to itd - 1.

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6 months ago, # |
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Problem C.

Solution
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6 months ago, # |
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    6 months ago, # ^ |
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    Because it is O(n^2)
    Think what happens if no one ever dies

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    6 months ago, # ^ |
    Rev. 5   Vote: I like it 0 Vote: I do not like it

    The time complexities of your two codes in the worst condition (i.e. No one dies) are both $$$O(Tn^2\log n)$$$. It will get TLE obviously. To improve, you can check my submission. We should avoid useless check of whether a monster dies when its left and right neighbours don't die.

    Spoiler
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6 months ago, # |
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When does the system testing start?

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6 months ago, # |
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Video solutions for A-C, D-E

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6 months ago, # |
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Really wondering if this solution for F is hackable (in both idea and execution time)? I know it had to be DP but the presented idea seems pretty messy and duct-taped, would actually be open to some revelations, and the complexity seems to be a pretty high-constant-factor $$$O \left( n^3 x \right)$$$.

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6 months ago, # |
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Finally Expert !!!

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6 months ago, # |
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Strange, I got TLE for my submission for B despite it being O(nlogn) in worst case.

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6 months ago, # |
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242432261 Can anyone explain why this code is giving WA on testcase 5

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6 months ago, # |
  Vote: I like it 0 Vote: I do not like it

can anuone help me. I have written the code for problem D with linkedlist and its working properly for test1. But, I am getting runtime error: "-1073741819 exit code" for test2 . what should i do.

The test case details are in https://mirror.codeforces.com/contest/1922/submission/242434985

my code:


// object of linked list class obj{ public: int a, d; // attack and defend values obj* next{nullptr}; obj* prev{nullptr}; //next and previous monsters obj(int at, int de, obj* p, obj* n=nullptr){ a=at, d=de; prev=p;next=nullptr; } }; bool check(obj* curr){ // checks if current monster will be deleted if(curr->prev==nullptr && curr->next==nullptr) return false; else if(curr->prev==nullptr){ if(curr->d<curr->next->a) return true; else return true;} else if(curr->next==nullptr){ if(curr->d < curr->prev->a) return true; else return false; }else{ if(curr->d < curr->prev->a + curr->next->a) return true; return false; } } int main(int argc, char const *argv[]) { int T; cin>>T; while(T--){ int n;cin>>n; vector<int> A(n, 0), D(n, 0); // A for attack and D for defend for(int i = 0;i<n;i++) cin>>A[i]; for(int i = 0;i<n;i++) cin>>D[i]; obj* head = new obj(A[0], D[0], nullptr); obj* l = head; for(int i = 1;i<n;i++){ l->next=new obj(A[i], D[i], l); l=l->next; } deque<obj*> next_del;// contains all monsters that are going to be deleted in next round for(auto t = head; t!=nullptr; t=t->next){ if(t->prev==nullptr && t->next==nullptr){ }else if(t->prev==nullptr){ if(t->d < t->next->a) next_del.push_back(t); }else if(t->next==nullptr){ if(t->d < t->prev->a) next_del.push_back(t); }else{ if(t->d < t->prev->a + t->next->a) next_del.push_back(t); } } vector<int> ans; while(next_del.size()){ ans.push_back(next_del.size()); set<obj*> altered; // contains all monsters that needs to be checked. while(next_del.size()){ obj* front = next_del.front(); next_del.pop_front(); obj* prev = front->prev; obj* next = front->next; while(next_del.size() && next_del.front()==next){ next_del.pop_front(); next=next->next; } for(obj* ob = front;ob && ob!= next;){ obj* ppp = ob; if(ob)ob=ob->next; if(ppp) delete ppp; } if(prev) altered.insert(prev); if(next) altered.insert(next); if(prev) prev->next=next; if(next) next->prev=prev; } next_del.clear(); for(auto &i: altered){ if(check(i)) next_del.push_back(i); } } while(ans.size()<n) ans.push_back(0); for(auto &i: ans) cout<<i<<" "; cout<<"\n"; } }
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    6 months ago, # ^ |
      Vote: I like it +8 Vote: I do not like it

    Try not to use pointers. In CP, using arrays prev and next will be better than using pointers. You can check my submission to see how to use arrays instead of pointers..

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    6 months ago, # ^ |
      Vote: I like it +8 Vote: I do not like it

    I found where's wrong in your code now. Here's the updated version.

    There are two modifies. The first is to remove the block:

    while(next_del.size() && next_del.front()==next){
        next_del.pop_front(); next=next->next;
    }
    for(obj* ob = front;ob && ob!= next;){
        obj* ppp = ob;
        if(ob)ob=ob->next; if(ppp) delete ppp; 
    }
    

    and add the following:

    front->prev=nullptr;
    front->next=nullptr;
    

    The second is correcting the check function. If curr->prev==nullptr satisfies, we should execute if(curr->d<curr->next->a) return true; else return false; but not if(curr->d<curr->next->a) return true; else return true;. (Pay attention to the last word)

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6 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Auto comment: topic has been updated by awoo (previous revision, new revision, compare).

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6 months ago, # |
  Vote: I like it 0 Vote: I do not like it
void solve() {
    ll n;
    cin>>n;
    vector<ll> arr(n+1);
    for(ll i=1;i<=n;i++){
        cin>>arr[i];
    }
    vector<ll> close(n+1);
    close[1]=2;
    close[n]=n-1;
    for(ll i=2; i<n; i++){
        if(abs(arr[i]-arr[i-1])<abs(arr[i]-arr[i+1])){
            close[i]=i-1;
        }
        else{
            close[i]=i+1;
        }
    }

    vector<ll> dist;
    for(ll i=1;i<n;i++){
        if(close[i]==i+1){
            dist.pb(1);
        }
        else{
            dist.pb(abs(arr[i]-arr[i+1]));
        }
    }

    vector<ll> backDist;
    for(ll i=n;i>1;i--){
        if(close[i]==i-1){
            backDist.pb(1);
        }
        else{
            backDist.pb(abs(arr[i]-arr[i-1]));
        }
    }

    reverse(backDist.begin(),backDist.end());

    vector<ll> prefixSum(dist.size()+1);
    vector<ll> suffixSum(dist.size()+1);

    prefixSum[0]=0;
    prefixSum[1]=dist[0];

    suffixSum[dist.size()]=0;
    suffixSum[dist.size()-1]=backDist[dist.size()-1];




    for(ll i=2;i<=dist.size();i++){
        prefixSum[i]=prefixSum[i-1]+dist[i-1];
    }

    for(ll i=dist.size()-2;i>=0;i--){
        suffixSum[i]=suffixSum[i+1]+backDist[i];
    }

    ll q;
    cin>>q;
    vector<ll> res;
    while(q--){
        ll x,y;
        cin>>x>>y;
        if(x<y){
            ll ans = prefixSum[y-1]-prefixSum[x-1];
            res.pb(ans);
        }
        else{
            ll ans = suffixSum[y-1]-suffixSum[x-1];
            res.pb(ans);
        }
    }
    for(ll i=0;i<res.size();i++){
        cout<<res[i]<<endl;
    }

}

Can anyone help me find why this code is giving runtime error in testcase-2

Problem-C

PLease help !!!

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    6 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    When $$$n=2$$$, dist only contains 1 element. Then, prefixSum[1]=dist[0]; and suffixSum[dist.size()-1]=backDist[dist.size()-1]; will cause an RE.

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      6 months ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      when n=2

      suffixSum[dist.size()-1] = backDist[dist.size()-1];
      

      that is

      suffixSum[0]=backDist[0]
      

      i dont get how it will cause RE can you explain a little

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        6 months ago, # ^ |
        Rev. 4   Vote: I like it 0 Vote: I do not like it

        My mistake. Actually it's here:

        for(ll i=dist.size()-2;i>=0;i--){
            suffixSum[i]=suffixSum[i+1]+backDist[i];
        }
        

        dist.size() will return a size_t (a.k.a unsigned long long or unsigned long, depend on your machine and compiler) type. Then, minus it with $$$2$$$, we'll get a very large number ($$$18446744073709551615$$$ or $$$4294967295$$$). Then let i be that and try visit suffixSum[i]. It will cause the RE here.

        To correct, just change dist.size()-2 to (int)dist.size()-2. This convert the unsigned long long to int.

        I've modified your solution and it's AC now.

        BTW it's very weird that you've submitted an AC solution but then submitted an RE one! And that two codes are so similar.

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          6 months ago, # ^ |
            Vote: I like it 0 Vote: I do not like it

          Thanks ohhh i was trying to figure out why my code is giving RE that's why i was trying to submit different codes

          just got a little pissed as i wasn't able to submit in the contest

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6 months ago, # |
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For problem C, I can't understand I just can get AC in contest even though I finish n queries in each test case rather than m queries :(

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    6 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    That's because the pretests are too weak. In fact, in this round, C's pretests are the weakest one as there's more than 100 hacks about problem C.

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6 months ago, # |
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Problem A Statement Was Too Confusing

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6 months ago, # |
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downvote, if you liked editorial

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6 months ago, # |
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Good luck to everyone!

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6 months ago, # |
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Why do you only takes 3 problems to the problem list?

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6 months ago, # |
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W