great

ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤ

New: Educational Rounds on Codeforces!

You didn't even give the contest

No please god no!

you're right

Often, how much harder are Educational Rounds compared to normal rounds? I want to know whether to compete or not. This is because my last educational round gave me some serious trauma: Educational Codeforces Round 153 (Rated for Div. 2)

for pupil i think 2 is enough.

we share the same rating, I hope we share the promotion too.

Upd: Solved 3. Did C Overcomplicated :(

true !!

and you are also using an alt, so shut up, do not complain of stuff you are personally doing

What happened to "romanian diversion strat"?

thats apio2022 perm

D is a lot easier than E, E has just appeared before

Yes! Problem statements were clear and precise. But I must be blind... :(

i pre-calculate cost of go forward and backward

This problem maybe doesn't require the shortest path algorithm

C was a modified version of prefix sums

It can be shown that in order to go from point x to a point y, going back is never a good choice.

I mean yeah. Like I did too many wrong subs in A :(

I think E was way harder :(

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

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

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();
    }

bro u fell of

Still do think its overrated after writing this piece of art which even i can't read now Submission

Bro just use set

you can create a Generic function that decrease the number of those caseworking.

242254308

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.

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.

I can feel your pain :(

No, I used two std::sets as well and it passed in 700ms

How did you solve it by segment trees? Like what was each Node storing, can you please share.

I used linked list — 265 ms.

I used 2 vectors and a static linked list. It's supposed to be O(n)...

Используй long long абсолютно везде :)

Notice that this problem appeared in APIO and copy paste

maybe 10^18 gave me a hint towards set bits....

10^18 says it should be done in O(logn).

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

When will I be expert like you?

A div four F had comparably equal ACs to today's D and lower ACs than E.

no bro

Struggled with both A and B :)

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.

i was struggling with B for 1 hour cause of overflow case. t-t

Cheaters cheat what can we do :( I spent all my time on D

OH NO!

Just noticed it by reading your comment :/

Its 2^ai

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.

Same here, i too thought the same and trying to figure out the sliding window solution!

noo it was the first E i did :(

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.

I totally agree with you

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.

how do u solve for < 90? I can understand upto 118

Us

ohh yes I was thinking that too, I realised after a while..

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

nvm

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.