" This ensures that any two vertices of the same color have a difference that is a multiple of 4, so their XOR is a multiple of 4"

Can anyone explain for me this line

First, thank for very good contest and fast editorial I am going to share my O(n) solution for D We can notice that in optimal solution, monster in node v can be kill in at mode |v| round (|v| is number of node have the same edge with D, round number from 0) So we can call dp[v][round]: minimum number of health decreases when kill monster in node v after "round" dp[v][round]= ∑_(u ∈chill of v) (min)⁡(dp[u][j]) (j ≠round) (sorry, i don't know how to write beautiful fomular)

we can use pref_min and suf_min to quickly calculate Node that we must update parent value from chill due to tle Here is my code 270747003

Your contest is very good. But what i do when i have stuck (except ask other, because is take much time), I don't find any editorial or can view code of other. Can you public your code for each problem and few hint?

In problem D, it seems that there is no answer for n which is even, can anyone prove this?

There are a lot of online problems, so inventing new problems without duplication is very hard for setter

And sometimes "homework" is harder than the actual problem, so we just "prove it by AC"

I agree. Some people will use two accounts, and submit in the account which uses the unrate option until it has AC. Then change some code (like the variable name), and submit it to the main account.

How can I do research like this for my account (i.e what code u use to make this)

Can u add your code following this explanation?

problems about mex is quite fun. That contest is so hard but I have learned so much from it. Thank you to take time for preparing such a contest

If max(a)>max(b), MAX=max(a). If min(a)=max(a), then min(a)>max(b), Jellyfish will do nothing. Otherwise, Jellyfish won't swap the apple with value MAX. In the tutorial for 1874A — Jellyfish and Game, can anyone explain to me what Otherwise means? Is do nothing and won't swap is different?

Do anyone know 42th subtest of test 2 of problem D. I can not pass this anyway

It is the greatest contest I have ever seen

I don't know how to explain this clearly. But the key is the minus and the add extend in a different direction

Sorry for my mistake. I had to fix it and now this code uses only 156KB of memory 220627524

for some x, say 19 whose binary representation is 10011 = 2^4 + 2^1 + 2^0. You can see that 2^1 divide 2^4, 2^0 divide 2^4, 2^0 divide 2^4. So the answer is you just minus the rightmost bit to x, in this case, is 2^0 and 2^1 until x has only one bit in binary representation.

P/S: Sorry for my bad English. Hope you understand this explanation

Another implement for D whose time complexity is O(n^2) but only O(n) space 220627524. The logic in this code may cause you confused, in this case, ask me(and vote)

The word "graph" appears only once in this tutorial, for 9 problem

about x13837 participants can solve C but only x5448 participants can solve B

Great idea. You can use setw() to make it more beautiful like this code

void print() {
    int layer = n; // n smallest power of 2 that great than or equal to the size of seg_tree
    for (int i = 0; i < 2*n - 1; i++) {
        cout << setw(layer) << seg_tree[i] << " ";
        if (__builtin_popcount(i + 2) == 1) {
            layer /= 2;
            cout << "\n";
        }

    }
    cout << "\n";
}