Hope everyone success! ~

i'm so excited to join my first educational Round wish me Luck Guys

sunday cf rounds (●'◡'●)

Contest week <3

I'm so excited to join this educational Round. Codeforces has been an instrumental platform for honing our coding skills and tackling intriguing algorithmic challenges. Among its diverse range of contests, the educational rounds stand out as a golden opportunity for learners and enthusiasts to delve into the depths of algorithms, data structures, and problem-solving techniques.

Wish everyone successfully take part of this round!

CM Time?

Not once did I get a good score in EDU :(

Hi, is it me or is problem 1821E opening in a PDF? Last time I checked, it was opening normally. Problems D and F in the same contest also open normally. Same thing with 1741F. Did something change?

I've have just started giving contests and my rating is around 600. Should I take part in this ? anyone ?

Hope I can learn something new from this round, good luck everyone ^^ (do not downvote me pls??)

3 contests in 3 consecutive days, Amazing!

I'm used to losing my rating in the edu round, come on, there's nothing to be afraid of anymore.(｀□′)╯┴┴ ┴—┴

$$$998244353$$$-forces.

Another horribly written Edu round.

Thanks for nice problems. Enjoyed solving D. Good one.

D was pretty good, idk how this many people solved it

D was pretty good, idk how this many people solved it

Maybe good problems, but with trash samples.

Passed D but failed to debug C T_T

What is the 4th test case of problem E?

Why E WA test 4??

How to solve D

The logic for problem B? Thanks in advance.

Had the correct solution for B, but spent literally an hour on a "wrong" answer because I used 0 instead of 0ll. RIP.

what is the approach for D ?

I hate C!!!!!!!!!

D was just a modified version of an existing problem with just a few changes needed while implementing it.

what was test 4 of E

Anyone else got wrong ans on test case 2 for C?

C seemed pretty easy at first but after getting 3 WA and getting AC finally I realized that I was right

What's wrong with #224889872:

    int suma=0,sumb=0;
    for (int i=0;i<n;i++) {
        cin >> a[i];
        suma+=a[i];
    }
    for (int i=0;i<n;i++) {
        cin >> b[i];
        sumb+=b[i];
    }
    sort(a.begin(),a.end());
    sort(b.begin(),b.end());
    cout << min(n*a[0]+sumb,n*b[0]+suma) << endl;
}
return 0;

}

Sucks to have had the answer to problem E but failed because my outputs didnt flush properly :( edit, nvm test 4 was a coutnerexample

Maybe there are too many details in E :(

Any hint on problem F? I did a naive sqrt solution and that TLE'd, Im guessing there's some sort of heuristic involved in cutting some possibilities away?

E is bad

i gonna say one thing C test 2

Link to the streaming of today's contest solutions anyone?

why can't u spare 1024MB for F QAQ.I was suffering with my solution with a memory complexity of $$$O(n\sqrt {n})$$$

I think E is not good, it makes us crazy. My classmates is shouting now.

A: If e[i]>=e[1], in order to keep the 1st player to win, we need to set w>=s[i]+1. Therefore, we need to set w=max(i:i>1, e[i]>=e[1])(s[i]+1), and if w>s[1] this value is invalid.

B: We need to choose the cell with minimum cost from every column or from every row. If we choose for rows, the answer is sum(a[i])+n*min(b[i]). If we choose for columns, the answer is sum(b[i])+n*min(a[i]).

C: For each maximal substring with same character of length L, we need to do L-1 operations. So the minimum number of operations is sum(L-1). For each substring we have L choices of the remaining char, and after choosing remaining elements, we have (sum(L-1))! ways to remove other chars. So the answer is prod(L)*(sum(L-1))!.

D: We can calculate the contribution from each bit separately, and now assume a[i]=0 or 1. Therefore, If we let sum[i]=a[1]^a[2]^...^a[i], the answer will be sum(0<=L<R<=n,sum[L]!=sum[R])(R-L), we can get it by simple dp.

E: The maximum number of k is 3. Proof: if we let color[i]=depth[i]%3+1, then for each node, the color of edge to the parent is different from all edges to the childs, and we can identify them by (color[edge to parent]+1)%3=color[edge to child]. Therefore, we only need to check if k<=2. First k==1 is equivalent to parent[i]=1, 2<=i<=n. Otherwise, we can observe that the color of edge to parent must be different from edge to childs, so we have color[parent[i]]=3-color[i] if i>=2. So we can do bipartite coloring for the tree (except node 1), and for i where deg[i]!=2, we can identify the edge to parent (since the count of color[i] will be 1, and the count of 3-color[i] will not be 1). But for i where deg[i]==2, we can't identify edges up and down by just count of colors. So we need to let color[i] be the same for all deg[i]==2. Therefore, we can assume color[i]=1 for them, and find a valid bipartite coloring for each subtree of node 1. If we find someone, we have k<=2.

On E, assigning random values to the children of the root enough times passes tests. if anyone wants to hack, should be pretty easy: https://mirror.codeforces.com/contest/1879/submission/224976624

upd: hacked

There are so many resources about the solution for D on the internet (for example). I doubt if there were any cheaters.

What I did wrong in third problem please anyone explain I have done this: first calculated factorial till 2e5 then for each consecutive characters that are same counted them and taken factorial for the same value counted and multiplied this value to original ans which i initialised by 1. Here is the code