I will be live with post contest discussion stream here

As an sc3developer, hope my computer doesn't fall into a bottomless pit this round

wow!

Why not "6 or 7 or 67 problems" this time?

Good luck to all participants!

Time to get humbled by c again. -:lol:-

How many questions should I aim for in a div 2 contest to reach pupil. In a 2 hour, 3 hour long contest, does the number or timing expected from one change?

Ahh another edu... goodbye purple

.

yay, another contest! lets have fun!!

Hello,

This will be my first CF contest, i hit usaco plat in march and wanted to do this for fun to see how it is in comparison

any tips?

Fight for Master go go go

yes, good luck to all participants!

I hope A and B are easy like old problems, there is no point of making A and B harder but still solvable by AI

Thanks for the round! Good luck, everyone!

Good luck!!

Good luck!

I love problem C.

seven submissions for last problem

As a graytist (graytist != ratist btw), nooo I'm getting back to newbies after this round.

Can F be solved with persistent treaps?

$$$A, B, C$$$ were kinda standard, as expected for a Educational round. But $$$D$$$ was a good problem!

D is a cool problem, sad that I solved it 3 minutes before the contest ended

.

so close to solve c, wasted a lot of time on b, i dont know why :(

D is a nice DP problem!

Damn, i bugged my solution for C but still, great contest!

First time I saw where we actually do the operation mentioned in the problem (B) and it works, otherwise mostly its some mathematical equation which solves the problem in O(1) or log. C seems out of my way, need to upsolve now, but loved the problems.

got 3 minute late to submit E!

B was a disaster

I thought my solution for D would be fast enough but it TLEs and I'm not sure why :(

My idea is that it doesn't matter in which order you apply the merges, and every merge can be represented by a mask. So for example if a = 12345 and mask = 1101 then we would get 69.

Then I loop over all masks for a and b, which should be small amount, and check if merge(a) == merge(b), and track the longest merge.

I thought I might've been too slow because I was working with strings when merging, but even after changing to working with ints I TLE on test case 10. Is it possible to make my solution work, or is this approach completely wrong? :P thanks for reading

wasted 1 hour thinking constraint was 1e5 for both the strings in D :((

Just wondering about C, there is any solution for which having a sorted array makes things any easier? My solution works by generalizing contiguos equal numbers as blocks so i really don't use that sorted fact at all, i wonder if i've made things more complicated by not using that fact

The problem F made me get Master!

I felt like the first four questions in this match are too easy. I solved them in 40 minutes. They don't seem like Div.2 level difficulty.

And among them, B is much harder than the other three.

Editorial of D+E

https://codeforces.com/contest/2242/submission/381527708 D. Say after operations the first digit is merged from A[i1:i2], second digit from A[i2:i3], etc. and similarly for B[j1:j2], ... Then if P are the prefix sums of A mod 10, and Q of B mod 10, we want P[i1] = Q[j1], P[i2] = Q[j2], etc. This is simply the longest common subsequence of P and Q, which is a known dp.

https://codeforces.com/contest/2242/submission/381527102 E. Let P = 2**k the highest power of 2 <= R. If we can do y = P, we should and x = max(L, P/2). This is because the ones digits of y are multiples of k+1, which is 1, 2, 3, ... modulo k, which is the same order as x. Therefore we just want the smallest value of x that is one less digit (bit-length).

If we can't, then x and y have the same bit-length. Every 1 in the common prefix of L and R must be on x and y, and the rest of the bits will be zero. To attain this, choose x = (common prefix) + 01111.... and y = (common prefix) + 10000....

Screencast with commentary

Problem B can be solved with the same idea as this problem: https://codeforces.com/problemset/problem/2103/C

Q:D I solved this problem by generating all possible strings using DFS. For every string, I merge one adjacent pair of digits ("i" and "i+1") into (a[i] + a[i+1]) % 10, remove leading zeros, store the new string in a map (to avoid duplicates), and push it into a stack to continue generating all possibilities. I repeat the same process for both input strings and then find the longest common generated string. Is this approach correct? If not, where is the flaw in my logic or implementation?

```cpp void solve() { stack st; string a, b; cin >> a >> b;

st.push(a);
unordered_map<string, int> mpp;
mpp[a] = 1;

while (!st.empty())
{
    string tt = st.top();
    st.pop();

    int n = tt.size();
    for (int i = 0; i < n - 1; i++)
    {
        int val1 = tt[i] - '0';
        int val2 = tt[i + 1] - '0';
        int f = (val1 + val2) % 10;

        string str = tt.substr(0, i);
        str += char(f + '0');
        str += tt.substr(i + 2);

        int pos = str.find_first_not_of('0');
        if (pos == string::npos)
            continue;

        str = str.substr(pos);

        if (!mpp.count(str))
        {
            mpp[str] = 1;
            st.push(str);
        }
    }
}

stack<string> st2;
st2.push(b);
unordered_map<string, int> mpp2;
mpp2[b] = 1;

while (!st2.empty())
{
    string tt = st2.top();
    st2.pop();

    int n = tt.size();
    for (int i = 0; i < n - 1; i++)
    {
        int val1 = tt[i] - '0';
        int val2 = tt[i + 1] - '0';
        int f = (val1 + val2) % 10;

        string str = tt.substr(0, i);
        str += char(f + '0');
        str += tt.substr(i + 2);

        int pos = str.find_first_not_of('0');
        if (pos == string::npos)
            continue;

        str = str.substr(pos);

        if (!mpp2.count(str))
        {
            mpp2[str] = 1;
            st2.push(str);
        }
    }
}

int maxi = INT_MIN;

for (auto &x : mpp2)
{
    string s = x.first;
    if (mpp.count(s))
    {
        maxi = max(maxi, (int)s.size());
    }
}

if (maxi == INT_MIN)
    cout << "-1\n";
else
    cout << maxi << "\n";

} ```

I have a question regarding the last problem.

So it's established by now that when $$$n\leq 100\,000$$$, a quadratic solution exists. My solution was essentially "go right to left, maintain the function $$$x\mapsto$$$ how much we'll have in the end if we start with $$$x$$$, we only need to know its values for $$$x\leq 2n$$$, we'll have to do like 3 memcpys every time". But I couldn't squeeze those memcpy(to, from, cnt * sizeof(int)) (it took about 1.6s on my maxtest in custom invocations). I managed to squeeze this solution by making each function value take 3 bytes instead of 4, so that we memcpy less stuff, but do people maybe know how to memcpy faster? I found some like intrinsics and stuff blog from @sslotin, is it the best known way to memcpy?

Perfect, it is my first time to join it.I feel funny and I enjoy to solve it

plz release the tutorials soon

Amazing round

How can I become a trusted participant?

for those who are ranting about B it was such an easy greedy just get smallest 1st and 2nd part but yeah remember while getting 1st part if you have more 1s then 2 and 3 and the next no. is 3 inlcude it in your set of 1st part(this was the only idea that was to be thought) ~~~~~

include <bits/stdc++.h>

using namespace std; int main(){ int t; cin >> t; while(t--){ int n; cin >> n; vector inp(n); for(int &x:inp) cin >> x; int cn1=0,cn2=0,cn3=0; int prt1=-1,prt2=-1; for(int i=0;i<n;i++){ if(inp[i]==1) cn1++; else if(inp[i]==2) cn2++; else cn3++; if(cn1>cn2+cn3 && i<n-1){ if(inp[i+1]==3){ prt1=i+1; break; }} if(cn1>=cn2+cn3 && cn1>0){ prt1=i; break; }} cn1=0; cn2=0; cn3=0; for(int i=prt1+1;i<n;i++){ if(inp[i]==1) cn1++; else if(inp[i]==2) cn2++; else cn3++; if(cn1+cn2>=cn3 && (cn1>0 || cn2>0)){ prt2=i; break; } } if(prt1>=0 && prt2>=0 && prt2<n-1) cout << "YES" << '\n'; else cout << "NO" << '\n'; }} ~~~~~

Problem C was interesting. Couldn't solve D, E or F, though. This CF website looks awesome though, just tried it a few days ago

Another FST T_T, why me (:)|( ಥ_ಥ

When will the editorial get posted? My solution got TLE for $$$F$$$ and I'm so interested to see and read the solution for it...

I am new to this, But this contest was rated right...?

about how long does it take for Ratings to come out?

Is there an issue with rating changes? I think unofficial participants got rated (for example, current div. 2 first place). To be honest, I'm biased since I'm second now but the point stands. Or maybe zhaoyuebo (current P1) registered as rated but is a new account so not counted as trusted but still able to get rating?

Q:C Input: 4 5 1 1 2 2 Output: 0

If I mark and remove the elements at indices 0 and 2, the resulting array becomes [1, 2]. According to the question, after applying the operation, all marks are reset. Then, the element at index 0 is marked automatically. I duplicate all the marked elements, so the array becomes [1, 1, 2]. Again, the element at index 0 is marked automatically, and I repeat the same operation. Continuing this process, the final array becomes [1, 1, 1, 1, 2]-->> according to me output is 1

Am I the only weird one who thought that in D O(NM) would TL because (10^3*5)^2 = 10^7*2.5 and was looking only for linear or smth(that doesn't exist ig)? I hoped there would be more losers like me here at least.

Regarding the plagiarism flag for submission 381498284 (Problem 2242E)

I am writing to clarify the system flag regarding my solution for Problem E, which matches other participants. I want to state clearly that I authored this code completely independently from scratch.

During the round, I used ideone.com to compile and test my logic, but I inadvertently left the visibility settings on public. It appears my code was scraped or copied by others due to this unintentional leakage.

While I understand this is a security oversight on my part, I did not intentionally share or cheat. I apologize for the careless slip, and I have already updated my online IDE settings to ensure all future code remains strictly private. I kindly request the coordinators to take this timeline into account.

Appeal regarding automated plagiarism flag for Problem 2242D

My submission [381510406] was flagged for similarity with users Dustu_103 and Debajyoti23je0291.

We are part of the same university group and use an identical base template for fast I/O, macros, and standard shortcut definitions. Because the core logic for problem 2242D is relatively concise, the structural similarity is entirely due to our shared template setup rather than any code sharing or collaboration during the live round.