As a tester, Dominater069 orz.

As A participant I hope Enjoy.

hope to solve ABCD in 90 mins :)

What happened to the contest which was gonna take place tomorrow?

OMG round $$$(1111111111)_2$$$ as a tester!

Fun fact: The previous contest that had 250 points on A was Global Round 28, which was on December 2024

As a tester, I tested

As a tester, I loved the problems

250 for A and 1500 for C.

ok.. surprising that the first problem has a score of 250 .... what does this mean?

People should consider solving C earlier in this round.

Explanation: If somebody can solve A in 10 min, B in 20min, and C in 40 min; The solving order A,B,C gives (250-10)+(500-60)+(1500-420)=1760 round score. On the other hand, C,B,A gives (1500-160)+(500-120)+(250-70)=2000 round score. But, if you aren't able to solve C, this would work negatively.

One of the tester is banned now lol

finally , i'll become newbie

With this score distribution problem A is going to be cout << a+b << endl;

luuia's testing hopefully means the round is GPT proof

Yeah I'm out of competition now

So Imma prepare a template for odd-even and sum of two no.s for A.:)

RGB contest

I hope the first problem is not just "print 2^10-1"

I hope to not get minus on A

I Wish to solve D; else will get cooked..

I know there will be an unordered_map solution which will get hacked, which makes rewhile happy.

I know it from past experience 287756206

As a tester, have fun and good luck!

why is hyman00 so strong?

As a participant, I hope a non-negative delta.

how many problems do i need to solve to reach pupil in this round.Also how much time should i invest

What a hard contest!

CASE_WORK_forces and WRONG_ANSWER_forces on C for me... One day I will properly test my code before submission.

I have some comments in my code but it is not AI generate I promise.

Is there any solution for D that don't require centroid, I don't want to implement centroid so I am very curious.

Does D solution use a tree structure with heap? What an interesting problem!

I made an assumption that breaking the diameter away will always make tree less than half in size .. so that my solution will not become O(n^2) .. but I don't know the proof for this.

if this is wrong I guess I might get TLE in system tests.

I wish next time I could know more than I am cooked when I see the Time constraint is 5s for Problem D.

please tell me there is a simple trick to solve C .. I spent so much time ...

I was able to solve D faster ( although my D can fail system tests )

nice d2. solved A and C with tons of penalties :(

How to solve B?

In "C," there is no mention that the sum must be greater than -1e18.[updated]

Why does finding diameter trivial solution still pass time limit for this test???

n
1 2
1 3
1 4
...
1 n

I am Cooked.I wasn't able to solve neither A nor B. Urgently Need to practice

What is wrong in my code for B it is giving WA5

void solve(){
   int n , k;cin>>n>>k;
   vi v(n);
   int mini = 1e18 , maxi = -1e18;
   fr(i,n){
      cin>>v[i];
      mini = min(mini , v[i]);
      maxi = max(maxi , v[i]);
   }
   int tot = accumulate(v.begin() , v.end() , 0);
   sort(v.begin() , v.end());
   if( (maxi - mini) <= k){
      if(tot % 2 == 1){
         cout << "Tom\n";
      }
      else{
         cout << "Jerry\n";
      }
   }
   else if((maxi - mini) == (k+1)){
      if(v[n-2] == maxi){
         cout << "Jerry\n";
      }
      else{
         if(tot % 2 == 1){
            cout << "Tom\n";
         }
         else{
            cout << "Jerry\n";
         }
      }
   }
   else{
      cout << "Jerry\n";
   }
}

My solution for D:

For the original tree, we find an optimal diameter and remove it from the tree. Afterward, the tree is split into several connected components, and we recursively solve the problem for each component. This process is somewhat similar to the centroid decomposition of a tree. However, I believe it is essentially a "center decomposition" of the tree.

How to compute the time complexity? We only need to consider the following problem: Given a tree T, define $$$f(T)$$$ as $$$max(f(comp[i])) + 1$$$, where $$$comp[i]$$$ are the connected components obtained by removing the diameter of $$$T$$$. An important observation is that removing the diameter of $$$T$$$ will inevitably delete the center of $$$T$$$, which ensures that the diameter of $$$comp[i]$$$ must be smaller than that of $$$T$$$. From this, we can deduce that $$$f(T)$$$ is $$$O(\sqrt(n))$$$ (for a tree $$$T$$$ satisfying $$$f(T) = i$$$, the diameter of $$$T$$$ must be $$$\ge i$$$).

I believe my implementation is $$$O(n \sqrt(n) log n)$$$, and fortunately, it passed. Unfortunately, I submitted it incorrectly multiple times.

AC code:318524754

E is very cool, I like it very much! F is cool too

how did you guys do c?