When I first saw this problem, I felt that this problem is somewhat similar to 1068 B. Niko's Tactical Cards in which DP was used , as I haven't studied DP I was unable to solve it .

But After the contest I learn't the basic trick that first we define states , we apply operation in every state and then take min or max as the problem tells and then define the next state for further interations .

Now , I applied the same logic during solving the online mirror of 2025-2026 ICPC, NERC, Northern Eurasia Finals Problem M. Medical Parity which gave me AC 353928464 .

Similarly , I had the same intuition for going for this trick in yesterday's Good Bye 2025 C problem as there were two operations held at position 1 and 2 so I tried to define two states .

1st state — it tried to maximise the outcome of the leftmost operation by taking the best X of 1st state so far , then applying a left operation on it and a right operation on it , then reseting it to the best of the left operation of 1st state and best left operation from the 2nd state .

2nd state — it tried to maximise the outcome of the rightmost operation by taking the best X of 2st state so far , then applying a left operation on it and a right operation on it , then reseting it to the best of the right operation of 1st state and best right operation from the 2nd state .

I did a dry run and it It perfectly worked on pretests 1 but somehow fails on pretests 2 . Is there a flaw in my logic or did someone implemented a similar approach and got AC . Can anyone Help ..

Submission — 355435711

vector<int> a(n) ;
for(int i = 0 ; i < n ; i++)cin >> a[i] ;
  
vector<int> l = {0 , 0 , 1} , r = {0 , 0 , 1} ; //l , r = {best X , cur left pointer , cur right pointer}
vector<int> ll(3) , lr(3) , rl(3) , rr(3) ;

while(l[2] < n)
{
  ll[0] = l[0] + a[l[1]] ; 
  ll[1] = l[2] ;
  ll[2] = l[2] + 1 ;

  lr[0] = l[0] - a[l[2]] ;
  lr[1] = l[1] ;
  lr[2] = l[2] + 1 ;

  rl[0] = r[0] + a[r[1]] ;
  rl[1] = r[2] ;
  rl[2] = r[2] + 1 ;

  rr[0] = r[0] - a[r[2]] ;
  rr[1] = r[1] ;
  rr[2] = r[2] + 1 ;

  if(ll[0] > rl[0])l = ll ;
  else if(ll[0] == rl[0])
  {
    if(a[ll[1]] >= a[rl[1]]) l = ll ;
    else l = rl ;
  }
  else l = rl ;

  if(lr[0] > rr[0])r = lr ;
  else if(lr[0] == rr[0])
  {
    if(a[lr[1]] >= a[rr[1]])r = lr ;
    else r = rr ;
  }
  else r = rr ;
}

  cout << max(l[0] , r[0]) << endl ;

Plus , It's my first blog :D