Just curious, are the prizes based on full ranking or just official/Div.2 ranking only?

I think this contest will be so interesting and exciting! Thanks in advance to them who organizated this contest, and of course thanks to company NEAR for supporting this contest!

Could someone please explain what's Ⓝ in the prizes?

Is anyone going to pull a rainboy?

Hello everyone, hope to reach CM in this round

As a tester vodacbaoan is genius

December 2023: "AlphaCode 2 is worth a gazillion specialists, competitive programming is done for!"

Fast-forward to present day six months later: "Sup nerds, wanna annotate some problems?"

This round will be rated for all participants with a rating of 2100 or below.

I am 2100 and I can’t register officially :(

Hey Authors

I'm curious to know all your ages/experience? since it's great to find people with 2-3 years of experience building contests!!

As a tester problems are really interesting and challenging!

As a tester, I literally tested just now because the coordinator invited me to test a few hours ago.

I will come back after a long time not participating in contests

OMG! Another FelixArg round.

The expression on the face of boys explain how dangerous their mind is.

What is meant by blue testing, purple testing, etc...?

i remember you guys removed 24-word seed restore in the default web wallet and my near coins were gone, very funny. then someone told me i have to paste it in the url to restore it. please be careful about backwards-compatibility, would hate for the new ai to suddenly vanish. looking forward to nearcrowd round 2!

weird score distribution, I've never seen 1000 point div2c

thinking speedforce, unless upsolve D of course.

64 random participants from the top 4096 in the overall ranking will each receive Ⓝ 4, that means we have a 1.5625% chances of becoming rich! All the Best!!

interesting score ditribution

Can someone help me ? How am i supposed to change my compiler from c++14 to c++20 (in my own laptop) ?

is there any penalty?

Contest is NEAR!

Me after solving D : (^_^) Checking standings afterwards : (o_o)

nice tasks, 2k submissions on D wow

B > C

Animated Video editorial for problems [A-C]:

https://youtu.be/-btzjr-u4dM

problem D must have something to do with 2D prefix and maybe some gcd yes? I believe

C<<<B

Can someone tell how to solve D?

My thought process for D : Find net difference between two types of mountains, this will give the value to increase. Now in each kxk submatrix, find difference between the number of two types of mountains : this will give the amount we can increase or decrease. And then I couldn't proceed further.

How to do B?? Spent more than an hour and couldn't solve it.

So you've really decided to use the $$$3 \cdot n + 1$$$ problem in a contest? I'm really salty right now if you didn't notice.

even though i couldn't solve it, i feel that C is much easier than B. From the One's I've read, A-C. I can say that the problems were very interesting and I think i would enjoying upsolving them.

B was very hard for people who aren't that good at math. :(

Why are all the A's getting hacked..?

Thanks for the contest. I got my lesson. I wont give any contest from now on which has the word TrAkToR in its description.

Why so many hacks on A? (observing those on page 1 of standings)

A: If the sign of (x-y) changed, then (x-y) must be 0 at some time, so answer is NO. Otherwise, the answer is YES.

B: If x<y, then the value of x will change among {1, 2, ..., y-1} periodically, so the answer will depend on k%(y-1). Otherwise, after (y-x%y) operations x will become (x+y-x%y)/y, which is not greater than (x/y)+1. Because y>=2 this can happen not more than log(x) times until x<y, so we can simulate the process in O(log(x)).

C: Let dp[i] be the anwser for first to i-th cards. We can binary search for the largest j such that sum(j+1, i)>=l, if sum(j+1, i)<=r we have dp[i]=max(dp[i-1], dp[j]+1), otherwise dp[i]=dp[i-1].

D: Let the balance of heights B=(sum of snowy heights)-(sum of not-snowy heights). If B==0 initially the answer is YES. If we let c[i][j]=1 when (i, j) is snowy, c[i][j]=-1 otherwise, then when we do operation on a k*k square, the change of balance will be sum(i,j)(c[i][j]) where (i,j) is in the square. So we can calculate the sum of c[i][j] over all k*k submatrices using 2D prefix-sum, and by the bezout's theorem, the change of balance can be any multiple of GCD of these sums. In order to make B=0, the initial value of B must be the multiple of GCD.

E: First think about case for n=2**d. If k>=d the answer is n*(n+1)/2, and if k==d-1 the answer is (n-1)*n/2, let's assume k<=d-2. We can see the answer is sum(1<=r<=k+1)((2**r-1)*(2**r)/2 * C(d-r-1,k+1-r)). Denote this as f(d, k).

Now think about arbitrary n. We can split [0, n) into several intervals with length of 2**j, where n has the j-th bit. If we let L=floor(n/2**(j+1))*(2**(j+1)), R=L+2**j, the answer on interval [L, R) will be f(j, k-popcount(L)). If popcount(R-1)>k, there will be no valid interval containing both R-1 and R, so we can add this to the answer and consider for [R, n) seperately. Otherwise, when popcount(R-1)<=k, we can see that for any R<=x<n we have popcount(x)<=k, so we can add (n-L)*(n-L+1)/2 to the answer.

F: Let L be the minimum i such that a[i]>a[i+1], R be the maximum i such that a[i-1]>a[i]. Then at least we need to sort subarray [L, R]. Let X=min(L<=i<=R)(a[i]), Y=max(L<=i<=R)(a[i]), if there are any L1<L such that a[L1]>X or any R1>R such that a[R1]<Y, then we also need to sort them. To find L,R we can maintain a set S={1<=i<n: a[i]>a[i-1]} which can be modified in O(log(n)) each operation. To find X,Y we can use a segment tree, and to find L1, R1 we can use binary search on a[i].

Who can prove D?

Wrote a overcomplicated DP + Segment Tree solution for C. D was a good problem.

In fact, I think that question C is much less difficult than question B, and I was thwarted by question B for forty minutes but passed C in ten minutes.

Can anyone teach me how to past B problem?

Lol, i just read F and isn't it just much much easier, than D and E? It took me like five minutes to come up with the solution.

Can anybody please help me understand the problem with my code for problem C(Boring days) ~~~~~

signed main() { ll t; cin >> t; while(t--) { ll n,l,r; cin >> n >> l >> r;

vector<ll> a;
  for(ll i=0;i<n;i++) {
    ll x;
    cin >> x;
    a.push_back(x);
  }

  ll i = 0;
  ll j = 0;
  ll sum = 0;
  ll count = 0;

  while(j<n){
    sum+=a[j];
    if(sum<l){
      j++;
    }
    else if(sum<=r and sum>=l){
      count+=1;
      i = j+1;
      j++;
      sum = 0;
    }
    else if(sum>r){
      while(sum>r){
        sum-=a[i];
        i++;
      }
      if(sum<=r and sum>=l){
        count+=1;
        i = j+1;
        j++;
        sum = 0;
      }
      else{
        i = j+1;
        j++;
        sum = 0;
      }
    }


  }

  cout << count << endl;

}

}

~~~~~