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

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 :(

Good luck to all the participants! Codeforces Round 955 is sure to be a rewarding experience.

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!!

Dashboard with 6 problems in Div-2 is a standard.

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!

How score distribution is related to problem rating? I mean how do I predict what can be the difficulty of the problem?

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

thinking speedforce, unless upsolve D of course.

Is it different from div2?

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

it's time to be expert

speedforces incoming!

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?

33!

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

nice tasks, 2k submissions on D wow

most annoying problemset ever. coupled with prizes to increase cheating.

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.

Unable to pass the second problem

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.

I thought in A for almost 20 minutes but then I activated guessing mode and was able to figure it out in less than 10 second

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

Is C possible with monotonic queue DP that define f[i] as the answer range 1 to i? It seems that it only need a simply greedy.

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.

D was a great problem, shame I couldn't code it up in time. Need to speed up.

Wow , first time got TLE just by using a % operatoion,

int solve(int i,auto& psum,int* dp,int n,int l,int r){
    if(i>=n) return 0;
    if(dp[i]>=0) return dp[i];
    int take=psum[i]+l;
    int nextpos=lower_bound(psum.begin()+i,psum.end(),take)-psum.begin();
    int val = 1;
    if (nextpos >= n || (nextpos < n && psum[nextpos] > (psum[i]+r)))
        val = 0;
    return dp[i]=max(val+solve(nextpos+1,psum,dp,n,l,r),solve(i+1,psum,dp,n,l,r));
}


void helper(){
    int n,l,r;
    cin>>n>>l>>r;
    vi a(n);scan(a);
    vector<int> pxsum(n);
    pxsum[0]=a[0];
    for(int i=1;i<n;i++){
        pxsum[i]=pxsum[i-1]+a[i];
    }

    int dp[n];
    fill(dp,dp+n,-1);
    cout<<solve(0,pxsum,dp,n,l,r);
    
}

what is wrong in the above code for c? Wtf is wrong...

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;

}

}

~~~~~

267409135 Can anyone tell me why is this not working? Like is my logic completely wrong or is there some edge case that the code misses?

My submission Can anyone pls tell me why this fails.... It shows wrong ans on 4288 no of test case 2

267392213 Can anyone pls tell me why this fails.... It shows wrong ans on 4288 no of test case 2

Hello everyone, I need somebody to hack my friend's code: https://codeforces.net/contest/1982/submission/267413232

I am so sure that His code will overflow since he just used 32bits integer (int / i32). Hurry up! Don't let him discover it! Just hack the submission!

PS: His computer's battery has run out and I submitted his code after contest. But why it AC!!!!! I cannot accept it!!!!!

Thanks!

Update:

Sorry, his code is correct. Just sum up all the "-1" and "+1" will not exceed i32. My bad! Actually, he cannot submit his code since the battery problem, I want to easy him that his code maybe cannot pass the test (So that he will not lose this big chance to increase his rating). Okay...... Now he will be definitely very sad and regretful.

Is filling "TON wallet" field (and "Secret Code") with account id from mynearwallet in settings enough for getting reward in NEAR cryptocurrency for current round? Or what should I do to get it?

I did surprisingly well in this round. This was a really fun one, questions were not very straightforward, I liked it a lot.

Why is this contest unrated for me? Any reason codeforces for this action

In c, how it is sure that x will become 1 in less than or equal to k operations?

My rating is 1110.I participated in this contest. But my rating not changed.Why?

Thank you

Till now i was happy that i became specialist but all of a sudden ratings changes is disappeared, can i know what's the reason behind it?

Subject: Issue with Judgement in Problem C

Hi,

I encountered a possible error in the judgement of my submission for Problem C during Codeforces Round 955 (Div. 2). It seems there was a bug during the judging process.

Details:

My Submission During Contest: Submission 267358679 My Submission After Contest: Submission 267429867 During the contest, in the 2nd pretest, 6th test case, my submission was judged incorrectly. The actual output should be 0, but it was judged as 1, resulting in a wrong submission.

After the contest, I resubmitted the same solution, and it was judged correctly with the output being 0.

I would appreciate it if the issue could be reviewed and corrected as it affected my contest standing unfairly.

Thank you for your attention.

Best regards, Deshik conan45

I am awaiting to be in random 64 contestants!!!

I have one question for the cf community, why to give the score distribution beforehand, people can initially guess the difficulty of the contest, isn't it non exciting it would be more fun if we are not known that and may be we can keep it constant in every contest