127.0.0.1's blog

By 127.0.0.1, history, 9 months ago, translation, In English

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9 months ago, # |
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Thanks for the fast editorial!

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    9 months ago, # ^ |
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    I am having a little problem in D. Its working on nearly all the test cases and logic is to point. But I think there is some problem with modulus and its giving wrong answer on one test case-v 179 1000000000000000000 Can any body help I cant seem to figure it our. This is my code where maxi is modulus ll l,r; cin>>l>>r;

    ll te=l; ll ans=0; while(te<=r) { ll p=log2(te); ll up; // if(p==63) // up=r; // else up=min((ll)(pow(2,p+1)-1LL),r); //cout<<up<<endl; ll ct=0; ll x=1;

    while(x*p<=te) { x*=p; ++ct; } // cout<<te<<" "<<x<<endl; while(true) { ll nx; //cout<<x*p<<e ndl; x*=p;nx=min(x,up+1); ll nxm=nx%maxi; ll tem=te%maxi; ans=(ans+(ct*((nx-te)%maxi))%maxi)%maxi; if(nx>up) break; ct=(ct+1)%maxi; te=nx; } if(up==r) break; te=up+1;

    } cout<<ans<<endl;

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      9 months ago, # ^ |
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      I had kinda similar problems, try using unsigned long long instead if you're not using that

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9 months ago, # |
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Why the approach for F with euler tour+Lazy prop is giving tle on tc 21? here is the code-https://mirror.codeforces.com/contest/1891/submission/230589277

any help will be appreciated.

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9 months ago, # |
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I came up with the author's solution F, but I didn't have enough time to debug((

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9 months ago, # |
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F can be solved without reversing the queries. It is offline though.

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    9 months ago, # ^ |
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    It would be online if the final tree was known beforehand. Otherwise — yes, the online solution would be quite nasty with maybe some link-cut trees or treaps.

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    9 months ago, # ^ |
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    What do you mean by reversing the queries? I just run on the queries and update to 0 when adding a node

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9 months ago, # |
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What does the "transition" mean in D?

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9 months ago, # |
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For the B problem what i did was that i just added condition that any element in x should be taken one time but according to the editorial this condition can fail in some cases.So is my code correct https://mirror.codeforces.com/contest/1891/submission/230580104

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    9 months ago, # ^ |
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    But still you can have at most 30 distinct operations which can include the useless ones but it won't change the array and time complexity remains linear, so it works.

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9 months ago, # |
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achha contest

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9 months ago, # |
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nice round for a beginner like me!!

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In C problem I did as proposed in editorial, but in one test case answers differ. Here is sequence of 14 hits from my implementation:

[1, 1, 2, 5, 6, 6]
=1h small stack=1 0/6
=2h small stack=1 1/6
=4h small stack=2 2/6
=6h big stack=5 4/6
crushing 6/6
=7h -> [3, 6] combo=0
[3, 6]
=10h small stack=3 0/6
crushing 3/6
=11h -> [3] combo=0
1 stack with 3 left. hit by 1
[3]=14h

Can someone provide a sequence to win with just 13 hits?

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    9 months ago, # ^ |
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    On the final stack of 6 use the combo at the end (2 small attacks (4) -> combo(-1)) instead of doing combo(3) -> 3 small attacks(0).

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      9 months ago, # ^ |
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      Thanks. Premature use of combo is the root of evil

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    9 months ago, # ^ |
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    If you want to stick to this approach, try using std::deque, erase() from the beginning of vector is expensive operation. Or take a look how beautifully contest leaders solved it.

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9 months ago, # |
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Here is an alternative solution for F, using Fenwick Tree.

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    9 months ago, # ^ |
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    Same, I think this is more direct.

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    9 months ago, # ^ |
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    What's the logic?

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      9 months ago, # ^ |
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      I believe that my solution 230576374 is the same as in this comment, so I'll describe it.
      Let's build our tree and for each vertex save its creation time and updates with current time and value.
      What is answer for some vertex? It's sum of all requests which were made on this vertex or any of parents LATER than creation time of current vertex. So let's create BIT for sum of requests by time and dfs our tree. We perform updates before getting ans and rollback them before dfs exit thus for each vertex only relevant updates remain:

      dfs(u, p):
          for (time, value) in updates[u]:
              bit.update(time, value)
          ans[u] = bit.get(createdAt[u], n)
          for (int v: g[u]): if v != p:
              dfs(v, u)
          for (time, value) in updates[u]:
              bit.update(time, -value)
      

      Each update is made exactly twice, so it's $$$O((n + q) \log(n))$$$

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        9 months ago, # ^ |
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        Thanks for the explanation! Very creative solution. I like that you no longer need range updates, only point updates and range queries.

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9 months ago, # |
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What was the reasoning for putting problem F at that position?

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Anyone have an answer for D in python?

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9 months ago, # |
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This is probably the easiest F I've ever seen :(

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9 months ago, # |
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C can be implemented not using two pointers. But during the contest i forgot to check if n == 1 and a[i] == 1 :\

void solve() {
    scanf("%lld", &n);
    int s = 0;
    for (int i = 1; i <= n; i ++) {
        scanf("%lld", &a[i]);
        s += a[i];
    }
    if (n == 1 && a[1] == 1) {
        puts("1");
        return;
    }
    int sp = s / 2;
    sort(a + 1, a + 1 + n);
    int ss = 0, cnt = 0;
    for (int i = n; i >= 1; i --) {
        ss += a[i];
        cnt ++;
        if (ss >= sp){
            break;
        }
    }
    printf("%lld\n", cnt + s - s / 2);
}
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9 months ago, # |
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I have a problem on C . In 230619057 the code id accepted , but in 230619275 , the code is wrong . The only difference between them is merely in "lower_bound(pres,prew+n+1)" or "lower_bound(pres+1,pres+n+1)" . But the pres is the prefix sum of a which indicates that the sum divided by 2 (floor,except for n = 1) must be positive . So "+1" should not make a difference to my solution . I am quite confused . Please help we .

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    9 months ago, # ^ |
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    I reverse the order , sorry . The former one is wrong , while the latter one is correct .

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9 months ago, # |
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I think swap E and F is a good idea

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9 months ago, # |
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Can someone please tell me why is my submission for Problem D giving TLE?

Code
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    9 months ago, # ^ |
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    You recalculate intervals every time you run "func" function. Doing it once and storing it somewhere could help. My solution working as I said: 230684869

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      9 months ago, # ^ |
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      Yeah, I got that later. Even this solution is correct if I directly calculate from L to R rather 1 to R and 1 to L.

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9 months ago, # |
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This round was too easy, though people still couldn't solve problems, I don't know why. I managed to solve D in time and saw my rank go way up. A handful of 20 minutes was still left for the contest but I convinced myself I wouldn't be able to solve E. I saw E later after the contest ended and it was the easiest Division 2 E I've ever seen.

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9 months ago, # |
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Mathforces

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    9 months ago, # ^ |
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    Nope

    Spoiler
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9 months ago, # |
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why is E is much difficult than F. Or why is F much easier than E.

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9 months ago, # |
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Maybe the simplest implementation for C:

sort(a+1,a+1+n);
for(int i=1;i<=n;i++) sum[i]=sum[i-1]+a[i];
int tmp=(sum[n]+1)/2;
int c=0;
for(int i=1;i<=n;i++) if(sum[i]>tmp) c++;
printf("%lld\n",tmp+c);

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    9 months ago, # ^ |
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    can you explain the reason behind this approach?

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    9 months ago, # ^ |
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    Please explain your solution!

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      9 months ago, # ^ |
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      Lets call $$$A$$$ = total number of monsters.
      Suppose we make $$$s = ceil(A/2)$$$ moves of type-1 (killing single monsters). Then at the end we will have accumulated total power of $$$(A - s)$$$. This much power is certainly enough to kill all the remaining hordes of monsters, no matter what they might be. Making more than $$$s$$$ moves of type-1 is certainly wasteful.

      Now, it only remains to use the accumulated power, intelligently. Which is simply to kill the biggest hoardes. (since cost of killing small hordes is also $$$1$$$.)

      (Also, making lesser than $$$s$$$ moves of type-1, cannot lead to a better answer, atleast not in a single step. And if we do multiple rounds, like some recursive solution that could lead to a better solution, then the values could be simply added to the initial answer straightaway. So, there is no-real benefit of doing multiple rounds.)

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9 months ago, # |
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can anyone explain why the greedy solution works in C? also for the case when i==j and the last number is an odd number and x=0, then there is no way we can use the 2nd method on this horde

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    9 months ago, # ^ |
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    It's possible to use 2nd method when there is only 1 horde left with odd size and x = 0.

    For example left horde size = 7 and x = 0 Use 3 attacks of first type. After that horde size = 4 and x = 3. Use second type (ultimate) attack. After that horde size = 1 and x = 0. Use 1 attack of first type. After that horde size = 0.

    So it takes 5 attacks to destroy horde size including second attack. Without second attack it would take 7 attacks.

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      9 months ago, # ^ |
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      thanks, and it was my bad, I took the 2nd method as it can be used only when there are exactly x monsters left in a horde :-(

      it's also clear now why the greedy solution works

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9 months ago, # |
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I came up with using offline query in problem F. My idea is first save the query and build the completed tree which is presented in the end. Then, I just need to loop from q to 1 and update normally using Euler's tour when type 2 is meet otherwise if type 1 is the current query, output it's node value. But unfortunately, I got WA immediately at 2nd test case although this idea is kind of nature (or maybe it's wrong in some case), anyone suggests me why am i wrong here?

My submissions: 230648660

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9 months ago, # |
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can anyone help in f...getting wa on 2..230657380

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    9 months ago, # ^ |
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    are u wrong answer in 4927th or 799th

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      9 months ago, # ^ |
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      4927

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        9 months ago, # ^ |
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        It is wrong in answer of node. Node in tree is different from node in euler tour (my english is bad sry about it)

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          9 months ago, # ^ |
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          sorry for being such stupid....replaced a[i] with a[tin[i]] and accepted....thanks

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9 months ago, # |
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#

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9 months ago, # |
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Can someone give a simple idea about F? Also may I know what are the prerequisites to learn to solve problem F?

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    9 months ago, # ^ |
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    1. Queries are offline, you already know how your tree will look like
    2. Can you solve this problem if first queries are always making the tree and other queries are adding (I mean, try solve this: you have a tree and $$$q$$$ queries add some $$$value$$$ on subtree, how solve this?)
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      9 months ago, # ^ |
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      Thank you :)

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      9 months ago, # ^ |
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      I guess there are multiple ways of doing the 2. one you have mentioned. Can you suggest which method is good to opt.

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        9 months ago, # ^ |
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        You can do $$$dfs$$$ and go down from root to leafes.

        If in node $$$x$$$ you get some sum from above, you also get same sum into all nodes in $$$x$$$ subtree.

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          9 months ago, # ^ |
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          ok thank for your valuable idea.

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    9 months ago, # ^ |
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    Prerequisites: "offline" approach, DFS pre-order, fenwick tree (range add/point get interpretation)

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      9 months ago, # ^ |
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      ok thank you for your valuable reply.

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Is there any online solution for F?

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9 months ago, # |
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In problem D

and on each segment there are at most O(logn) transitions

shouldnt there be like at most 2 transitions per segment because if we make 3 transitions that would mean jump to next segment?

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My logic is similar to one in editorial but I am having issues with MOD. What changes should be done in my code ?

https://mirror.codeforces.com/contest/1891/submission/230655160

Also what are some good sources for topics like modular arithmetic and other topics that will help me remain stable expert. Thanks

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Hi Codeforces, this was my second ever contest and this time i was able to solve problem A my submission: 230544275 i found this to be interesting because it was difficult for me to comprehend what the question was asking, then i just read the problem and analyzed the sample testcases for about 15-20 mins, I found that if the index that is not sorted (a[i]>a[i+1]) lies in the powers of 2(0,1,3,7,15) then it is possible to sort it otherwise not, I just implemented this and got it through. I was elated after the first submission passed the pretests so my confidence boosted for the second problem, I just implemented a brute force solution 230561923 and it passed the sample cases so I decided to submit it, Obviously it gave a TLE at pretest 3 but overall it was a great experience, looking forward to solving more and growing. Please if possible, share your helpful feedbacks on my solutions. happy coding!!

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For problem C I kept on overthinking that the solution would involve DP, however I did notice that the constraints on $$$x$$$ was too big and that it would most probably lead to TLE. However, can anyone think if DP is feasible for problem C if $$$x$$$ was actually some small number?

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Out of curiosity following problem F , how will this problem be solved? Initially i have only one vertex which is numbered 1.There are 1e5 qeuries where in one type of query i can add a vertex to the tree with number sz+1 (sz is the no of nodes in the tree currently).Can i answer other type of query where i need to tell the size of subtree rooted at a given vertex in logn time or what will be the most optimized algorithm for this.

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    9 months ago, # ^ |
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    Euler Tour Technique

    Note that you don't need to answer the queries online. So you can create the whole tree and process queries later.

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      9 months ago, # ^ |
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      can you explain a bit more . Like when a node is added i need to change the size of subtree of all the ancestors of the node . How will i do that?

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        9 months ago, # ^ |
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        https://mirror.codeforces.com/contest/1891/submission/230704812

        here is my solution using exactly the same idea as in the analysis of the contest. maybe it will help you :D

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        9 months ago, # ^ |
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        From the page:

        Notice that after running dfs, each range [start[i], end[i]] contains all ranges [start[j], end[j]] for each j in the subtree of i.

        So, to add something to the subtree of i, you just update the values from start[i] to end[i]. This can be done with any range update technique.

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      9 months ago, # ^ |
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      what does online means here? I saw it in other comments also but don't know it's meaning

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        9 months ago, # ^ |
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        u dont know the queries in advance

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For Problem B, I wrote my code on Python, and I am getting correct answers when I run it on my system, but wrong answers when I run on CF. Here is my code — https://mirror.codeforces.com/contest/1891/submission/230716694

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I understood the editorial algorithm for C, but can someone help me with why the greedy algorithm is correct? I am unable to prove it. Thanks alot!

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PROBLEM D

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9 months ago, # |
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Here is the implementation of sqrt decomposition for F. It is giving TLE as warned by author. Any optimisation in provided code is welcomed.

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For D problem can someone provide me simple explanation ? (Even its implementation is complicated even tried) https://mirror.codeforces.com/contest/1891/problem/D

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9 months ago, # |
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is there a way to solve problem c with binary search

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Can F be solved online?

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https://mirror.codeforces.com/contest/1891/submission/230866197

can anyone help me with problem F? cant find any error

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9 months ago, # |
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I think it is better with spoilers, and code.

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Can someone explain the solution of problem B? I didn't understand the idea of the editorial.

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    9 months ago, # ^ |
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    Hello,bro I saw your profile You have solved a good amount of questions but your rating is not increasing because you are not solving/practicing questions of 1200 — 1500 rating you are only solving questions of below 1000 rating

    use this website https://clist.by/problems/?resource=1 for codeforces question and try to solve more than 1200 rating questions :-) Thank You

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9 months ago, # |
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Thank you, 127.0.0.1

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9 months ago, # |
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127.0.0.1 I am having a little problem in D. Its working on nearly all the test cases and logic is to point. But I think there is some problem with modulus and its giving wrong answer on one test case-v 179 1000000000000000000 Can any body help I cant seem to figure it our. This is my code where maxi is modulus ll l,r; cin>>l>>r;

ll te=l;
 ll ans=0;
 while(te<=r)
 {
    ll p=log2(te);
    ll up;
    // if(p==63)
    // up=r;
    // else
    up=min((ll)(pow(2,p+1)-1LL),r);
    //cout<<up<<endl;
    ll ct=0;
    ll x=1;

   while(x*p<=te)
   {
      x*=p;
      ++ct;
   }
  // cout<<te<<" "<<x<<endl;
   while(true)
   {
    ll nx;
    //cout<<x*p<<e ndl;
    x*=p;nx=min(x,up+1);
    ll nxm=nx%maxi;
    ll  tem=te%maxi;
     ans=(ans+(ct*((nx-te)%maxi))%maxi)%maxi;
     if(nx>up)
     break;
     ct=(ct+1)%maxi;
      te=nx;
   }
   if(up==r)
   break;
   te=up+1;



 }
 cout<<ans<<endl;
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9 months ago, # |
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https://mirror.codeforces.com/contest/1891/submission/231287946[submission:231287946] why i am getting tle in this approach can anyone pls help

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can someone explain for me more about how to solve problem D , please ?

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can someone help me,i am getting WA on test 2 of F https://mirror.codeforces.com/contest/1891/submission/232563023

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8 months ago, # |
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Did anyone try (probably the dumbest) solution to F by reversing queries and doing segtree with range updates on the euler tour of the final tree? That's the first thing that comes to mind

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8 months ago, # |
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I have a problem in D. I got a wrong anwer on test 8, and I have trouble finding where I went wrong. My submission is Your text to link here... . Can someone help me figure out what the problem is? Thanks a lot if you can help!

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    8 months ago, # ^ |
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    I am having the same issue, failed on the same test case. Did you manage to figure it out? if so, what was the issue?

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      8 months ago, # ^ |
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      I filed on test case 8 because I used the function std::log2(), which led to a accuracy issue. I switched to using binary search to avoid the accuracy issue.

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6 months ago, # |
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Can anyone please tell me if the error is in the steps of modular operations or in the map in the following solution. It would be a big help. My submission for Problem D

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6 months ago, # |
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I am quite stuck on the 1891D task. I have checked the solution explanation, and I can't figure out one key element. The explanation says, that on each part with f(i)==f(i+1) we will have at max O(logn) transitions, such that g(i)!=g(i+1).
My question: how can g(i)!=g(i+1) be on the segment where f(i)==f(i+1), when f(x)=log2(x), while g(x) = logf(x)x, which means that it's base is bigger than 2, and that it should change its values significantly more rare that log2(x).

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6 months ago, # |
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In Problem C, can someone explain why we are taking the attack iteratively rather than the subarray smaller to the largest horde (in this case the current j pointer)? Why not distributing attacks (of type I) on the remaining hordes beneficial since we can keep the largest hordes larger? and also, why are we minimizing the attacks of type II, wouldn't they be more effective if use them when the combo is higher than 1?