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TOYOTA SYSTEMS Programming Contest 2024（AtCoder Beginner Contest 377) Announcement
By atcoder_official, history, 3 days ago, In English

We will hold TOYOTA SYSTEMS Programming Contest 2024（AtCoder Beginner Contest 377).

We are looking forward to your participation!

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2 days ago, # |
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Good luck & Have fun!

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2 days ago, # |
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lzn_tops

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46 hours ago, # |
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These point values are just right for the beginners.

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46 hours ago, # |
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我AC了第三题！可是我第四题不会 qwq

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    45 hours ago, # ^ |
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    I know you are happy, but please speak English.

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    45 hours ago, # ^ |
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    It's not good to talk about problems during the contest.

    And do not send Chinese!!!

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45 hours ago, # |
  Vote: I like it +8 Vote: I do not like it

any Hint on how to solve $$$E$$$? Only hint

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    45 hours ago, # ^ |
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    Also why binary exponentiation of the permutation https://cp-algorithms.com/algebra/binary-exp.html#applying-a-permutation-k-times doesn't work? Still couldn't figure it out.

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      45 hours ago, # ^ |
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      You misunderstood the statement. You don't have to apply the given $$$p$$$ $$$k$$$ times (i.e. compute $$$p^{k + 1}$$$), you have to square $$$p$$$ $$$k$$$ times (i.e. compute $$$((p^2)^2)^{...}$$$ and so on $$$k$$$ times).

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        44 hours ago, # ^ |
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        then how its different from this problem: link

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          44 hours ago, # ^ |
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          The difference is that we need to compute not p^(k+1) but p^(2^k) which seems to be a harder task.

          But at the first glance I also thought that the problem from the current round and https://atcoder.jp/contests/abc367/tasks/abc367_e are almost identical.

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          44 hours ago, # ^ |
          ← Rev. 2   Vote: I like it +3 Vote: I do not like it

          I think because in this problem we are applying p to itself each time. But in that problem applying X to A , where X doesnt change at all.

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        44 hours ago, # ^ |
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        What does squaring a permutation mean ? Can you give some understanding on that , I get confused when I try to think these kinds of stuff in my head. I guess I lack some understanding on "applying a permutation".

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          44 hours ago, # ^ |
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          You can look at the code from here https://cp-algorithms.com/algebra/binary-exp.html#applying-a-permutation-k-times to understand what applying a permutation means:

          vector<int> applyPermutation(vector<int> sequence, vector<int> permutation) {
    vector<int> newSequence(sequence.size());
    for(int i = 0; i < sequence.size(); i++) {
        newSequence[i] = sequence[permutation[i]];
    }
    return newSequence;
}
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          44 hours ago, # ^ |
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          Since permutations form a group under composition it is sometimes convenient to use the multiplication notation for composition. So with $$$p^2$$$ I mean $$$p \circ p$$$ and in general with $$$p^n$$$ I mean $$$p \circ p \circ \ldots \circ p$$$, composing $$$n - 1$$$ times.

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        44 hours ago, # ^ |
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        Thank you! Now it's clear that I indeed interpreted the statement incorrectly.

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      45 hours ago, # ^ |
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      yes , i also tried binary exponentiation but didnt worked

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      45 hours ago, # ^ |
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      I guess because in binary exponentiation we don't change the permutation while in the task permutation changes after every operation

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    45 hours ago, # ^ |
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    Think about the cycles of the permutation.

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45 hours ago, # |
← Rev. 2   Vote: I like it +11 Vote: I do not like it

What was the intuition for E so that to move k steps we do 2^k-1 i = p[i] ?

I solved it but I kinda guessed it from the sample case and it took me a long time. The idea of 2^k-1 came to my mind from this :

When we do 1 move p[i] = p[p[i]] and p[p[i]] = p[p[p[i]]] , it seemed like number of moves would double each move. However this was a very vague idea and the problem kinda teased my brain , thats why it took me so long to solve the problem.

Here is my submission : https://atcoder.jp/contests/abc377/submissions/59187007

My Solution
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    45 hours ago, # ^ |
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    The 2^k part is clear from the start, but how did you find the actual number of steps since 2^k is too big? I split p into cycles and solved for each of those independently, but it looks too complex, is there more elegant solution?

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      45 hours ago, # ^ |
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      How is it clear ? Can you elaborate a bit more on that. Also my solution was same but I dont think it is that complex. I found the idea in the last 5 mins of the contest and it took me 3 mins to implement it.

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        44 hours ago, # ^ |
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        How is it clear?

        I've just generated several examples with straightforward op implementation ASAP and saw the pattern from there. I have no idea or strict proof why it's like that though.

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      44 hours ago, # ^ |
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      well you just do $$$2^k - 1$$$ $$$mod$$$ $$$m$$$ ,$$$m$$$ is the number of nodes in the cycle it is in

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    44 hours ago, # ^ |
    ← Rev. 3   Vote: I like it +3 Vote: I do not like it

    Basically the problem asks you to square the permutation $$$p$$$ $$$k$$$ times, so you are actually computing $$$p^{2^k}$$$. A cycle of size $$$s$$$ shifts by one each time you apply $$$p$$$ and after applying $$$p$$$ $$$s$$$ times you get back to the start, so applying $$$p$$$ $$$2^k$$$ times you have effectively to shift by $$$2^k \mod s$$$.

    Clarification about the notation: Permutations form a group under composition, so it is often convenient to use product notation for compostion of permutations. So with $$$p^n$$$ I mean $$$p \circ p \circ \ldots \circ p$$$, composing $$$n - 1$$$ times.

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      44 hours ago, # ^ |
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      Can you please share some sources with a more detailed explanation on how applying a permutation works how you have described?

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        44 hours ago, # ^ |
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        The notation is introduced for example here. There is a lot of theory about permutations, the group of all permutations of a fixed size $$$n$$$ for example is often denoted by $$$S_n$$$ and called a symmetric group. It should not be too hard to find more information if you are interested.

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          44 hours ago, # ^ |
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          Thanks for your response electron. I think I intuitively understand how we get the shift factor of 2 ^ k — 1 now when you mentioned it.

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45 hours ago, # |
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How E? I used binary lifting but got wrong

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44 hours ago, # |
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Can someone please help with the mistake in following solution for problem D? Note: My idea was just inverse of what is done in editorial.

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

int main() {
    ll N, M, L, R;
    cin >> N >> M;
    ll answer = (M * (M + 1)) / 2;
    
    unordered_map<ll, ll> RtoL;
    for (ll i = 0; i < M; i += 1) {
        cin >> L >> R;
        if (RtoL.find(R) == RtoL.end()) RtoL[R] = L;
        else RtoL[R] = max(RtoL[R], L);
    }
    
    map<ll, ll> LtoR;
    for (auto& [R, L] : RtoL) {
        if (LtoR.find(L) == LtoR.end()) LtoR[L] = R;
        else LtoR[L] = min(LtoR[L], R);
    }
    
    deque<pair<ll, ll>> Q;
    for (auto& [L, R] : LtoR) {
        if (Q.empty()) {
            Q.push_back({L, R});
            continue;
        }
        
        if (Q.back().second > R) Q.pop_back();
        Q.push_back({L, R});
    }
    
    while (!Q.empty()) {
        L = Q.front().first;
        R = Q.front().second;
        Q.pop_front();
        
        ll Rnext = Q.empty() ? (M + 1) : Q.front().second;
        answer -= (L * (Rnext - R));
    }
    
    cout << answer;
}
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44 hours ago, # |
  Vote: I like it +10 Vote: I do not like it

Someone look at the first place person's F submission: https://atcoder.jp/contests/abc377/submissions/59171780

They import sys like 8 times, comments in code, wrote 140 lines in 5 minutes.

Surely they will get banned right?

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44 hours ago, # |
  Vote: I like it +6 Vote: I do not like it

I enjoyed the problems, and I'm amazed at how surprising they turned out to be!

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44 hours ago, # |
← Rev. 3   Vote: I like it -8 Vote: I do not like it

hard problems qwq

seems to be A < B <= C < F <= G < D < E

hint: someone seems angery so here's why I think that:
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44 hours ago, # |
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Problem E is great! At first, I thought it is just binary-lift, but later find that in fact we need find p[p[p[.....p[i]]]], where there are 2^k p[ ] operations. Thus, we can use dfs to find the period T of each i, and calculate R = 2^k mod T. Meanwhile, we also compute the "binary-lift" during the above dfs. Finally, the answer is p[p[p....p[i]]], where there are R p[ ] operations, which can be solved based on the above "binary-lift".

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42 hours ago, # |
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solved A-C

Tried to solve F using N-Queens logic, but ultimately got TLE

E was some disjoint graph cycle i think, will try to upsolve.

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37 hours ago, # |
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Is problem G well known, or classical?

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    32 hours ago, # ^ |
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    It's just based on a data structure named Trie, which is one of the "more known" advanced data structure.

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27 hours ago, # |
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Can someone explain how E [https://atcoder.jp/contests/abc377/tasks/abc377_e] can be solved.

I am not able to visualise it. I understood that if we apply permutation p on a sequence k times then how its changing. But if the permutation itself is changing then how the sequence is gonna change. Can someone help me understand it or point me to some relevant resource.

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26 hours ago, # |
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Does anybody have the derivation of that $$$P^{2^{k}}$$$ in problem $$$E$$$?

(I'm asking about derivation, not the proof by induction)

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