maroonrk's blog

By maroonrk, history, 3 months ago, In English

We will hold AtCoder Regular Contest 152.

The point values will be 400-500-600-700-800-1000.

We are looking forward to your participation!

 
 
 
 
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3 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Will participate. Give me 2 Dan.

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    3 months ago, # ^ |
      Vote: I like it +8 Vote: I do not like it

    May not get 2 Dan, but the problems are very interesting!!! Thank you, writer!!!

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      3 months ago, # ^ |
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      I'm so proud that I've solved ABD and I'm sure to get positive delta :)

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3 months ago, # |
  Vote: I like it -8 Vote: I do not like it

rp++!

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3 months ago, # |
  Vote: I like it +13 Vote: I do not like it

What the hell are the samples? They're so weak

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3 months ago, # |
Rev. 2   Vote: I like it -15 Vote: I do not like it

I'm so sad right now. Here is why.

Why
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3 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Any hints for C?

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    3 months ago, # ^ |
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    +1

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    3 months ago, # ^ |
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    I think the idea behind problem C is quite inspiring, although I still can not understand the editorial. I have realized that selecting some value s means a mirror symmetry, but can't go further. Waiting for some other hints too.

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3 months ago, # |
  Vote: I like it +24 Vote: I do not like it

What a weird contest. Difficulty to me was A<E<D=C<<<<B.... I did not prove B too, just convinced myself that its ARC B so guessed some lower bound and it worked.... was also very scared to submit E. Thought I misread but it is strangely that easy.

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    3 months ago, # ^ |
      Vote: I like it +16 Vote: I do not like it

    E is more of a statement parsing problem but in a good sense. I don't think it's an easy problem for most participants because you need good skills in constructing math models and understanding what is important.

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    2 months ago, # ^ |
      Vote: I like it +28 Vote: I do not like it

    B is quite easily solvable if you notice that if they need to pass each other twice, they might as well start at the point where they passed each other for the first time.

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3 months ago, # |
  Vote: I like it +5 Vote: I do not like it

That was so hard that I only solved A. :(

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3 months ago, # |
Rev. 2   Vote: I like it -39 Vote: I do not like it

What the hell was that ??? I'm 1667 Elo on codechef and 1132 Elo on codeforces and unable to solve problem A. It's ridiculous.

And Editorial provide no explanation of the formula.

I basically solve nothing and learn nothing.

I Will no more do Regular Contest for a long long time...

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    3 months ago, # ^ |
      Vote: I like it +12 Vote: I do not like it

    Go and do more ABC problems. They are much easier.

    When you are likely to solve problem E~F in ABC, then back to ARC. I am sure you can enjoy the problems.

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3 months ago, # |
  Vote: I like it -8 Vote: I do not like it

Can someone please explain problem B approach ?? I didn't get clearly the idea even with the tutorial.

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3 months ago, # |
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Can anyone explain what needed to be done in problem A? and why was it to be done this way!?

I don't understand the editorial to a good extent, maybe someone can help here.

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    3 months ago, # ^ |
    Rev. 3   Vote: I like it 0 Vote: I do not like it

    Ordering of the group matters like they will come in the order given there so you just need to let the group sit like upcoming can't seat so here only two kind of group make exists (i.e. size of 1 and 2) so group of size 1 can sit anywhere, ans can be "NO" only when a group of size 2 arrives! with some condition.

    So we just need to let groups sit so that it can occupy one or two more space than their own sizes which will lead to not make a sit for upcomers.

    Remember group of 2 only will face difficulty always!

    My Code

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    3 months ago, # ^ |
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    My approach is to simulate the worst case that each arrived group will pick a position one unit away from previous group if possible. i.e. x group1 x group2 x group3 ... where x is empty seats.

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3 months ago, # |
  Vote: I like it +47 Vote: I do not like it

Thank you for your participation!

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3 months ago, # |
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Can anyone prove the solution of B problem, I solved but I can't prove it.

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3 months ago, # |
Rev. 2   Vote: I like it 0 Vote: I do not like it

why is this not working my approch -> https://atcoder.jp/contests/arc152/submissions/36673139

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3 months ago, # |
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Can someone explain C in detail? I can't understand the solution...

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    2 months ago, # ^ |
      Vote: I like it +32 Vote: I do not like it

    Consider a sorted sequence and arbitrary two operations with $$$s_1$$$ and $$$s_2$$$. The sequence changes by $$$+2(s_1-s_2)$$$, the order of elements is unchanged. Here $$$s_1 = a_p$$$ and $$$s_2 = 2s_1-a_r$$$ for some $$$p,r$$$, so $$$2(s_1-s_2) = 2(a_r-a_p)$$$. Obviously then, if $$$g$$$ is the GCD of all elements, then their remainders modulo $$$2g$$$ can't change. In an even number of operations, it's clear that the smallest possible value of the first element is $$$a_1$$$ modulo $$$2g$$$.

    The remainders modulo $$$2g$$$ won't change in one operation either, since $$$2a_p-a_i = 2(a_p-a_i)+a_i \equiv a_i$$$ modulo $$$2g$$$. If the number of operations is odd, all that matters is that the order of elements in the sorted sequence is reversed, so the smallest possible value of the first element could be $$$a_N$$$ modulo $$$2g$$$ instead. (Since $$$g$$$ is the GCD of differences, each $$$a_i = b_i g + r$$$, only parities of $$$b_1, b_N$$$ matter.)

    Finally, it's always possible to construct a sequence in which the smallest element is $$$a_1 \% 2g$$$ by first adding and then subtracting some 2*differences, see Bezout's identity. The rule on non-negative elements isn't broken then.

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3 months ago, # |
  Vote: I like it -10 Vote: I do not like it

Why this is wrong for B?

constexpr int N = 200005;
int a[N];
set<int> s;
signed main(){
    int n, l;
    cin>>n>>l;
    for(int i = 1; i <= n; i++) cin>>a[i];
    int ans = inf;
    for(int i = 1; i <= n; i++){
        s.insert(l - a[i]);
    }
    int mi = inf;
    for(int i = 1; i <= n; i++){
        set<int>::iterator it=s.lower_bound(a[i]);
        set<int>::iterator it2=it;
        if(it!=s.begin()) it--;
        if(it2==s.end()) it2--;
        mi = min(mi, abs(a[i] - *it));
        mi = min(mi, abs(a[i] - *it2));
    }
        ans = min(ans, mi * 2 + l * 2);
    cout<<ans<<endl;
    return 0;
}
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2 months ago, # |
  Vote: I like it 0 Vote: I do not like it

in problem A "Seat Occupation"

I see that my submission 36675947 send during contest was correct except "Yes" (resp; "No") replace by "yes" (resp. "no").

If the evaluatiuon is CASE SENSITIVE, it should clarify in statement.

CodeChef and CodeForces are not case sensitive.

Sorry, but i can only blame the author.

submission with wrong case: https://atcoder.jp/contests/arc152/submissions/36675947

submission with correct case: https://atcoder.jp/contests/arc152/submissions/37023059

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5 weeks ago, # |
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It seems that output values for the first 23 tests for problem D in dropbox (here) are all empty.