-emli-'s blog

AtCoder Regular Contest 071 / Beginner Contest 058
By -emli-, history, 8 years ago, In English

Two contests AtCoder Regular Contest 071 and AtCoder Beginner Contest 058 will be held at the same time.

You can participate in whichever contest you want. (However, you can't register to both competitions.) The last two tasks in ABC are the same as the first two tasks in ARC.

Time： April 8th (Saturday), 21:00 JST

Duration: 100 minutes

Number of Tasks: 4

writer： nuip

Rating： 0-2799 for ARC, 0-1199 for ABC

The point value are:

ABC: 100 — 200 — 300 — 500

ARC: 300 — 500 — 600 — 1000

We are looking forward to your participation!

UPD: 19 hours to begin.

UPD: 50 minutes to begin.

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Comments Comments (10)
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8 years ago, # |
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Do you need to register for the contest?

I can't see anything like this and I remember that previously it was quite obvious (like automatic redirect).

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8 years ago, # |
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Are beginners contests rated?

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8 years ago, # |
← Rev. 2   Vote: I like it +2 Vote: I do not like it

How to solve D?

edit: Thanks for answers + English editorial!

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

    sum of differences of all pairs in X multiplied by sum of differences of all pairs in Y

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

    Ex:- 3 3 a1=1 3 6 a2=1 4 6

    Let a = (3-1)+(6-1)+(6-3) which is equal to

    For(i=1;i<n;i++) a+=i*a1[i]-cum1[i-1];

    where cum1 is its cumulative frequency. Caluculate same for other array(a2)and multiply both of them to get the final answer

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    8 years ago, # ^ |
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    It is easy to use prefix and know how many units per unit length corresponds to the width

    The same re-use of the prefix and how many units have calculated how much contribution has been made

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8 years ago, # |
← Rev. 2   Vote: I like it 0 Vote: I do not like it

That moment when you realized you thought that 2, 3, 3, 4, 4, 4, ... (and similar sequences) is valid for some reason and thus fail to solve F T_T

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    8 years ago, # ^ |
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    2, 3, 3, 4, 4, 4, ... , this sequence isn't valid.

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      8 years ago, # ^ |
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      Yes, I know, but I thought it was valid during contest and wasted more than 1 hour solving something that's completely unrelated to the problem :(

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