atcoder_official's blog

AtCoder Regular Contest 210 Announcement
By atcoder_official, history, 6 months ago, In English

We will hold AtCoder Regular Contest 210.

The point values will be 300-600-700-800-1000-1100.

We are looking forward to your participation!

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5 months ago, hide # |
 
Vote: I like it -42 Vote: I do not like it

I was automatically registered as unrated, how to change it to rated?

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5 months ago, hide # |
 
Vote: I like it +4 Vote: I do not like it

Is it div.1 or div.2?

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5 months ago, hide # |
 
Vote: I like it +2 Vote: I do not like it

wow 6 prob arc!

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5 months ago, hide # |
 
Vote: I like it -25 Vote: I do not like it

So hard!!!

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

ARC_T[1,2,3]=ABC_T[3,5,5]

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5 months ago, hide # |
← Rev. 4  
Vote: I like it +1 Vote: I do not like it

How to solve problem C? I don't know at all greedy (as an atcoder maxrating>=2100 user)

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

Can't believe another nonsensical ARC appears.

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5 months ago, hide # |
← Rev. 2  
Vote: I like it -28 Vote: I do not like it

Why is my code wrong? link

I generated 5 million cases at random but all of the answers are correct.

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    5 months ago, hide # ^ |
     
    Vote: I like it 0 Vote: I do not like it

    I have the same problem, what troubles me is why the errors are concentrated in the test points of 03_mid.

    It's possible that the data in 03_mid has been specially constructed, reflecting some potentially flawed practices in our code.

    Coincidentally, I also generated many sets of data, but none of them had any issues.

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    5 months ago, hide # ^ |
     
    Vote: I like it 0 Vote: I do not like it

    It seems there's an issue with your approach. The process of simulating the combination of smaller coins into larger ones should start from the smallest denomination and merge upwards, rather than only considering values 20 less than the current coin.

    A direct implementation of the above solution would be $$$O(n^2)$$$. Therefore, you need to merge the coins from the smallest to the largest denomination in one go, and then apply a greedy approach from the largest to the smallest denomination, breaking down the previously simulated merged coins. This would bring the complexity down to $$$O(n)$$$. Referring to my code might be helpful.

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

Just curious, when will rating get updated on kenkoooo. Most of the time it's pretty fast.

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