MikeMirzayanov's blog

By MikeMirzayanov, history, 6 years ago, translation, In English

Hello, Codeforces!

I want to test the system before the rated rounds based on Technocup stage, to be sure that everything works as expected.

I invite you to take part in Testing Round 14 (Unrated). It will start soon, on December, 22, 11:05:00 (UTC). It will be unofficial unrated round. The duration is 90 minutes.

Pretests are unusually weak to trigger more hack.

Thank you,
MikeMirzayanov

UPD: Thank you for participation. I didn't notice any bugs. It seems we are ready for new rounds!

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6 years ago, # |
  Vote: I like it +19 Vote: I do not like it

it's showing div2 contestants are not out of contest

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6 years ago, # |
  Vote: I like it -15 Vote: I do not like it

"Pretests are unusually weak" , nothing new.

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6 years ago, # |
Rev. 2   Vote: I like it +26 Vote: I do not like it

A hackforces round? Great! :)

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6 years ago, # |
  Vote: I like it +2 Vote: I do not like it

Excellent move to make sure there is nothing bugging during the important contests ahead.

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6 years ago, # |
  Vote: I like it +4 Vote: I do not like it

This must be an exciting hackforces round! :p Good luck to everyone!

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6 years ago, # |
  Vote: I like it +22 Vote: I do not like it

Wonderful!

The time is fantastic for Chinese, and we can enjoy hacking time~

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6 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Why I cannot view other people source codes? I have locked the problem and enabled Flash Player.

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    6 years ago, # ^ |
      Vote: I like it +17 Vote: I do not like it

    Are you trying to open your roommate's code?

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

    I have locked the problem and enabled Flash Player too, I can view other people source codes but I can't hack more than once (Hacking windows try to reload)

    But when I changed web browser from Google Chrome to Microsoft Edge, that issue was fixed. May someone help me with fixing that issue for Google Chrome? Thanks.

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6 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Is there a simple way to solve problem C other than naive 10*10! solution

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    6 years ago, # ^ |
    Rev. 4   Vote: I like it +2 Vote: I do not like it

    There is. You can count the total significance of each digit (for each number, the last digit has the significance of 1, its adjacent digit to the left has the significance of 10, then 100, 1000, etc.).

    You can use greedy here to find the minimum. Assign digit 0 to the letter that never appears first in a number and has the highest significance possible.

    The other digits, from 1 to 9, can be assigned greedily based on their total significances only.

    UPD: highest, not smallest. I made this critical mistake while writing the comment (though the code's idea was correct though). Very sorry :<

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      6 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      suppose if the number is ajdddd, bddddd, cddddd, ... , iddddd except dddddd. here i may put 0 for j or d. but i should put it for d. I thought what you said, but i thought there are many test cases. so i left that method

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6 years ago, # |
  Vote: I like it +2 Vote: I do not like it

Will there be an editorial as well? :D