Привет, Codeforces!
У меня появилось желание перед этапом Технокубка и рейтинговыми раундами по нему хорошенько проверить, что всё работает как надо после недавних нововведений.
Приглашаю вас принять участие в Testing Round 14 (Unrated). Старт состоится 22-го декабря в 14:05. Раунд будет неофициальным, нерейтинговым. Продолжительность: 90 минут.
Претесты будут необычно слабыми, чтобы спровоцировать побольше взломов.
Спасибо,
MikeMirzayanov
UPD: Спасибо! Раунд прошел без замеченных багов. Все системы работают исправно.
it's showing div2 contestants are not out of contest
"Претесты будут необычно слабыми" , ничего нового.
"Pretests are unusually weak" , nothing new.
codeforces should take it as a compliment.
It is very cool that make a big challenge in hacking
A hackforces round? Great! :)
Excellent move to make sure there is nothing bugging during the important contests ahead.
This must be an exciting hackforces round! :p Good luck to everyone!
Wonderful!
The time is fantastic for Chinese, and we can enjoy hacking time~
Why I cannot view other people source codes? I have locked the problem and enabled Flash Player.
Are you trying to open your roommate's code?
Yes...
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.
Same issue happended with me.
Official unrated round to be precise.
Is there a simple way to solve problem C other than naive 10*10! solution
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 :<
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
Will there be an editorial as well? :D
Будем надеяться, что завтра все пройдет так же гладко и без лагов! Всем удачи в грядущем отборе, кто будет его писать, а так же удачи участникам Div.1 и Div.2 раундов!