After taking a peek at the editorial, it looks like pretty much the same thing they are saying so I am not sure what I am missing at the moment.
A quick summary of my current logic as attached above:
Logic
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1 | tourist | 3993 |
2 | jiangly | 3743 |
3 | orzdevinwang | 3707 |
4 | Radewoosh | 3627 |
5 | jqdai0815 | 3620 |
6 | Benq | 3564 |
7 | Kevin114514 | 3443 |
8 | ksun48 | 3434 |
9 | Rewinding | 3397 |
10 | Um_nik | 3396 |
Страны | Города | Организации | Всё → |
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1 | cry | 167 |
2 | Um_nik | 163 |
3 | maomao90 | 162 |
3 | atcoder_official | 162 |
5 | adamant | 159 |
6 | -is-this-fft- | 158 |
7 | awoo | 155 |
8 | TheScrasse | 154 |
9 | Dominater069 | 153 |
10 | djm03178 | 152 |
Help on WA for problem Moscow Gorilla
After taking a peek at the editorial, it looks like pretty much the same thing they are saying so I am not sure what I am missing at the moment.
A quick summary of my current logic as attached above:
Count number of ways to make MEX i for i in range [1,n+1]:
a) Only one way to make n+1
b) To make MEX 1, if l and r are min and max of locations of 1 in the two arrays, you must choose from range [1,l), (l,r), or (r , n] ( 1 indexed )
c) For every MEX ( other than 1 or n+1 ), keep track of the range of the previous MEXes (l and r unions of previous mexes) and find points such that l and r of previous mexes do no coincide with l and r of current number. Then merge the ranges!
Rev. | Язык | Кто | Когда | Δ | Комментарий | |
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en3 | SriniV | 2023-06-24 06:51:37 | 26 | Tiny change: 'ary of my current logic as attached above:\n\n<spoi' -> 'ary of my logic:\n\n<spoi' | ||
en2 | SriniV | 2023-06-24 06:51:02 | 63 | Tiny change: 'ubmission:[submissio' -> 'ubmission: [submissio' | ||
en1 | SriniV | 2023-06-24 06:50:24 | 816 | Initial revision (published) |
Название |
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