№ | Пользователь | Рейтинг |
---|---|---|
1 | tourist | 4009 |
2 | jiangly | 3823 |
3 | Benq | 3738 |
4 | Radewoosh | 3633 |
5 | jqdai0815 | 3620 |
6 | orzdevinwang | 3529 |
7 | ecnerwala | 3446 |
8 | Um_nik | 3396 |
9 | ksun48 | 3390 |
10 | gamegame | 3386 |
Страны | Города | Организации | Всё → |
№ | Пользователь | Вклад |
---|---|---|
1 | cry | 167 |
2 | Um_nik | 163 |
3 | maomao90 | 162 |
3 | atcoder_official | 162 |
5 | adamant | 159 |
6 | -is-this-fft- | 158 |
7 | awoo | 157 |
8 | TheScrasse | 154 |
9 | Dominater069 | 153 |
9 | nor | 153 |
Название |
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Indeed, multiplying everything by 2 helps because you can make each person enter at an even time and leave at an odd time, so you don't have to worry about what to count first.
can you give me a case where marking [a +1] [b+1 -1] fails. And thank you tenshi_kanade you always help people out :)
There are many tricky cases in this problem. For example, try the following out...
7 7 4 4 6 4 7 4 8 4 9
Answer in this case is 3 3.
Another one...
10 15 3 10 15 10 15 10 15.
Here, the answer is 3 3 too.