I have a non linear recurrence relation a(n)=(n-1)*(a(n-1)+a(n-2)).How can I calculate a(n)%MOD in O(log(n)) time? Base conditions:-a(0)=0,a(1)=1
→ Обратите внимание
→ Трансляции
→ Лидеры (рейтинг)
| № | Пользователь | Рейтинг |
|---|---|---|
| 1 | ecnerwala | 3649 |
| 2 | Benq | 3581 |
| 3 | jiangly | 3578 |
| 4 | orzdevinwang | 3570 |
| 5 | Geothermal | 3569 |
| 5 | cnnfls_csy | 3569 |
| 7 | tourist | 3565 |
| 8 | maroonrk | 3531 |
| 9 | Radewoosh | 3521 |
| 10 | Um_nik | 3482 |
| Страны | Города | Организации | Всё → |
→ Лидеры (вклад)
| № | Пользователь | Вклад |
|---|---|---|
| 1 | maomao90 | 174 |
| 2 | awoo | 164 |
| 3 | adamant | 161 |
| 4 | TheScrasse | 159 |
| 5 | nor | 158 |
| 6 | maroonrk | 156 |
| 7 | -is-this-fft- | 152 |
| 8 | SecondThread | 147 |
| 9 | orz | 146 |
| 10 | pajenegod | 145 |
→ Найти пользователя
→ Прямой эфир
Codeforces (c) Copyright 2010-2024 Михаил Мирзаянов
Соревнования по программированию 2.0
Время на сервере: 27.04.2024 23:52:06 (k2).
Десктопная версия, переключиться на мобильную.
При поддержке
This is basically a recurrence for n!. (The base conditions are not mentioned but even with those, the solution will not be very different). And I dont think there is a simple way to do it in sub-linear time. :(
Actually, the base conditions matter a lot. For example, if a0 = 1, a1 = 0, then it's a recurrence for !n (number of derangements).
Xellos,since you are saying that base condition matters a lot,I take a(0)=1 and a(1)=0 .Can I calculate the recurrence relation in O(log(n))?
Maybe there's some obscure algorithm that does it, I don't know. But for the purpose of solving contest problems, assume you can't.
Sorry! By solution I meant "solving the task in logN time", because there is at least one base condition in which solution takes linear time. I guess I messed up with the wordings :P