Numbers 1....N are written in table (3 <= N <= 1000), at each step one player choose some number and mark it as visited, player wins if he can make 3 consecutive numbers visited in his turn. Who will win the game if both play optimally? game is 2-player game with alternating turns.
| № | Пользователь | Рейтинг |
|---|---|---|
| 1 | tourist | 3690 |
| 2 | jiangly | 3647 |
| 3 | Benq | 3581 |
| 4 | orzdevinwang | 3570 |
| 5 | Geothermal | 3569 |
| 5 | cnnfls_csy | 3569 |
| 7 | Radewoosh | 3509 |
| 8 | ecnerwala | 3486 |
| 9 | jqdai0815 | 3474 |
| 10 | gyh20 | 3447 |
| Страны | Города | Организации | Всё → |
| № | Пользователь | Вклад |
|---|---|---|
| 1 | maomao90 | 174 |
| 2 | awoo | 164 |
| 3 | adamant | 163 |
| 4 | TheScrasse | 159 |
| 5 | nor | 157 |
| 6 | maroonrk | 156 |
| 7 | -is-this-fft- | 152 |
| 8 | Petr | 146 |
| 8 | orz | 146 |
| 10 | BledDest | 145 |
It's obvious that if we mark some number x, an opponent instantly loses if he takes any number y : |x - y| ≤ 2.
Let's define f(n) the answer for table of "length" n. We know that f(3) = 1 (1st player wins on such board anyways). Trying all possible moves we can easily calculate f(n): by taking some number x (1 ≤ x ≤ n), we split the table into two parts of lengths max(0, x - 3) and max(0, n - x - 2), since there's no point of keeping neighbouring to x unavailable numbers on the table.
If you still have some difficulties, googling "sprague grundy theory" shuold help.