Problem Link: https://uva.onlinejudge.org/external/125/12546.pdf
Solution Link: https://ideone.com/q2SX1x
| № | Пользователь | Рейтинг |
|---|---|---|
| 1 | jiangly | 3810 |
| 2 | Benq | 3676 |
| 3 | Kevin114514 | 3655 |
| 4 | maroonrk | 3463 |
| 5 | strapple | 3390 |
| 6 | Um_nik | 3387 |
| 7 | tourist | 3384 |
| 8 | heuristica | 3322 |
| 9 | turmax | 3319 |
| 10 | jiangbowen | 3291 |
| Страны | Города | Организации | Всё → |
| № | Пользователь | Вклад |
|---|---|---|
| 1 | Qingyu | 155 |
| 2 | nik_exists | 150 |
| 2 | maspy | 150 |
| 4 | Um_nik | 143 |
| 5 | AmShZ | 142 |
| 6 | Errichto | 139 |
| 7 | adamant | 137 |
| 8 | maroonrk | 133 |
| 9 | BledDest | 132 |
| 10 | qwexd | 129 |
Problem Link: https://uva.onlinejudge.org/external/125/12546.pdf
Solution Link: https://ideone.com/q2SX1x
| Название |
|---|



Finally I solved it . From seeing your code I can guess that you are constructing all pairs of integers whose LCM is n . But the problem is that there are an exponential number of possibilities .
My approach was the following , first notice that p and q are always the divisors of n also you can create equivalence classes for p . If n = p1pow1 * ...pkpowk is the prime factorisation of n , then we partition all the divisors into equivalence classes such that a particular equivalence class contains a subset of prime divisors to their highest power and other prime divisors can have smaller powers . For each of the equivalence class in p we consider how many numbers q can be created such that lcm(p, q) = n .
This is the abstract of my approach , Take a look at my CODE