Problem Link: https://uva.onlinejudge.org/external/125/12546.pdf
Solution Link: https://ideone.com/q2SX1x
# | User | Rating |
---|---|---|
1 | tourist | 3985 |
2 | jiangly | 3814 |
3 | jqdai0815 | 3682 |
4 | Benq | 3529 |
5 | orzdevinwang | 3526 |
6 | ksun48 | 3517 |
7 | Radewoosh | 3410 |
8 | hos.lyric | 3399 |
9 | ecnerwala | 3392 |
9 | Um_nik | 3392 |
# | User | Contrib. |
---|---|---|
1 | cry | 169 |
2 | maomao90 | 162 |
2 | Um_nik | 162 |
4 | atcoder_official | 161 |
5 | djm03178 | 158 |
6 | -is-this-fft- | 157 |
7 | adamant | 155 |
8 | awoo | 154 |
8 | Dominater069 | 154 |
10 | luogu_official | 150 |
Problem Link: https://uva.onlinejudge.org/external/125/12546.pdf
Solution Link: https://ideone.com/q2SX1x
Name |
---|
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