| # | 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 |
this should be a classical problem: We have a group of numbers named A with the size of n. Does A have a subgroup such that the sum of it is divisible by m?
m,n>=10^5
I couldn't find any solution with an order better than O(n*m).
thanks for helping :)
There is a randomized tilde-O(n+m) solution for prime
mout there, but I'm not sure if it's feasible with your constraints since its constants are fairly high. I'm pretty sure there's a way to adapt the solution to non-primem, though I haven't spent too much time trying to do so.