| № | Пользователь | Рейтинг |
|---|---|---|
| 1 | tourist | 4009 |
| 2 | jiangly | 3823 |
| 3 | Benq | 3738 |
| 4 | Radewoosh | 3633 |
| 5 | jqdai0815 | 3620 |
| 6 | orzdevinwang | 3529 |
| 7 | ecnerwala | 3446 |
| 8 | Um_nik | 3396 |
| 9 | ksun48 | 3390 |
| 10 | gamegame | 3386 |
| Страны | Города | Организации | Всё → |
| № | Пользователь | Вклад |
|---|---|---|
| 1 | cry | 167 |
| 2 | Um_nik | 163 |
| 3 | maomao90 | 162 |
| 3 | atcoder_official | 162 |
| 5 | adamant | 159 |
| 6 | -is-this-fft- | 158 |
| 7 | awoo | 157 |
| 8 | TheScrasse | 154 |
| 9 | Dominater069 | 153 |
| 9 | nor | 153 |
Has anyone solved any of these two problems using Dominator Tree?
https://cses.fi/problemset/task/1703
https://cses.fi/problemset/task/1203
Unfortunately my code is failing on a couple tests.
I'd really appreciate it if anyone provides an accepted code using Dominator Tree.
How to find Product of Divisors in this CSES Problem
for each divisor i there exist a divisor n/i such that i*(n/i)=n,so we can put the divisors into pairs of the form (i,n/i) and the product of each pair is equal to n,so the answer is (n power s/2) where s is the number of divisors,there is a case where n is perfect square then the number of divisors is odd so we just multiply the answer by sqrt(n)