| № | 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 | 166 |
| 3 | adamant | 163 |
| 4 | TheScrasse | 160 |
| 5 | nor | 158 |
| 6 | maroonrk | 156 |
| 7 | -is-this-fft- | 152 |
| 8 | SecondThread | 147 |
| 9 | orz | 146 |
| 10 | pajenegod | 145 |
Today in 1557C - Moamen and XOR I wanted to compute $$${(2^{k-1})}^{n}$$$. But instead of writing binpow(binpow(2, k — 1), n), I simplified it to $$$2^{(k-1)*n}$$$ and wrote binpow(2, mul(k — 1, n)) and failed to pass pretest 2.
Why are not the two equivalent?
