| # | User | Rating |
|---|---|---|
| 1 | ecnerwala | 3648 |
| 2 | Benq | 3580 |
| 3 | orzdevinwang | 3570 |
| 4 | cnnfls_csy | 3569 |
| 5 | Geothermal | 3568 |
| 6 | tourist | 3565 |
| 7 | maroonrk | 3530 |
| 8 | Radewoosh | 3520 |
| 9 | Um_nik | 3481 |
| 10 | jiangly | 3467 |
| # | User | Contrib. |
|---|---|---|
| 1 | maomao90 | 174 |
| 2 | adamant | 164 |
| 2 | awoo | 164 |
| 4 | TheScrasse | 160 |
| 5 | nor | 159 |
| 6 | maroonrk | 156 |
| 7 | -is-this-fft- | 150 |
| 7 | SecondThread | 150 |
| 9 | orz | 146 |
| 10 | pajenegod | 145 |
Doubt in Modular Arithmetic
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?
