| № | 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 | 166 |
| 2 | maomao90 | 163 |
| 2 | Um_nik | 163 |
| 4 | atcoder_official | 161 |
| 5 | adamant | 160 |
| 6 | -is-this-fft- | 158 |
| 7 | awoo | 157 |
| 8 | TheScrasse | 154 |
| 9 | nor | 153 |
| 9 | Dominater069 | 153 |
Number Theory Problem
I found the following problem here.
f(x, n) = x21 + x22 + x23 + ... + x2n
Example: f(2, 10) mod 1000000007 = 180974681
Calculate 
mod 1000000007.
We know that abc mod p = 
mod p.
The period of the sequence 2n mod 1000000006 is φ(1000000006) = 500000002.
The period is too large to calculate the sum. Help please?
| Rev. | Язык | Кто | Когда | Δ | Комментарий | |
|---|---|---|---|---|---|---|
| en1 |
|
ironman7453 | 2015-07-02 19:26:01 | 520 | Initial revision (published) |
