problem:
given at most 400 values of C. for each C,find an integer X (X<=10^18) such that 2^x mod (10^9 + 7) = c
i know that a value x must exist (X<10^9 + 7).but finding X is a problem for me. is there an efficient way to find x?
problem:
given at most 400 values of C. for each C,find an integer X (X<=10^18) such that 2^x mod (10^9 + 7) = c
i know that a value x must exist (X<10^9 + 7).but finding X is a problem for me. is there an efficient way to find x?
| № | Пользователь | Рейтинг |
|---|---|---|
| 1 | Benq | 3792 |
| 2 | VivaciousAubergine | 3647 |
| 3 | Kevin114514 | 3603 |
| 4 | jiangly | 3583 |
| 5 | turmax | 3559 |
| 6 | tourist | 3541 |
| 7 | strapple | 3515 |
| 8 | ksun48 | 3461 |
| 9 | dXqwq | 3436 |
| 10 | Otomachi_Una | 3413 |
| Страны | Города | Организации | Всё → |
| № | Пользователь | Вклад |
|---|---|---|
| 1 | Qingyu | 157 |
| 2 | adamant | 153 |
| 3 | Um_nik | 147 |
| 4 | Proof_by_QED | 146 |
| 5 | Dominater069 | 145 |
| 6 | errorgorn | 141 |
| 7 | cry | 139 |
| 8 | YuukiS | 135 |
| 9 | TheScrasse | 134 |
| 10 | chromate00 | 133 |
U can solve using shank baby step gaint step algorithm
To elaborate on teja349's answer, the idea is that there are at most m = 109 + 6 different values that 2x can take on, and if there is a solution to 2x = c, then you can write
where 
and 
(modulo off-by-one errors).
All you have to do is compute all possible 2x values where
and throw them in a hashtable, and then for each value of a, check if there is some element in the hashtable having value 
.