hi, i think all of divisors of n(1 to 10^18) with in 9 sec is quit immpossible. if any idea or solution.please share it :) thank you
| # | User | Rating |
|---|---|---|
| 1 | Benq | 3676 |
| 2 | jiangly | 3670 |
| 3 | Kevin114514 | 3657 |
| 4 | maroonrk | 3355 |
| 5 | Um_nik | 3353 |
| 6 | squareOf105 | 3346 |
| 7 | tourist | 3343 |
| 8 | strapple | 3324 |
| 9 | heuristica | 3322 |
| 10 | jiangbowen | 3291 |
| # | User | Contrib. |
|---|---|---|
| 1 | Qingyu | 158 |
| 2 | nik_exists | 150 |
| 2 | maspy | 150 |
| 4 | AmShZ | 143 |
| 4 | Um_nik | 143 |
| 6 | Errichto | 139 |
| 7 | adamant | 137 |
| 8 | BledDest | 135 |
| 9 | maroonrk | 133 |
| 10 | qwexd | 130 |
hi, i think all of divisors of n(1 to 10^18) with in 9 sec is quit immpossible. if any idea or solution.please share it :) thank you
| Name |
|---|



Iterating from 1 to sqrt(n) and using divisor pairs will give you the answer. In 9 secs, I think you can do 10^9 operations, even with the constant of the modulo operation.
Good day to you,
well firstly you can start with "Pollard Rho" (you will also need something like miller-rabin), which could factorize the number in O(N^(1/4)) [it is somehow long coding with many possible mistakes, but the core is not that bad].
As you have it, you can generate all divisors by "simple recursion" in O(#NUMBER_OF_DIVISORS)
Hope I guessed rightly what you asked for ^_^
Good Luck & Have nice day!!
thanks i got it...yeah there have some mistake.. for large number it's quite impossible there is no exact solution for this ques with 9 sec :) thank you :)
Well imho there is something more exact, but this is a lesser pain ^_^
Also the probability is "soo big" you can treat is as exact :)
Have nice day ^_^