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 | tourist | 4009 |
2 | jiangly | 3839 |
3 | Radewoosh | 3646 |
4 | jqdai0815 | 3620 |
4 | Benq | 3620 |
6 | orzdevinwang | 3612 |
7 | Geothermal | 3569 |
7 | cnnfls_csy | 3569 |
9 | ecnerwala | 3494 |
10 | Um_nik | 3396 |
# | User | Contrib. |
---|---|---|
1 | Um_nik | 164 |
2 | maomao90 | 160 |
3 | -is-this-fft- | 159 |
4 | atcoder_official | 158 |
4 | awoo | 158 |
4 | cry | 158 |
7 | adamant | 155 |
8 | nor | 154 |
9 | TheScrasse | 153 |
10 | maroonrk | 152 |
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 ^_^