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 | ecnerwala | 3649 |
2 | Benq | 3581 |
3 | orzdevinwang | 3570 |
4 | Geothermal | 3569 |
4 | cnnfls_csy | 3569 |
6 | tourist | 3565 |
7 | maroonrk | 3531 |
8 | Radewoosh | 3521 |
9 | Um_nik | 3482 |
10 | jiangly | 3468 |
# | User | Contrib. |
---|---|---|
1 | maomao90 | 174 |
2 | awoo | 164 |
3 | adamant | 162 |
4 | TheScrasse | 159 |
5 | nor | 158 |
6 | maroonrk | 156 |
7 | -is-this-fft- | 151 |
8 | SecondThread | 147 |
9 | orz | 146 |
10 | pajenegod | 145 |
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 ^_^