| № | 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 | 164 |
| 1 | maomao90 | 164 |
| 3 | Um_nik | 163 |
| 4 | atcoder_official | 160 |
| 5 | -is-this-fft- | 158 |
| 6 | adamant | 157 |
| 6 | awoo | 157 |
| 8 | TheScrasse | 154 |
| 8 | nor | 154 |
| 10 | djm03178 | 153 |
Современные процессоры выполняют умножения намного быстрее делений, потому что разработчики процессоров больше уделяют этому внимания, да и просто легче распарралеливаются в кремнии. Вот способ замены двух делений на одно деление и несколько умножений с небольшой потерей точности:
http://stereopsis.com/2div.html
Подобным образом можно заменить и три, и четыре деления на одно...
Auto comment: topic has been translated by beatoriche(original revision, translated revision, compare)
Nice idea...
But I think this will be useful just when we have lots of divides so the program will be slow...But in this situation we can't use any variable to put 1/(a[0]*a[1]*...) in it...(if we want to use just one divide)
So we can just decrease the number of divides(but not to 1)...But does it help in speed a lot?
I'm a beginner and it was just a question...If I'm wrong you can tell me...
The proper way to determine is TESTING. It depends on particular compiler and CPU.
Ok...
I did it...
Test
I didn't except slower program...But why?
cout is slow. Printing anything is slow, don't do it when comparing speed of programs.
Ok thank you for that point...
I did it without printing too and just put results in array...
For N=1e5 not any differences...
But it was 1s better for N=1e6...(3.95 vs 4.5)
I know it should have better time but I still think it won't be useful(exactly in CF cause time limits are usually 1s and integer limits are not 1e6 for this problems)
It's still useful, even if this only keeps your brain thinking :D
which type is 'll'? long long? it's for floating point only, unfortunately
"ll" is long long but I used "(ld)" in my code...It's long double...
Ans also in my test without printing I used a "long double" array for result so not any problem...
Автокомментарий: текст был обновлен пользователем beatoriche (предыдущая версия, новая версия, сравнить).