@[waOooo](/user/211144) n 是啥?
by xutongwei @ 2020-09-20 13:41:57
@[灵空](/user/91204) 噢噢
by waOooo @ 2020-09-20 13:43:10
如果只是要求质数个数的话有一个算法是 $ O(n^{\frac 2 3}) $ 的
by Prean @ 2020-09-20 13:43:28
有$< \mathcal O(n)$的"筛"法(严格来说其实不是筛法):$\mathrm{min\_25}$筛与洲阁筛基础就是筛质数个数
by zzw4257 @ 2020-09-20 13:44:13
[如果你只要求素数个数的话,可以康康这个](https://loj.ac/problem/6235)
by impuk @ 2020-09-20 13:46:11
@[灵空](/user/91204) 可能
by panyf @ 2020-09-20 13:46:54
@[panyf](/user/221955) 怎么搞啊/fad
by Fatalis_Lights @ 2020-09-20 13:47:49
听说是 $O(n/\log \log n)$ 的
by panyf @ 2020-09-20 13:48:59
/fad/fad
by Fatalis_Lights @ 2020-09-20 13:49:33
[Sieve of Atkin ](https://encyclopedia.thefreedictionary.com/Sieve+of+Atkin )
by panyf @ 2020-09-20 13:50:10