目录
如何编程最快速度求出两百万以内素数个数(不限语言和算法)?
回复内容:
x贴一个优化得比较过分的线性筛。用位模式保存状态,直接绕过2和3的倍数。<span class="cp">#include <stdio.h></span> <span class="cp">#include <time.h></span> <span class="cp">#include <string.h></span> <span class="cp">#include <stdlib.h></span> <span class="cp">#define PRIME_LIM 10000000</span> <span class="cp">#define N 100000000</span> <span class="kt">int</span> <span class="n">primes</span><span class="p">[</span><span class="n">PRIME_LIM</span><span class="p">]</span> <span class="o">=</span> <span class="p">{</span><span class="mi">0</span><span class="p">};</span> <span class="kt">int</span> <span class="n">flags</span><span class="p">[</span><span class="n">N</span><span class="o">/</span><span class="mi">96</span> <span class="o">+</span> <span class="mi">1</span><span class="p">]</span> <span class="o">=</span> <span class="p">{</span><span class="mi">0</span><span class="p">};</span> <span class="kt">int</span> <span class="nf">get_prime</span><span class="p">()</span> <span class="p">{</span> <span class="kt">int</span> <span class="n">nu</span> <span class="o">=</span> <span class="mi">5</span><span class="p">,</span> <span class="n">to</span> <span class="o">=</span> <span class="mi">0</span><span class="p">;</span> <span class="n">primes</span><span class="p">[</span><span class="mi">0</span><span class="p">]</span> <span class="o">=</span> <span class="mi">2</span><span class="p">;</span> <span class="n">primes</span><span class="p">[</span><span class="mi">1</span><span class="p">]</span> <span class="o">=</span> <span class="mi">2</span><span class="p">,</span> <span class="n">primes</span><span class="p">[</span><span class="mi">2</span><span class="p">]</span> <span class="o">=</span> <span class="mi">3</span><span class="p">;</span> <span class="k">for</span><span class="p">(</span><span class="kt">int</span> <span class="n">i</span> <span class="o">=</span> <span class="mi">0</span><span class="p">;</span> <span class="n">nu</span> <span class="o"><=</span> <span class="n">N</span><span class="p">;</span> <span class="n">i</span><span class="o">++</span><span class="p">)</span> <span class="p">{</span> <span class="k">if</span><span class="p">(</span><span class="o">!</span><span class="p">(</span><span class="n">flags</span><span class="p">[</span><span class="n">i</span><span class="o">>></span><span class="mi">5</span><span class="p">]</span><span class="o">&</span><span class="p">(</span><span class="mi">1</span><span class="o"><<</span><span class="p">(</span><span class="n">i</span><span class="o">&</span><span class="mi">31</span><span class="p">))))</span> <span class="n">primes</span><span class="p">[</span><span class="o">++</span><span class="n">primes</span><span class="p">[</span><span class="mi">0</span><span class="p">]]</span> <span class="o">=</span> <span class="n">nu</span><span class="p">;</span> <span class="k">for</span><span class="p">(</span><span class="kt">int</span> <span class="n">j</span> <span class="o">=</span> <span class="mi">3</span><span class="p">;</span> <span class="n">j</span> <span class="o"><=</span> <span class="n">primes</span><span class="p">[</span><span class="mi">0</span><span class="p">]</span> <span class="o">&&</span> <span class="n">primes</span><span class="p">[</span><span class="n">j</span><span class="p">]</span> <span class="o">*</span> <span class="n">nu</span> <span class="o"><=</span> <span class="n">N</span><span class="p">;</span> <span class="n">j</span><span class="o">++</span><span class="p">)</span> <span class="p">{</span> <span class="n">to</span> <span class="o">=</span> <span class="p">(</span><span class="n">nu</span> <span class="o">*</span> <span class="n">primes</span><span class="p">[</span><span class="n">j</span><span class="p">]</span> <span class="o">-</span> <span class="mi">5</span><span class="p">)</span> <span class="o">>></span> <span class="mi">1</span><span class="p">;</span> <span class="n">to</span> <span class="o">-=</span> <span class="n">to</span><span class="o">/</span><span class="mi">3</span><span class="p">;</span> <span class="n">flags</span><span class="p">[</span><span class="n">to</span><span class="o">>></span><span class="mi">5</span><span class="p">]</span> <span class="o">|=</span> <span class="mi">1</span><span class="o"><<</span><span class="p">(</span><span class="n">to</span><span class="o">&</span><span class="mi">31</span><span class="p">);</span> <span class="k">if</span><span class="p">(</span><span class="o">!</span><span class="p">(</span><span class="n">nu</span> <span class="o">%</span> <span class="n">primes</span><span class="p">[</span><span class="n">j</span><span class="p">]))</span> <span class="k">break</span><span class="p">;</span> <span class="p">}</span> <span class="n">nu</span> <span class="o">+=</span> <span class="mi">2</span> <span class="o">+</span> <span class="p">((</span><span class="n">i</span><span class="o">&</span><span class="mi">1</span><span class="p">)</span> <span class="o"><<</span> <span class="mi">1</span><span class="p">);</span> <span class="p">}</span> <span class="k">return</span> <span class="n">primes</span><span class="p">[</span><span class="mi">0</span><span class="p">];</span> <span class="p">}</span> <span class="kt">int</span> <span class="nf">main</span><span class="p">()</span> <span class="p">{</span> <span class="kt">clock_t</span> <span class="n">t</span> <span class="o">=</span> <span class="n">clock</span><span class="p">();</span> <span class="n">printf</span><span class="p">(</span><span class="s">"%d</span><span class="se">\n</span><span class="s">"</span><span class="p">,</span> <span class="n">get_prime</span><span class="p">());</span> <span class="n">printf</span><span class="p">(</span><span class="s">"Time:%f</span><span class="se">\n</span><span class="s">"</span><span class="p">,</span> <span class="mf">1.0</span> <span class="o">*</span> <span class="p">(</span><span class="n">clock</span><span class="p">()</span> <span class="o">-</span> <span class="n">t</span><span class="p">)</span> <span class="o">/</span> <span class="n">CLOCKS_PER_SEC</span><span class="p">);</span> <span class="k">for</span><span class="p">(</span><span class="kt">int</span> <span class="n">i</span> <span class="o">=</span> <span class="mi">1</span><span class="p">;</span> <span class="n">primes</span><span class="p">[</span><span class="n">i</span><span class="p">]</span> <span class="o"><</span> <span class="mi">100</span><span class="p">;</span> <span class="n">i</span><span class="o">++</span><span class="p">)</span> <span class="p">{</span> <span class="n">printf</span><span class="p">(</span><span class="s">"%d</span><span class="se">\n</span><span class="s">"</span><span class="p">,</span> <span class="n">primes</span><span class="p">[</span><span class="n">i</span><span class="p">]);</span> <span class="p">}</span> <span class="k">return</span> <span class="mi">0</span><span class="p">;</span> <span class="p">}</span>
登录后复制
看到各位大神用各种语言写的代码,我这个外行人也跃跃欲试了。
鉴于大家已经给出了C,C++,Python,Mathmatica等的实现过程,那我就用Java吧。
我不会流氓地直接用各种Prime函数(那样对问题讨论毫无意义),还是给出完整实现过程吧。
算法一般,还有待改进,欢迎各位大神指正:
我用的是筛法,稍稍做了优化(把偶数单独列出来筛),代码如下:
1、初始版代码:
<span class="kd">class</span> <span class="nc">Prime</span><span class="o">{</span> <span class="kd">public</span> <span class="kd">static</span> <span class="kt">int</span> <span class="nf">calculateNumber</span><span class="o">(</span><span class="kt">int</span> <span class="n">Nmax</span><span class="o">){</span> <span class="kt">boolean</span><span class="o">[]</span> <span class="n">isPrime</span><span class="o">=</span><span class="k">new</span> <span class="kt">boolean</span><span class="o">[</span><span class="n">Nmax</span><span class="o">+</span><span class="mi">1</span><span class="o">];</span> <span class="k">for</span><span class="o">(</span><span class="kt">int</span> <span class="n">i</span><span class="o">=</span><span class="mi">3</span><span class="o">;</span><span class="n">i</span><span class="o"><=</span><span class="n">Nmax</span><span class="o">;</span><span class="n">i</span><span class="o">+=</span><span class="mi">2</span><span class="o">)</span> <span class="n">isPrime</span><span class="o">[</span><span class="n">i</span><span class="o">]=</span><span class="kc">true</span><span class="o">;</span> <span class="n">isPrime</span><span class="o">[</span><span class="mi">2</span><span class="o">]=</span><span class="kc">true</span><span class="o">;</span> <span class="k">for</span><span class="o">(</span><span class="kt">int</span> <span class="n">i</span><span class="o">=</span><span class="mi">3</span><span class="o">;</span><span class="n">i</span><span class="o"><=</span><span class="n">Math</span><span class="o">.</span><span class="na">sqrt</span><span class="o">(</span><span class="n">Nmax</span><span class="o">);</span><span class="n">i</span><span class="o">+=</span><span class="mi">2</span><span class="o">){</span> <span class="k">if</span><span class="o">(</span><span class="n">isPrime</span><span class="o">[</span><span class="n">i</span><span class="o">]==</span><span class="kc">true</span><span class="o">){</span> <span class="k">for</span><span class="o">(</span><span class="kt">int</span> <span class="n">j</span><span class="o">=</span><span class="n">i</span><span class="o">*</span><span class="n">i</span><span class="o">;</span><span class="n">j</span><span class="o"><=</span><span class="n">Nmax</span><span class="o">;</span><span class="n">j</span><span class="o">+=</span><span class="mi">2</span><span class="o">*</span><span class="n">i</span><span class="o">)</span> <span class="n">isPrime</span><span class="o">[</span><span class="n">j</span><span class="o">]=</span><span class="kc">false</span><span class="o">;</span> <span class="o">}</span> <span class="o">}</span> <span class="kt">int</span> <span class="n">primeNum</span><span class="o">=</span><span class="mi">0</span><span class="o">;</span> <span class="k">for</span><span class="o">(</span><span class="kt">int</span> <span class="n">i</span><span class="o">=</span><span class="mi">1</span><span class="o">;</span><span class="n">i</span><span class="o"><=</span><span class="n">Nmax</span><span class="o">;</span><span class="n">i</span><span class="o">++){</span> <span class="k">if</span><span class="o">(</span><span class="n">isPrime</span><span class="o">[</span><span class="n">i</span><span class="o">]==</span><span class="kc">true</span><span class="o">)</span> <span class="n">primeNum</span><span class="o">++;</span> <span class="o">}</span> <span class="k">return</span> <span class="n">primeNum</span><span class="o">;</span> <span class="o">}</span> <span class="kd">public</span> <span class="kd">static</span> <span class="kt">void</span> <span class="nf">main</span><span class="o">(</span><span class="n">String</span><span class="o">[]</span> <span class="n">args</span><span class="o">){</span> <span class="kd">final</span> <span class="kt">int</span> <span class="n">Nmax</span><span class="o">=</span><span class="mi">2000000</span><span class="o">;</span> <span class="kt">double</span> <span class="n">startTime</span><span class="o">=</span><span class="n">System</span><span class="o">.</span><span class="na">currentTimeMillis</span><span class="o">();</span> <span class="kt">int</span> <span class="n">primeNum</span><span class="o">=</span><span class="n">Prime</span><span class="o">.</span><span class="na">calculateNumber</span><span class="o">(</span><span class="n">Nmax</span><span class="o">);</span> <span class="kt">double</span> <span class="n">timeSpent</span><span class="o">=(</span><span class="n">System</span><span class="o">.</span><span class="na">currentTimeMillis</span><span class="o">()-</span><span class="n">startTime</span><span class="o">)/</span><span class="mi">1000</span><span class="o">;</span> <span class="n">System</span><span class="o">.</span><span class="na">out</span><span class="o">.</span><span class="na">println</span><span class="o">(</span><span class="s">"The prime numbers from 1 to "</span><span class="o">+</span><span class="n">Nmax</span><span class="o">+</span><span class="s">" is "</span><span class="o">+</span><span class="n">primeNum</span><span class="o">);</span> <span class="n">System</span><span class="o">.</span><span class="na">out</span><span class="o">.</span><span class="na">println</span><span class="o">(</span><span class="s">"Time spent : "</span><span class="o">+</span><span class="n">timeSpent</span><span class="o">+</span><span class="s">" s"</span><span class="o">);</span> <span class="o">}</span> <span class="o">}</span>
登录后复制
至于算法层面,从标题来看只是求质数个数,而并不需要枚举所有质数。于是存在比线性更优的算法,可以参考:素数计数函数。其时间复杂度为O(x^(2/3)/log(x)),空间复杂度为O(x^(1/3)log(x)^2)
具体代码实现可以围观:kimwalisch/primecount · GitHub 。
当然最后运行时间对于两百万这个“小”数据基本是可以忽略不计的(< 10 ms),甚至装载这软件用到的运行库文件都不止这么久,但对于上亿甚至更大的数,相比筛法的优势是很明显的。


线性筛法 我来终结此问题。
计算素数个数被数学家和ACMer玩烂了,没啥优化的空间了。
用C语言,计算200万以内素数个数,100次实验取平均
用埃氏筛法,0.035620 seconds
用欧拉筛法,0.025630 seconds
计算1亿以内素数个数,10次实验取平均
用埃氏筛法,2.890600 seconds
用欧拉筛法,1.473400 seconds
运行机器:32位XP

<span class="cp">#include <math.h></span> <span class="cp">#include <stdio.h></span> <span class="cp">#include <string.h></span> <span class="cp">#include <time.h></span> <span class="k">const</span> <span class="kt">int</span> <span class="n">N</span> <span class="o">=</span> <span class="mi">2000000</span><span class="p">;</span> <span class="kt">bool</span> <span class="n">b</span><span class="p">[</span><span class="n">N</span><span class="o">+</span><span class="mi">1</span><span class="p">];</span> <span class="kt">int</span> <span class="n">c</span><span class="p">[</span><span class="n">N</span><span class="o">+</span><span class="mi">1</span><span class="p">];</span> <span class="kt">void</span> <span class="nf">Era_select</span><span class="p">(){</span> <span class="c1">// Eratosthenes</span> <span class="n">memset</span><span class="p">(</span><span class="n">b</span><span class="p">,</span> <span class="mi">0</span><span class="p">,</span> <span class="n">N</span><span class="o">+</span><span class="mi">1</span><span class="p">);</span> <span class="kt">int</span> <span class="n">q</span> <span class="o">=</span> <span class="p">(</span><span class="kt">int</span><span class="p">)</span><span class="n">sqrt</span><span class="p">(</span><span class="n">N</span><span class="p">);</span> <span class="kt">int</span> <span class="n">i</span><span class="p">,</span> <span class="n">j</span><span class="p">,</span> <span class="n">cnt</span> <span class="o">=</span> <span class="mi">0</span><span class="p">;</span> <span class="k">for</span> <span class="p">(</span> <span class="n">i</span><span class="o">=</span><span class="mi">2</span><span class="p">;</span> <span class="n">i</span><span class="o"><=</span><span class="n">q</span><span class="p">;</span> <span class="o">++</span><span class="n">i</span> <span class="p">){</span> <span class="k">if</span> <span class="p">(</span> <span class="o">!</span> <span class="n">b</span><span class="p">[</span><span class="n">i</span><span class="p">]</span> <span class="p">){</span> <span class="k">for</span> <span class="p">(</span> <span class="n">j</span><span class="o">=</span><span class="n">i</span><span class="o"><<</span><span class="mi">1</span><span class="p">;</span> <span class="n">j</span><span class="o"><=</span><span class="n">N</span><span class="p">;</span> <span class="n">j</span><span class="o">+=</span><span class="n">i</span> <span class="p">){</span> <span class="n">b</span><span class="p">[</span><span class="n">j</span><span class="p">]</span> <span class="o">=</span> <span class="nb">true</span><span class="p">;</span> <span class="p">}</span> <span class="p">}</span> <span class="p">}</span> <span class="k">for</span> <span class="p">(</span> <span class="n">i</span><span class="o">=</span><span class="mi">2</span><span class="p">;</span> <span class="n">i</span><span class="o"><=</span><span class="n">N</span><span class="p">;</span> <span class="o">++</span><span class="n">i</span> <span class="p">){</span> <span class="k">if</span> <span class="p">(</span> <span class="o">!</span> <span class="n">b</span><span class="p">[</span><span class="n">i</span><span class="p">]</span> <span class="p">){</span> <span class="o">++</span><span class="n">cnt</span><span class="p">;</span> <span class="p">}</span> <span class="p">}</span> <span class="c1">//printf("%d\n", cnt);</span> <span class="p">}</span> <span class="kt">void</span> <span class="nf">Euler_select</span><span class="p">(){</span> <span class="n">memset</span><span class="p">(</span><span class="n">b</span><span class="p">,</span> <span class="mi">0</span><span class="p">,</span> <span class="n">N</span><span class="o">+</span><span class="mi">1</span><span class="p">);</span> <span class="kt">int</span> <span class="n">i</span><span class="p">,</span> <span class="n">j</span><span class="p">,</span> <span class="n">cnt</span> <span class="o">=</span> <span class="mi">0</span><span class="p">,</span> <span class="n">t</span><span class="p">;</span> <span class="k">for</span> <span class="p">(</span> <span class="n">i</span><span class="o">=</span><span class="mi">2</span><span class="p">;</span> <span class="n">i</span><span class="o"><=</span><span class="n">N</span><span class="p">;</span> <span class="o">++</span><span class="n">i</span> <span class="p">){</span> <span class="k">if</span> <span class="p">(</span> <span class="o">!</span> <span class="n">b</span><span class="p">[</span><span class="n">i</span><span class="p">]</span> <span class="p">){</span> <span class="n">c</span><span class="p">[</span><span class="n">cnt</span><span class="o">++</span><span class="p">]</span> <span class="o">=</span> <span class="n">i</span><span class="p">;</span> <span class="p">}</span> <span class="k">for</span> <span class="p">(</span> <span class="n">j</span><span class="o">=</span><span class="mi">0</span><span class="p">;</span> <span class="n">j</span><span class="o"><</span><span class="n">cnt</span><span class="p">;</span> <span class="o">++</span><span class="n">j</span> <span class="p">){</span> <span class="n">t</span> <span class="o">=</span> <span class="n">i</span><span class="o">*</span><span class="n">c</span><span class="p">[</span><span class="n">j</span><span class="p">];</span> <span class="k">if</span> <span class="p">(</span> <span class="n">t</span> <span class="o">></span> <span class="n">N</span> <span class="p">){</span> <span class="k">break</span><span class="p">;</span> <span class="p">}</span> <span class="n">b</span><span class="p">[</span><span class="n">t</span><span class="p">]</span> <span class="o">=</span> <span class="nb">true</span><span class="p">;</span> <span class="k">if</span> <span class="p">(</span> <span class="n">i</span><span class="o">%</span><span class="n">c</span><span class="p">[</span><span class="n">j</span><span class="p">]</span> <span class="o">==</span> <span class="mi">0</span> <span class="p">){</span> <span class="k">break</span><span class="p">;</span> <span class="p">}</span> <span class="p">}</span> <span class="p">}</span> <span class="c1">//printf("%d\n", cnt);</span> <span class="p">}</span> <span class="kt">int</span> <span class="nf">main</span><span class="p">(){</span> <span class="kt">int</span> <span class="n">i</span><span class="p">,</span> <span class="n">num</span><span class="o">=</span><span class="mi">100</span><span class="p">;</span> <span class="kt">clock_t</span> <span class="n">start</span><span class="p">;</span> <span class="n">start</span> <span class="o">=</span> <span class="n">clock</span><span class="p">();</span> <span class="k">for</span> <span class="p">(</span> <span class="n">i</span><span class="o">=</span><span class="mi">0</span><span class="p">;</span> <span class="n">i</span><span class="o"><</span><span class="n">num</span><span class="p">;</span> <span class="o">++</span><span class="n">i</span> <span class="p">){</span> <span class="n">Era_select</span><span class="p">();</span> <span class="p">}</span> <span class="n">printf</span><span class="p">(</span><span class="s">"%lf seconds</span><span class="se">\n</span><span class="s">"</span><span class="p">,</span> <span class="p">(</span><span class="kt">double</span><span class="p">)(</span><span class="n">clock</span><span class="p">()</span><span class="o">-</span><span class="n">start</span><span class="p">)</span> <span class="o">/</span> <span class="n">CLOCKS_PER_SEC</span> <span class="o">/</span> <span class="n">num</span><span class="p">);</span> <span class="n">start</span> <span class="o">=</span> <span class="n">clock</span><span class="p">();</span> <span class="k">for</span> <span class="p">(</span> <span class="n">i</span><span class="o">=</span><span class="mi">0</span><span class="p">;</span> <span class="n">i</span><span class="o"><</span><span class="n">num</span><span class="p">;</span> <span class="o">++</span><span class="n">i</span> <span class="p">){</span> <span class="n">Euler_select</span><span class="p">();</span> <span class="p">}</span> <span class="n">printf</span><span class="p">(</span><span class="s">"%lf seconds</span><span class="se">\n</span><span class="s">"</span><span class="p">,</span> <span class="p">(</span><span class="kt">double</span><span class="p">)(</span><span class="n">clock</span><span class="p">()</span><span class="o">-</span><span class="n">start</span><span class="p">)</span> <span class="o">/</span> <span class="n">CLOCKS_PER_SEC</span> <span class="o">/</span> <span class="n">num</span><span class="p">);</span> <span class="k">return</span> <span class="mi">0</span><span class="p">;</span> <span class="p">}</span>
登录后复制
<span class="k">def</span> <span class="nf">rwh_primes2</span><span class="p">(</span><span class="n">n</span> <span class="o">=</span> <span class="mi">10</span><span class="o">**</span><span class="mi">6</span><span class="p">):</span> <span class="c"># http://stackoverflow.com/questions/2068372/fastest-way-to-list-all-primes-below-n-in-python/3035188#3035188</span> <span class="sd">""" Input n>=6, Returns a list of primes, 2 <= p < n """</span> <span class="n">correction</span> <span class="o">=</span> <span class="p">(</span><span class="n">n</span><span class="o">%</span><span class="mi">6</span><span class="o">></span><span class="mi">1</span><span class="p">)</span> <span class="n">n</span> <span class="o">=</span> <span class="p">{</span><span class="mi">0</span><span class="p">:</span><span class="n">n</span><span class="p">,</span><span class="mi">1</span><span class="p">:</span><span class="n">n</span><span class="o">-</span><span class="mi">1</span><span class="p">,</span><span class="mi">2</span><span class="p">:</span><span class="n">n</span><span class="o">+</span><span class="mi">4</span><span class="p">,</span><span class="mi">3</span><span class="p">:</span><span class="n">n</span><span class="o">+</span><span class="mi">3</span><span class="p">,</span><span class="mi">4</span><span class="p">:</span><span class="n">n</span><span class="o">+</span><span class="mi">2</span><span class="p">,</span><span class="mi">5</span><span class="p">:</span><span class="n">n</span><span class="o">+</span><span class="mi">1</span><span class="p">}[</span><span class="n">n</span><span class="o">%</span><span class="mi">6</span><span class="p">]</span> <span class="n">sieve</span> <span class="o">=</span> <span class="p">[</span><span class="bp">True</span><span class="p">]</span> <span class="o">*</span> <span class="p">(</span><span class="n">n</span><span class="o">/</span><span class="mi">3</span><span class="p">)</span> <span class="n">sieve</span><span class="p">[</span><span class="mi">0</span><span class="p">]</span> <span class="o">=</span> <span class="bp">False</span> <span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">xrange</span><span class="p">(</span><span class="nb">int</span><span class="p">(</span><span class="n">n</span><span class="o">**</span><span class="mf">0.5</span><span class="p">)</span><span class="o">/</span><span class="mi">3</span><span class="o">+</span><span class="mi">1</span><span class="p">):</span> <span class="k">if</span> <span class="n">sieve</span><span class="p">[</span><span class="n">i</span><span class="p">]:</span> <span class="n">k</span><span class="o">=</span><span class="mi">3</span><span class="o">*</span><span class="n">i</span><span class="o">+</span><span class="mi">1</span><span class="o">|</span><span class="mi">1</span> <span class="n">sieve</span><span class="p">[</span> <span class="p">((</span><span class="n">k</span><span class="o">*</span><span class="n">k</span><span class="p">)</span><span class="o">/</span><span class="mi">3</span><span class="p">)</span> <span class="p">::</span><span class="mi">2</span><span class="o">*</span><span class="n">k</span><span class="p">]</span><span class="o">=</span><span class="p">[</span><span class="bp">False</span><span class="p">]</span><span class="o">*</span><span class="p">((</span><span class="n">n</span><span class="o">/</span><span class="mi">6</span><span class="o">-</span><span class="p">(</span><span class="n">k</span><span class="o">*</span><span class="n">k</span><span class="p">)</span><span class="o">/</span><span class="mi">6</span><span class="o">-</span><span class="mi">1</span><span class="p">)</span><span class="o">/</span><span class="n">k</span><span class="o">+</span><span class="mi">1</span><span class="p">)</span> <span class="n">sieve</span><span class="p">[(</span><span class="n">k</span><span class="o">*</span><span class="n">k</span><span class="o">+</span><span class="mi">4</span><span class="o">*</span><span class="n">k</span><span class="o">-</span><span class="mi">2</span><span class="o">*</span><span class="n">k</span><span class="o">*</span><span class="p">(</span><span class="n">i</span><span class="o">&</span><span class="mi">1</span><span class="p">))</span><span class="o">/</span><span class="mi">3</span><span class="p">::</span><span class="mi">2</span><span class="o">*</span><span class="n">k</span><span class="p">]</span><span class="o">=</span><span class="p">[</span><span class="bp">False</span><span class="p">]</span><span class="o">*</span><span class="p">((</span><span class="n">n</span><span class="o">/</span><span class="mi">6</span><span class="o">-</span><span class="p">(</span><span class="n">k</span><span class="o">*</span><span class="n">k</span><span class="o">+</span><span class="mi">4</span><span class="o">*</span><span class="n">k</span><span class="o">-</span><span class="mi">2</span><span class="o">*</span><span class="n">k</span><span class="o">*</span><span class="p">(</span><span class="n">i</span><span class="o">&</span><span class="mi">1</span><span class="p">))</span><span class="o">/</span><span class="mi">6</span><span class="o">-</span><span class="mi">1</span><span class="p">)</span><span class="o">/</span><span class="n">k</span><span class="o">+</span><span class="mi">1</span><span class="p">)</span> <span class="k">return</span> <span class="p">[</span><span class="mi">2</span><span class="p">,</span><span class="mi">3</span><span class="p">]</span> <span class="o">+</span> <span class="p">[</span><span class="mi">3</span><span class="o">*</span><span class="n">i</span><span class="o">+</span><span class="mi">1</span><span class="o">|</span><span class="mi">1</span> <span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">xrange</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span><span class="n">n</span><span class="o">/</span><span class="mi">3</span><span class="o">-</span><span class="n">correction</span><span class="p">)</span> <span class="k">if</span> <span class="n">sieve</span><span class="p">[</span><span class="n">i</span><span class="p">]]</span>
登录后复制
对于治病救人,我有时候有不同的理解,这很正常。
看到有人在吃屎,安全的做法是告诉他慢慢吃别噎着,加点油盐酱醋味精啥的,再端碗鸡汤,然后皆大欢喜,大家都很高兴。
虽然我倡导diversity,但我对这种行为异常鄙夷,我认为,你他妈应该立即打断啊。。。(当然,我知道有人是嗜粪的,所以因为我每次都打断所以有时候我也会犯错,但这种情况还是罕见的)
如果不是我必须表现得礼貌,我会说你写的这些东西还不如一坨屎
- 这是一段根本没法运行的代码,whese is is_prime()???
- 100000用1e6,我不知道你是什么心态?如果装逼,则可以叉出去了,如果不知道可以直接用200000.....尼玛32位数4294967296是常识吧?更应该叉出去了
- 乱取名字,竟敢覆盖list,真坑爹
- 好好的math.sqrt不用,用什么**0.5,什么心态
- 你那么喜欢所谓“黑科技”优化,为啥不用xrange?
- 乱用列表推倒,列表没推倒,你自己先倒了
- 你那么喜欢省代码行数讨厌回车,我推荐你用c语言,用那个可以写成一行
- 哪有这样写python的?
- 你觉得跑了21秒的程序有必要写清楚型号配置吗?
- 如果不是我必须表现得礼貌,我会问你觉得这种垃圾有必要双配置对比吗?
- 你们可以说他不懂,但这种眼高手低,搞两个名词,乱来一气的做法,跟《小时代》受众在性质上也差不离,人家郭敬明的粉丝也不懂啦
如果你随便玩玩,忽略这些话
如果想好好学,那么建议摆正心态,脚踏实地,不要走火入魔,误入歧途。想要成功,要牢记3点,1是切忌南辕北辙,2是不能说太多, 如果不要求准确值的话,用估算好了,x/ln(x)
以前为了算某个群号,曾经计算了八千万以内的素数,花了两秒钟。群里有个人0.2秒,觉得很牛逼,始终没明白是怎么做的。
我的做法很简单,给出八千万个bool(其实可以去掉偶数)。一开始拿出2,把2的倍数都true了。下一个false的是3,把3的倍数都true了。下一个false的是5,把5的倍数都true了。一只做到完。
本站声明
本文内容由网友自发贡献,版权归原作者所有,本站不承担相应法律责任。如您发现有涉嫌抄袭侵权的内容,请联系admin@php.cn

热AI工具

Undresser.AI Undress
人工智能驱动的应用程序,用于创建逼真的裸体照片

AI Clothes Remover
用于从照片中去除衣服的在线人工智能工具。

Undress AI Tool
免费脱衣服图片

Clothoff.io
AI脱衣机

Video Face Swap
使用我们完全免费的人工智能换脸工具轻松在任何视频中换脸!

热门文章
刺客信条阴影:贝壳谜语解决方案
3 周前
By DDD
Windows 11 KB5054979中的新功能以及如何解决更新问题
2 周前
By DDD
在哪里可以找到原子中的起重机控制钥匙卡
3 周前
By DDD
刺客信条阴影 - 如何找到铁匠,解锁武器和装甲定制
1 个月前
By DDD
<🎜>:死铁路 - 如何完成所有挑战
3 周前
By DDD

热工具

记事本++7.3.1
好用且免费的代码编辑器

SublimeText3汉化版
中文版,非常好用

禅工作室 13.0.1
功能强大的PHP集成开发环境

Dreamweaver CS6
视觉化网页开发工具

SublimeText3 Mac版
神级代码编辑软件(SublimeText3)

热门话题

Linux终端中查看Python版本时遇到权限问题的解决方法当你在Linux终端中尝试查看Python的版本时,输入python...

如何在10小时内教计算机小白编程基础?如果你只有10个小时来教计算机小白一些编程知识,你会选择教些什么�...

在使用Python的pandas库时,如何在两个结构不同的DataFrame之间进行整列复制是一个常见的问题。假设我们有两个Dat...

使用FiddlerEverywhere进行中间人读取时如何避免被检测到当你使用FiddlerEverywhere...

Uvicorn是如何持续监听HTTP请求的?Uvicorn是一个基于ASGI的轻量级Web服务器,其核心功能之一便是监听HTTP请求并进�...

在Python中,如何通过字符串动态创建对象并调用其方法?这是一个常见的编程需求,尤其在需要根据配置或运行...
