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    <![CDATA[<p>你好，欢迎访问个人主页！</p><p>擅长密码学，安全分析，数字水印等技术。</p><p>你可以联系我通过:findmykexin@gmail.com或者知乎私信。</p><p>我的知乎链接：<a href="https://www.zhihu.com/people/su-chi-dan-dao" rel="noopener noreferrer" target="_blank">苏迟但到 - 知乎 (zhihu.com)</a></p><p>我的github链接：<a href="https://github.com/kexinoh" rel="noopener noreferrer" target="_blank">kexinoh</a></p>]]>
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    <title>以m比特为密钥的RSA加密和解密n比特长的数据的时间复杂度分别为多少？</title>
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    <pubDate>Sat, 30 Sep 2023 09:05:23 GMT</pubDate>
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      <![CDATA[<p data-pid="-QhHCzJK">期望时间复杂度计算为：</p><p data-pid="stZG0eLc"><img src="https://www.zhihu.com/equation?tex=f%28n%2Cm%29%3D%5Bn%2Fm%5D%2A%28%5Cfrac%7Bm%7D%7B2%7D%2Af%28m%2C%2A%29%2Bm%2Af%28m%2C%2A%29%29" alt="f(n,m)=[n/m]*(\frac{m}{2}*f(m,*)+m*f(m,*))" eeimg="1"/> </p><p data-pid="GpooLoqI">其中 <img src="https://www.zhihu.com/equation?tex=f%28m%2C%2A%29" alt="f(m,*)" eeimg="1"/> 指对m比特的进行相乘的复杂度。</p><blockquote data-pid="TwgfFGU_">在计算机进行乘法计算的时候，传统的计算方法采用了类似竖式计算的方法——正如我们在小学课堂里面所学到的那样，因此时间复杂度是O(n ^ 2)。在进行大数（比如说数字位数超过2048）乘法计算的时候，目前的做法是Schönhage–Strassen算法，该算法利用快速傅里叶变换，将算法的复杂度降到O(n log n log log n)。<br/><br/>最近，Harvey 和Van Der Hoeven 在HAL贴出了两人合著的新论文“Integer<a href="https://www.zhihu.com/search?q=multiplication%20in%20time&amp;search_source=Entity&amp;hybrid_search_source=Entity&amp;hybrid_search_extra=%7B%22sourceType%22%3A%22article%22%2C%22sourceId%22%3A%2262143130%22%7D" class="internal">multiplication in time</a>O(n log n)”，宣称已经发现时间复杂度为O(n log n)的整数乘法算法。<br/>作者：留德华叫兽<br/>链接：<a href="https://zhuanlan.zhihu.com/p/62143130" class="internal"><span class="invisible">https://</span><span class="visible">zhuanlan.zhihu.com/p/62</span><span class="invisible">143130</span><span class="ellipsis"></span></a><br/>来源：知乎<br/>著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。</blockquote><p data-pid="s_t5YnaT">在通常意义下，为 <img src="https://www.zhihu.com/equation?tex=f%28n%2Cm%29%3D%5Bn%2Fm%5D%2A%281.5%2Am%5E3%29%3DO%28m%5E2%29" alt="f(n,m)=[n/m]*(1.5*m^3)=O(m^2)" eeimg="1"/> </p><p data-pid="1L9f1p16">在极限下，为 <img src="https://www.zhihu.com/equation?tex=f%28n%2Cm%29%3D%5Bn%2Fm%5D%2A%281.5%2Am%5E2+%5Clog+m%29%3DO%28n%2Am%5Clog+m%29" alt="f(n,m)=[n/m]*(1.5*m^2 \log m)=O(n*m\log m)" eeimg="1"/> </p><p data-pid="sTT7WrZY">空间复杂度包含：</p><p data-pid="hRlmxxUF"><img src="https://www.zhihu.com/equation?tex=f%28n%2Cm%29%3Dn%2Bk%2Am%3DO%28m%2Bn%29" alt="f(n,m)=n+k*m=O(m+n)" eeimg="1"/> </p><p data-pid="Z-Y6XJTS">其中k是常数项，具体取决于算法。</p><p></p>]]>
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