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 快速幂是一种效率极高，以至于题目中不得不取模防止数据越界的求幂算法。计算 \(a^n\) 的复杂度为 \(O(\log n)\)。"><meta property="og:type" content="article"><meta property="og:title" content="快速幂"><meta property="og:url" content="https://schtonn.github.io/blog/posts/quick-pow/index.html"><meta property="og:site_name" content="schtonn"><meta property="og:description" content="引入
 快速幂是一种效率极高，以至于题目中不得不取模防止数据越界的求幂算法。计算 \(a^n\) 的复杂度为 \(O(\log n)\)。"><meta property="og:locale" content="en_US"><meta property="article:published_time" content="2020-11-17T13:33:40.000Z"><meta property="article:modified_time" content="2022-10-19T15:02:06.666Z"><meta property="article:author" content="Alex"><meta property="article:tag" content="math"><meta name="twitter:card" content="summary"><link rel="canonical" href="https://schtonn.github.io/blog/posts/quick-pow/"><script id="page-configurations">CONFIG.page={sidebar:"",isHome:!1,isPost:!0,lang:"en"}</script><title>快速幂 | schtonn</title><noscript><style>.sidebar-inner,.use-motion .brand,.use-motion .collection-header,.use-motion .comments,.use-motion .menu-item,.use-motion .pagination,.use-motion .post-block,.use-motion .post-body,.use-motion .post-header{opacity:initial}.use-motion .site-subtitle,.use-motion .site-title{opacity:initial;top:initial}.use-motion .logo-line-before i{left:initial}.use-motion .logo-line-after 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itemprop="description" content="blog"></span><script type="text/javascript" src="/blog/js/md5.js"></script><script></script><script>document.oncopy=function(e){window.event&&(e=window.event);try{var t=e.srcElement;return"INPUT"==t.tagName&&"text"==t.type.toLowerCase()||"TEXTAREA"==t.tagName}catch(e){return!1}}</script><span hidden itemprop="publisher" itemscope itemtype="http://schema.org/Organization"><meta itemprop="name" content="schtonn"></span><header class="post-header"><h1 class="post-title" itemprop="name headline"> 快速幂</h1><div class="post-meta"><span class="post-meta-item"><span class="post-meta-item-icon"><i class="fa fa-calendar-o"></i></span> <span class="post-meta-item-text">Posted on</span> <time title="Created: 2020-Nov-17 21:33:40" itemprop="dateCreated datePublished" datetime="2020-11-17T21:33:40+08:00">2020-Nov-17</time></span><span class="post-meta-item"><span class="post-meta-item-icon"><i class="fa fa-calendar-check-o"></i></span> <span class="post-meta-item-text">Edited on</span> <time title="Modified: 2022-Oct-19 23:02:06" itemprop="dateModified" datetime="2022-10-19T23:02:06+08:00">2022-Oct-19</time></span><span class="post-meta-item"><span class="post-meta-item-icon"><i class="fa fa-comment-o"></i></span> <span class="post-meta-item-text">Valine:</span><a title="valine" href="/blog/posts/quick-pow/#valine-comments" itemprop="discussionUrl"><span class="post-comments-count valine-comment-count" data-xid="/blog/posts/quick-pow/" itemprop="commentCount"></span></a></span></div></header><div class="post-body" itemprop="articleBody"><h2 id="引入">引入</h2><p>快速幂是一种效率极高，以至于题目中不得不取模防止数据越界的求幂算法。计算 <span class="math inline">\(a^n\)</span> 的复杂度为 <span class="math inline">\(O(\log n)\)</span>。</p><a id="more"></a><h2 id="思想">思想</h2><p>想一想，如果让你用一种简便的方法计算 <span class="math inline">\(2^{22}\)</span>，你会怎么做？</p><p><span class="math inline">\(2^{22}\)</span> 可以转化成 <span class="math inline">\(2^1*2^2*2^4*2^8*2^7\)</span>, 其中的每一项都能用前项步骤直接转化而来，如 <span class="math inline">\(2^2=2^1*2^1\)</span>，<span class="math inline">\(2^7=2^4*2^2*2^1\)</span>，只用做 <span class="math inline">\(\log\)</span> 等级次数的计算。</p><h2 id="算法">算法</h2><p>使用上述思路，我们先考虑 <span class="math inline">\(n=2^k\)</span> 的情况。</p><p>则每次循环 <span class="math inline">\(a\)</span> 自乘，<span class="math inline">\(n\)</span> 除以二，有代码：</p><figure class="highlight cpp"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">int</span> ans=<span class="number">1</span>;</span><br><span class="line"><span class="keyword">while</span>(n)&#123;</span><br><span class="line">    a*=a;</span><br><span class="line">    n/=<span class="number">2</span>;</span><br><span class="line">&#125;</span><br><span class="line">ans=a;</span><br></pre></td></tr></table></figure><p><span class="math inline">\(n\neq2^k\)</span> 时，要考虑 <span class="math inline">\(n\)</span> 是单数的情况，这时应该 <code>ans*=a</code>，刚好削减 <span class="math inline">\(n\)</span> 单出来的余数，使我们惊叹该算法精妙的设计：</p><figure class="highlight cpp"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">int</span> ans=<span class="number">1</span>;</span><br><span class="line"><span class="keyword">while</span>(n)&#123;</span><br><span class="line">    <span class="keyword">if</span>(n%<span class="number">1</span>==<span class="number">1</span>)ans*=a;</span><br><span class="line">    a*=a;</span><br><span class="line">    n/=<span class="number">2</span>;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure><p>同时我们还可以使用位运算进一步优化：</p><figure class="highlight cpp"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">int</span> ans=<span class="number">1</span>;</span><br><span class="line"><span class="keyword">while</span>(n)&#123;</span><br><span class="line">    <span class="keyword">if</span>(n&amp;<span class="number">1</span>)ans*=a;</span><br><span class="line">    a*=a;</span><br><span class="line">    n&gt;&gt;=<span class="number">1</span>;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure><p>完结</p></div><div><ul class="post-copyright"><li class="post-copyright-author"> <strong>Post author:</strong> Alex</li><li class="post-copyright-link"> <strong>Post link:</strong> <a href="https://schtonn.github.io/blog/posts/quick-pow/" title="快速幂">https://schtonn.github.io/blog/posts/quick-pow/</a></li><li class="post-copyright-license"> <strong>Copyright Notice:</strong> All articles in this blog are licensed under<a href="https://creativecommons.org/licenses/by-nc-sa/4.0/" rel="noopener" target="_blank"><i class="fa fa-fw fa-creative-commons"></i> BY-NC-SA</a> unless stating additionally.</li></ul></div><footer class="post-footer"><div class="post-tags"><a href="/blog/tags/math/" rel="tag"><i class="fa fa-tag"></i> math</a></div><div class="post-nav"><div class="post-nav-item"><a href="/blog/posts/num-convert/" rel="prev" title="进制转换"><i class="fa fa-chevron-left"></i> 进制转换</a></div><div class="post-nav-item"> <a href="/blog/posts/KMP/" rel="next" title="KMP">KMP<i class="fa fa-chevron-right"></i></a></div></div></footer></article></div><div class="comments" id="valine-comments"></div><script>
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