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 中国剩余定理，
 引入
 从 \(n\) 个不同元素中取出 \(m\) 个元素的组合数，即 \(C_n^m\)，是从 \(n\) 个不同元素中，取出 \(m\) 个元素所形成的组合个数。
 我们需要使用各种方式求 \(C_n^m\)，每个方式难度，速度和空间都有不同。"><meta property="og:type" content="article"><meta property="og:title" content="求组合数"><meta property="og:url" content="https://schtonn.github.io/blog/posts/combination/index.html"><meta property="og:site_name" content="schtonn"><meta property="og:description" content="前置知识
 中国剩余定理，
 引入
 从 \(n\) 个不同元素中取出 \(m\) 个元素的组合数，即 \(C_n^m\)，是从 \(n\) 个不同元素中，取出 \(m\) 个元素所形成的组合个数。
 我们需要使用各种方式求 \(C_n^m\)，每个方式难度，速度和空间都有不同。"><meta property="og:locale" content="en_US"><meta property="article:published_time" content="2021-01-01T12:26:59.000Z"><meta property="article:modified_time" content="2022-10-19T15:02:06.673Z"><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/combination/"><script id="page-configurations">CONFIG.page={sidebar:"",isHome:!1,isPost:!0,lang:"en"}</script><title>求组合数 | 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 i{right:initial}</style></noscript></head><body itemscope itemtype="http://schema.org/WebPage"><div class="container use-motion"><div class="headband"></div><header class="header" itemscope itemtype="http://schema.org/WPHeader"><div class="header-inner"><div class="site-brand-container"><div class="site-nav-toggle"><div class="toggle" aria-label="Toggle navigation bar"><span class="toggle-line toggle-line-first"></span><span class="toggle-line toggle-line-middle"></span><span class="toggle-line toggle-line-last"></span></div></div><div class="site-meta"><a href="/blog/" class="brand" rel="start"><span class="logo-line-before"><i></i></span><h1 class="site-title">schtonn</h1><span class="logo-line-after"><i></i></span></a><p class="site-subtitle" itemprop="description">schtonn</p></div><div class="site-nav-right"><div class="toggle popup-trigger"></div></div></div><nav class="site-nav"><ul id="menu" class="menu"><li class="menu-item menu-item-home"><a href="/blog/" rel="section"><i 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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"> 求组合数</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: 2021-Jan-01 20:26:59" itemprop="dateCreated datePublished" datetime="2021-01-01T20:26:59+08:00">2021-Jan-01</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/combination/#valine-comments" itemprop="discussionUrl"><span class="post-comments-count valine-comment-count" data-xid="/blog/posts/combination/" itemprop="commentCount"></span></a></span></div></header><div class="post-body" itemprop="articleBody"><h2 id="前置知识">前置知识</h2><p>中国剩余定理，</p><h2 id="引入">引入</h2><p>从 <span class="math inline">\(n\)</span> 个不同元素中取出 <span class="math inline">\(m\)</span> 个元素的组合数，即 <span class="math inline">\(C_n^m\)</span>，是从 <span class="math inline">\(n\)</span> 个不同元素中，取出 <span class="math inline">\(m\)</span> 个元素所形成的组合个数。</p><p>我们需要使用各种方式求 <span class="math inline">\(C_n^m\)</span>，每个方式难度，速度和空间都有不同。</p><a id="more"></a><h2 id="朴素算法">朴素算法</h2><p>首先，求组合数有一个基本公式：</p><p><span class="math display">\[\large C_n^m=\dfrac{P_n^m}{P_m}=\dfrac{n!}{m!(n-m)!}\]</span></p><p><span class="math display">\[\large C_n^0=1\]</span></p><p>我们的朴素算法，就是每遇到一个组合数，都直接套用这个公式求阶乘。</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><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br></pre></td><td class="code"><pre><span class="line"><span class="function"><span class="keyword">int</span> <span class="title">fact</span><span class="params">(<span class="keyword">int</span> x)</span></span>&#123;</span><br><span class="line">    <span class="keyword">int</span> ans=<span class="number">1</span>;</span><br><span class="line">    <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">2</span>;i&lt;=x;i++)&#123;</span><br><span class="line">        ans*=i;</span><br><span class="line">    &#125;</span><br><span class="line">    <span class="keyword">return</span> ans;</span><br><span class="line">&#125;</span><br><span class="line"><span class="function"><span class="keyword">int</span> <span class="title">C</span><span class="params">(<span class="keyword">int</span> n,<span class="keyword">int</span> m)</span></span>&#123;</span><br><span class="line">    <span class="keyword">if</span>(m==<span class="number">0</span>)<span class="keyword">return</span> <span class="number">1</span>;</span><br><span class="line">    <span class="keyword">return</span> fact(n)/(fact(m)*fact(n-m));</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure><p>这个算法适合 <span class="math inline">\(n,m\leq15\)</span>，时间复杂度极高，但是实现难度为零。</p><h2 id="杨辉三角">杨辉三角</h2><p>如果把组合数转换成实际问题，可以考虑每一步选定一个元素，那么一个组合数一定可以由更小的组合数转移而来。</p><p>具体思路为，<span class="math inline">\(C_n^m\)</span> 是 <span class="math inline">\(n\)</span> 中选 <span class="math inline">\(m\)</span> 个，考虑从 <span class="math inline">\(n-1\)</span> 个中选的所有情况，一部分是选新加入的这第 <span class="math inline">\(n\)</span> 个，一部分是不选。</p><p>那么可以列出这样一个式子：</p><p><span class="math display">\[\large C_n^m=C_{n-1}^{m-1}+C_{n-1}^m\]</span></p><p>（当然，通过数学的方式推也能推出来）</p><p>这时我们发现，如果把这个式子列成表，竖着拎起来，那就正好是一个杨辉三角（每一位等于上面左右两位的和）。</p><p>那么根据这个式子，就可以写出一种动态规划算法。</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><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">int</span> C[N][M];</span><br><span class="line"><span class="function"><span class="keyword">void</span> <span class="title">calc</span><span class="params">(<span class="keyword">int</span> n,<span class="keyword">int</span> m)</span></span>&#123;</span><br><span class="line">    <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">0</span>;i&lt;=n;i++)&#123;</span><br><span class="line">        f[i][i]=<span class="number">1</span>;</span><br><span class="line">        f[i][<span class="number">0</span>]=<span class="number">1</span>;</span><br><span class="line">    &#125;</span><br><span class="line">    <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">1</span>;i&lt;=n;i++)&#123;</span><br><span class="line">        <span class="keyword">for</span>(<span class="keyword">int</span> j=<span class="number">1</span>;j&lt;i;j++)&#123;</span><br><span class="line">            f[i][j]=f[i<span class="number">-1</span>][j]+f[i<span class="number">-1</span>][j<span class="number">-1</span>];</span><br><span class="line">        &#125;</span><br><span class="line">    &#125;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure><p>这个算法适合 <span class="math inline">\(n,m\leq10000\)</span>，时间复杂度适中，需要一个数组，而且实现难度也很低。</p><h2 id="lucas-定理">Lucas 定理</h2><p>求 <span class="math inline">\(C_m^n\bmod p\)</span>，<span class="math inline">\(p\)</span> 是质数。</p><p>对质数 <span class="math inline">\(p\)</span>，有 Lucas 定理： <span class="math display">\[\large C_n^m\bmod p=C_{\lfloor n/p\rfloor}^{\lfloor m/p\rfloor}\cdot C_{n\bmod p}^{m\bmod p}\bmod p\]</span></p><h3 id="证明">证明</h3><p>（摘自 <a href="https://oi-wiki.org/math/lucas/#lucas_1" target="_blank" rel="noopener">OI-wiki</a>）</p><p>考虑 <span class="math inline">\(C_p^n\bmod p\)</span> 的取值，注意到 <span class="math inline">\(C_p^n=\dfrac{p!}{n!(p-n)!}\)</span>，分子的质因子分解中 <span class="math inline">\(p\)</span> 次项恰为 <span class="math inline">\(1\)</span>，因此只有当 <span class="math inline">\(n=0\)</span> 或 <span class="math inline">\(n=p\)</span> 的时候 <span class="math inline">\(n!(p-n)!\)</span> 的质因子分解中含有 <span class="math inline">\(p\)</span>。进而我们可以得出： <span class="math display">\[\begin{aligned}(a+b)^p&amp;=\sum_{n=0}^p C_p^n a^nb^{p-n}\\ &amp;\equiv\sum_{n=0}^p[n=0\vee n=p]a^nb^{p-n}\\ &amp;\equiv a^p+b^p\pmod p\end{aligned}\]</span></p><p>注意过程中没有用到费马小定理，因此这一推导不仅适用于整数，亦适用于多项式。因此我们可以考虑二项式 <span class="math inline">\(f(x)=(ax^n+bx^m)^p\bmod p\)</span> 的结果： <span class="math display">\[\begin{aligned}(ax^n+bx^m)^p&amp;\equiv a^px^{pn}+b^px^{pm}\\ &amp;\equiv ax^{pn}+bx^{pm}\\ &amp;\equiv f(x^p)\end{aligned}\]</span></p><p>考虑二项式 <span class="math inline">\((1+x)^n\mod p\)</span>，那么 <span class="math inline">\(C_n^m\)</span> 就是求其在 <span class="math inline">\(x^m\)</span> 次项的取值。使用上述引理，我们可以得到： <span class="math display">\[\begin{aligned}(1+x)^n&amp;\equiv(1+x)^{p\lfloor n/p\rfloor}(1+x)^{n\bmod p}\\&amp;\equiv(1+x^p)^{\lfloor n/p\rfloor}(1+x)^{n\bmod p}\end{aligned}\]</span></p><p>注意前者只有在 <span class="math inline">\(p\)</span> 的倍数位置才有取值，而后者最高次项为 <span class="math inline">\(n\bmod p\leq p-1\)</span>，因此这两部分的卷积在任何一个位置只有最多一种方式贡献取值，即在前者部分取 <span class="math inline">\(p\)</span> 的倍数次项，后者部分取剩余项，即 <span class="math inline">\(C_n^m\bmod p=C_{\lfloor n/p\rfloor}^{\lfloor m/p\rfloor}\cdot C_{n\bmod p}^{m\bmod p}\bmod p\)</span>。</p><h3 id="求解">求解</h3><p>观察 Lucas 定理，可知 <span class="math inline">\(n\bmod p\)</span> 和 <span class="math inline">\(m\bmod p\)</span> 一定是小于 <span class="math inline">\(p\)</span> 的数，可以直接求解 <span class="math inline">\(C_{\lfloor n/p\rfloor}^{\lfloor m/p\rfloor}\)</span>，递归使用 Lucas 定理。</p><p><span class="math inline">\(C_{n\bmod p}^{m\bmod p}\)</span> 则由于 <span class="math inline">\(p\)</span> 是质数，利用费马小定理转化成 <span class="math inline">\(n!\)</span> 乘 <span class="math inline">\(m!(n-m)!\)</span> 的逆元，也就是 <span class="math inline">\(m!(n-m)!^{p-2}\)</span>。</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><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br></pre></td><td class="code"><pre><span class="line"><span class="function"><span class="keyword">long</span> <span class="keyword">long</span> <span class="title">pow</span><span class="params">(<span class="keyword">long</span> <span class="keyword">long</span> a,<span class="keyword">long</span> <span class="keyword">long</span> b,<span class="keyword">long</span> <span class="keyword">long</span> m)</span></span>&#123;</span><br><span class="line">    <span class="keyword">long</span> <span class="keyword">long</span> ans=<span class="number">1</span>;</span><br><span class="line">    a%=m;</span><br><span class="line">    <span class="keyword">while</span>(b)&#123;</span><br><span class="line">        <span class="keyword">if</span>(b&amp;<span class="number">1</span>)ans=(ans%m)*(a%m)%m;</span><br><span class="line">        b/=<span class="number">2</span>;</span><br><span class="line">        a=(a%m)*(a%m)%m;</span><br><span class="line">    &#125;</span><br><span class="line">    ans%=m;</span><br><span class="line">    <span class="keyword">return</span> ans;</span><br><span class="line">&#125;</span><br><span class="line"><span class="function"><span class="keyword">long</span> <span class="keyword">long</span> <span class="title">inv</span><span class="params">(<span class="keyword">long</span> <span class="keyword">long</span> x,<span class="keyword">long</span> <span class="keyword">long</span> p)</span></span>&#123;</span><br><span class="line">    <span class="keyword">return</span> <span class="built_in">pow</span>(x,p<span class="number">-2</span>,p);</span><br><span class="line">&#125;</span><br><span class="line"><span class="function"><span class="keyword">long</span> <span class="keyword">long</span> <span class="title">C</span><span class="params">(<span class="keyword">long</span> <span class="keyword">long</span> n,<span class="keyword">long</span> <span class="keyword">long</span> m,<span class="keyword">long</span> <span class="keyword">long</span> p)</span></span>&#123;</span><br><span class="line">    <span class="keyword">if</span>(m&gt;n)<span class="keyword">return</span> <span class="number">0</span>;</span><br><span class="line">    <span class="keyword">long</span> <span class="keyword">long</span> up=<span class="number">1</span>,down=<span class="number">1</span>;</span><br><span class="line">    <span class="keyword">for</span>(<span class="keyword">int</span> i=n-m+<span class="number">1</span>;i&lt;=n;i++)up=up*i%p;</span><br><span class="line">    <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">1</span>;i&lt;=m;i++)down=down*i%p;</span><br><span class="line">    <span class="keyword">return</span> up*inv(down,p)%p;</span><br><span class="line">&#125;</span><br><span class="line"><span class="function"><span class="keyword">long</span> <span class="keyword">long</span> <span class="title">Lucas</span><span class="params">(<span class="keyword">long</span> <span class="keyword">long</span> n,<span class="keyword">long</span> <span class="keyword">long</span> m,<span class="keyword">long</span> <span class="keyword">long</span> p)</span></span>&#123;</span><br><span class="line">  <span class="keyword">if</span>(m==<span class="number">0</span>)<span class="keyword">return</span> <span class="number">1</span>;</span><br><span class="line">  <span class="keyword">return</span>(Lucas(n/p,m/p,p)*C(n%p,m%p,p))%p;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure><p>这个算法适合 <span class="math inline">\(n,m\leq10^8,p\leq10^4\)</span>，速度较快，理解后难度也不高。</p><h2 id="拓展-lucas-定理">拓展 Lucas 定理</h2><p>适用于求 <span class="math inline">\(C_m^n\bmod p\)</span>，<span class="math inline">\(p\)</span> 不一定是质数。 ### 分解 将 <span class="math inline">\(p\)</span> 分解质因数： <span class="math display">\[p=q_1^{a_1}\cdot q_2^{a_2}\cdots q_r^{a_r}=\prod_{i-1}^rq_i^{a_i}\]</span></p><p>对于任意 <span class="math inline">\(i,j\)</span>，有 <span class="math inline">\(p_i^{a_i}\)</span> 与 <span class="math inline">\(p_j^{a_j}\)</span> 互质，所以可以构造 <span class="math inline">\(r\)</span> 个同余方程：</p><p><span class="math display">\[ \left\{ \begin{aligned} a_1&amp;\equiv C_n^m\pmod {q_1^{a_1}}\\ a_2&amp;\equiv C_n^m\pmod {q_2^{a_2}}\\ &amp;\cdots\\ a_r&amp;\equiv C_n^m\pmod {q_r^{a_r}} \end{aligned} \right. \]</span></p><p>我们发现，在求出 <span class="math inline">\(a_i\)</span> 后，就可以用中国剩余定理求解出 <span class="math inline">\(C_n^m\)</span>。</p><h3 id="转化">转化</h3><p>根据同余的定义，<span class="math inline">\(a_i=C_n^m\bmod q_i^{a_i}\)</span>，问题转化成，求 <span class="math inline">\(C_n^m\bmod\)</span></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/combination/" title="求组合数">https://schtonn.github.io/blog/posts/combination/</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/ex-KMP/" rel="prev" title="拓展 KMP"><i class="fa fa-chevron-left"></i> 拓展 KMP</a></div><div class="post-nav-item"> <a href="/blog/posts/CRT/" rel="next" title="中国剩余定理">中国剩余定理<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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