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 矩阵快速幂，扩展欧几里得求逆元，基本的找规律能力。
 题面
 (luogu P2020)"><meta property="og:type" content="article"><meta property="og:title" content="兔农"><meta property="og:url" content="https://schtonn.github.io/blog/posts/rabbit/index.html"><meta property="og:site_name" content="schtonn"><meta property="og:description" content="前置知识
 矩阵快速幂，扩展欧几里得求逆元，基本的找规律能力。
 题面
 (luogu P2020)"><meta property="og:locale" content="en_US"><meta property="article:published_time" content="2021-04-13T14:27:26.970Z"><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/rabbit/"><script id="page-configurations">CONFIG.page={sidebar:"",isHome:!1,isPost:!0,lang:"en"}</script><title>兔农 | 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 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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"> 兔农</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-Apr-13 22:27:26" itemprop="dateCreated datePublished" datetime="2021-04-13T22:27:26+08:00">2021-Apr-13</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/rabbit/#valine-comments" itemprop="discussionUrl"><span class="post-comments-count valine-comment-count" data-xid="/blog/posts/rabbit/" 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><a href="/blog/posts/matrix-pow">矩阵快速幂</a>，扩展欧几里得求逆元，基本的找规律能力。</p><h2 id="题面">题面</h2><p>(<a href="https://www.luogu.com.cn/problem/P2020" target="_blank" rel="noopener">luogu P2020</a>)<a id="more"></a> ### 问题描述 农夫栋栋近年收入不景气，正在他发愁如何能多赚点钱时，他听到隔壁的小朋友在讨论兔子繁殖的问题。</p><p>问题是这样的：第一个月初有一对刚出生的小兔子，经过两个月长大后，这对兔子从第三个月开始，每个月初生一对小兔子。新出生的小兔子生长两个月后又能每个月生出一对小兔子。问第n个月有多少对兔子？</p><p>聪明的你可能已经发现，第n个月的兔子数正好是第n个Fibonacci(斐波那契)数。栋栋不懂什么是Fibonacci数，但他也发现了规律：第i+2个月的兔子对数等于第i个月的兔子对数加上第i+1个月的兔子对数。前几个月的兔子对数依次为：</p><blockquote><p>1 1 2 3 5 8 13 21 34 …</p></blockquote><p>栋栋发现越到后面兔子对数增长的越快，期待养兔子一定能赚大钱，于是栋栋在第一个月初买了一对小兔子开始饲养。</p><p>每天，栋栋都要给兔子们喂食，兔子们吃食时非常特别，总是每k对兔子围成一圈，最后剩下的不足k对的围成一圈，由于兔子特别害怕孤独，从第三个月开始，如果吃食时围成某一个圈的只有一对兔子，这对兔子就会很快死掉。</p><p>我们假设死去的总是刚出生的兔子，那么每个月的兔子对数仍然是可以计算的。例如，当k=7时，前几个月的兔子对数依次为：</p><blockquote><p>1 1 2 3 5 <strong>7</strong> 12 19 31 <strong>49</strong> 80 …</p></blockquote><p>给定n，你能帮助栋栋计算第n个月他有多少对兔子么？由于答案可能非常大，你只需要告诉栋栋第n个月的兔子对数除p的余数即可。 ### 输入格式 从文件rabbit.in中读入数据。</p><p>输入一行，包含三个正整数n, k, p。 ### 输出格式 输出到文件rabbit.out中。</p><p>输出一行，包含一个整数，表示栋栋第n个月的兔子对数除p的余数。 ### 样例 输入1：</p><figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br></pre></td><td class="code"><pre><span class="line">6 7 100</span><br></pre></td></tr></table></figure><p>输出1：</p><figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br></pre></td><td class="code"><pre><span class="line">7</span><br></pre></td></tr></table></figure><p>输入2：</p><figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br></pre></td><td class="code"><pre><span class="line">7 7 5</span><br></pre></td></tr></table></figure><p>输出2：</p><figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br></pre></td><td class="code"><pre><span class="line">2</span><br></pre></td></tr></table></figure><h2 id="题意">题意</h2><p>首先理解一下题意：</p><p><span class="math display">\[F_i=\left\{ \begin{array}{lr} F_{i-1}+F_{i-2}&amp; F_{i-1}+F_{i-2}\not\equiv1\pmod k\\ F_{i-1}+F_{i-2}-1&amp; F_{i-1}+F_{i-2}\equiv1\pmod k\\ \end{array} \right. \]</span></p><p>给定 <span class="math inline">\(n\)</span>，<span class="math inline">\(k\)</span>，<span class="math inline">\(p\)</span>，求 <span class="math inline">\(F(n)\bmod p\)</span>。</p><p>可以看出，它是一个略加修改的斐波那契数列，我们显然会想到使用矩阵快速幂，然而由于它会在不可预知的地方进行减一，简单地套用矩阵快速幂就无法达成目标。</p><h2 id="规律">规律</h2><p>遇到这种问题，可以先手推一下规律。</p><p>由于一个具有斐波那契数列性质地数列可以完全由一个二元组 <span class="math inline">\((a,b)\)</span> 固定，我们可以用一个二元组来表示当前状态。</p><p>例如从 <span class="math inline">\((1,1)\)</span> 开始，<span class="math inline">\(k=7\)</span>。</p><p><span class="math display">\[(1,1)-(1,2)-(2,3)-(3,5)-(5,1\rightarrow0)-(0,5)-(5,5)-\cdots\]</span></p><p>又如从 <span class="math inline">\((2,2)\)</span> 开始：</p><p><span class="math display">\[\begin{array}{lr}(2,2)-(2,4)-(4,6)-(6,3)-(3,2)-(2,5)-(5,0)-(0,5)-\\(5,5)-(5,3)-(3,1\rightarrow0)-(0,3)-(3,3)-\cdots\end{array}\]</span></p><p>可以得知，如果某个状态进入了 <span class="math inline">\((b,1)\)</span>，它就会被减一成为 <span class="math inline">\((b,0)\)</span>，接下来成为 <span class="math inline">\((0,b)-(b,b)\)</span>，这样我们就可以以 <span class="math inline">\((a,a)\)</span> 为节点，从某个 <span class="math inline">\((a,a)\)</span> 用简单矩阵快速幂推到下一个 <span class="math inline">\((b,1)\)</span>，然后减一并继续推进到 <span class="math inline">\((b,b)\)</span>，循环完成整个过程。</p><p>放上一个 <span class="math inline">\(k=7\)</span> 时的完整图表：</p><figure class="highlight plain"><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">(1,1)-(1,2)-(2,3)-(3,5)-(5,1);</span><br><span class="line">(2,2)-(2,4)-(4,6)-(6,3)-(3,2)-(2,5)-(5,0)-(5,5)-(5,3)-(3,1);</span><br><span class="line">(3,3)-(3,6)-(6,2)-(2,1);</span><br><span class="line">(4,4)-(4,1);</span><br><span class="line">(5,5)-(5,3)-(3,1);</span><br><span class="line">(6,6)-(6,5)-(5,4)-(4,2)-(2,6)-(6,1);</span><br></pre></td></tr></table></figure><p>注意到它没有那么友好，可能出现自环，两条路径的交集等情况，特别需要注意死循环，编写部分分代码时需要小心。</p><h2 id="k2-算法">k^2 算法</h2><p>令 <span class="math inline">\((i,i)\)</span> 经过 <span class="math inline">\(e_i\)</span> 步到达 <span class="math inline">\((g_i,1)\)</span>，可以在 <span class="math inline">\(O(k^2)\)</span> 的时间内维护 <span class="math inline">\(e_i\)</span> 和 <span class="math inline">\(g_i\)</span>。</p><p>还是以 <span class="math inline">\(k=7\)</span> 为例：</p><p><span class="math display">\[e:4,10,3,1,2,5\]</span> <span class="math display">\[g:5,3,2,4,3,6\]</span></p><p>即 <span class="math inline">\((1,1)\)</span> 经过 <span class="math inline">\(4\)</span> 步到达 <span class="math inline">\((5,1)\)</span>，<span class="math inline">\((2,2)\)</span> 经过 <span class="math inline">\(10\)</span> 步到达 <span class="math inline">\((3,1)\)</span>，etc.</p><p>求出这个之后我们就可以用矩阵快速幂来解决 <span class="math inline">\((a,a)\)</span> 到 <span class="math inline">\((b,b)\)</span> 的跳跃了。</p><p>这里面还有一些实现细节需要说明一下。由于原来斐波那契数列的递推矩阵实际上是做不到减一的，不过精通矩阵运算的人们都知道，我们只需要加一个常数项，然后稍微修改一下递推矩阵就好了。</p><p>原来的矩阵是这样的：</p><p><span class="math display">\[ \left[\begin{array}{lr}F_{n+1}\\F_n\end{array}\right]= \left[\begin{array}{lr}1&amp;1\\1&amp;0\end{array}\right]\times \left[\begin{array}{lr}F_n\\F_{n-1}\end{array}\right] \]</span></p><p>现在正常递推改成了这样： <span class="math display">\[ \left[\begin{array}{lr}F_{n+1}\\F_n\\1\end{array}\right]= \left[\begin{array}{lr}1&amp;1&amp;0\\1&amp;0&amp;0\\0&amp;0&amp;1\end{array}\right]\times \left[\begin{array}{lr}F_n\\F_{n-1}\\1\end{array}\right] \]</span></p><p>如果我们想减一，就可以这样： <span class="math display">\[ \left[\begin{array}{lr}F_n-1\\F_{n-1}\\1\end{array}\right]= \left[\begin{array}{lr}1&amp;0&amp;-1\\0&amp;1&amp;0\\0&amp;0&amp;1\end{array}\right]\times \left[\begin{array}{lr}F_n\\F_{n-1}\\1\end{array}\right] \]</span></p><p>如果想推一步的同时减一，就是这样： <span class="math display">\[ \left[\begin{array}{lr}F_{n+1}-1\\F_n\\1\end{array}\right]= \left[\begin{array}{lr}1&amp;1&amp;-1\\1&amp;0&amp;0\\0&amp;0&amp;1\end{array}\right]\times \left[\begin{array}{lr}F_n\\F_{n-1}\\1\end{array}\right] \]</span></p><p>将正常递推的那个矩阵设为 <span class="math inline">\(A\)</span>，推一步同时减一的矩阵设为 <span class="math inline">\(B\)</span>，最终的算式就是这样：</p><p><span class="math display">\[\cdots BA^{e_{g_1}-1}A^2BA^{e_1-1} \left[\begin{array}{lr}F_2\\F_1\\1\end{array}\right]\]</span></p><p>然后为了符号方便，我们记 <span class="math inline">\(F_0=0\)</span>，<span class="math inline">\(F_{-1}=1\)</span>，就可以将递推式写为：</p><p><span class="math display">\[\cdots BA^{e_{g_1}+1}BA^{e_1+1}\left[\begin{array}{lr}F_0\\F_{-1}\\1\end{array}\right]\]</span></p><p>这样，<span class="math inline">\(O(k^2)\)</span> 算法就妥妥地实现了。</p><h2 id="k-log-k-算法">k log k 算法</h2><p>原来求 <span class="math inline">\(e\)</span> 和 <span class="math inline">\(g\)</span> 地算法其实有很大的提升空间，如果你进行了深入的调试就会发现，对于大点的数据，它有 90% 以上的 <span class="math inline">\((a,a)\)</span> 都陷入了死循环，这说明我们做了大量无用的计算，何不换个角度想想？</p><p>如果我们不用二元组，而用正常方式，把斐波那契数列取模 <span class="math inline">\(k\)</span>，不管题目中的减一，写出来看看有什么发现？</p><figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br></pre></td><td class="code"><pre><span class="line">1 1 2 3 5 1 6 0 6 6 5 4 2 6 1 0 1 1</span><br></pre></td></tr></table></figure><p>发现它在这里进入了循环。</p><p>那么我们把数列乘上某个系数，并继续对 <span class="math inline">\(k\)</span> 取模，只写到第一次出现1的地方，再列出来看看有什么发现？</p><figure class="highlight plain"><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">1 1 2 3 5 1 6 0 6 6 5 4 2 6 1 0 1 1</span><br><span class="line">2 2 4 6 3 2 5 0 5 5 3 1</span><br><span class="line">3 3 2 6 1</span><br><span class="line">4 4 1</span><br><span class="line">5 5 3 1</span><br><span class="line">6 6 5 4 2 6 1</span><br></pre></td></tr></table></figure><p>这跟前面的二元组好像没什么区别，但是如果仔细观察，其实每一行的值都是它所对应的真正的数列值，乘上这行的系数，再取模k，也就是那个列最上面的数乘那行最左边的数。</p><p>如果我们对第一行的每个数求逆元，求得的就正是它所对应的 <span class="math inline">\(1\)</span> 的位置所在的行数！</p><p>这时我们只需要用扩展欧几里得在 <span class="math inline">\(O(k\log k)\)</span> 的时间内求逆元，就可以随之求出所有 <span class="math inline">\(e\)</span> 和 <span class="math inline">\(g\)</span> 的值！</p><p>虽然这个数列进入循环节之前可能不止有 <span class="math inline">\(k\)</span> 项，但是可以证明它必然有循环节（抽屉原理），而且根据<a href="https://en.wikipedia.org/wiki/Pisano_period" target="_blank" rel="noopener">皮萨诺周期</a>可以证明它不会超过 <span class="math inline">\(6k\)</span> 项，所以复杂度确实是 <span class="math inline">\(O(k\log k)\)</span> 的。</p><p>当然，这只是第一部分，求出 <span class="math inline">\(e,g\)</span> 后要需要做一些优化，因为对于矩阵的跳跃，如果 <span class="math inline">\(n\)</span> 足够大，很有可能会进入一个循环，这时记录下一整个循环的递推矩阵之乘积 <span class="math inline">\(C\)</span>，然后再套用一次矩阵快速幂，可以省去大量时间，具体实现细节都在代码里了。</p><h2 id="代码">代码</h2><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><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br><span class="line">41</span><br><span class="line">42</span><br><span class="line">43</span><br><span class="line">44</span><br><span class="line">45</span><br><span class="line">46</span><br><span class="line">47</span><br><span class="line">48</span><br><span class="line">49</span><br><span class="line">50</span><br><span class="line">51</span><br><span class="line">52</span><br><span class="line">53</span><br><span class="line">54</span><br><span class="line">55</span><br><span class="line">56</span><br><span class="line">57</span><br><span class="line">58</span><br><span class="line">59</span><br><span class="line">60</span><br><span class="line">61</span><br><span class="line">62</span><br><span class="line">63</span><br><span class="line">64</span><br><span class="line">65</span><br><span class="line">66</span><br><span class="line">67</span><br><span class="line">68</span><br><span class="line">69</span><br><span class="line">70</span><br><span class="line">71</span><br><span class="line">72</span><br><span class="line">73</span><br><span class="line">74</span><br><span class="line">75</span><br><span class="line">76</span><br><span class="line">77</span><br><span class="line">78</span><br><span class="line">79</span><br><span class="line">80</span><br><span class="line">81</span><br><span class="line">82</span><br><span class="line">83</span><br><span class="line">84</span><br><span class="line">85</span><br><span class="line">86</span><br><span class="line">87</span><br><span class="line">88</span><br><span class="line">89</span><br><span class="line">90</span><br><span class="line">91</span><br><span class="line">92</span><br><span class="line">93</span><br><span class="line">94</span><br><span class="line">95</span><br><span class="line">96</span><br><span class="line">97</span><br><span class="line">98</span><br><span class="line">99</span><br><span class="line">100</span><br><span class="line">101</span><br><span class="line">102</span><br><span class="line">103</span><br><span class="line">104</span><br><span class="line">105</span><br><span class="line">106</span><br><span class="line">107</span><br><span class="line">108</span><br><span class="line">109</span><br><span class="line">110</span><br><span class="line">111</span><br><span class="line">112</span><br><span class="line">113</span><br><span class="line">114</span><br><span class="line">115</span><br><span class="line">116</span><br><span class="line">117</span><br><span class="line">118</span><br><span class="line">119</span><br><span class="line">120</span><br><span class="line">121</span><br><span class="line">122</span><br><span class="line">123</span><br><span class="line">124</span><br><span class="line">125</span><br><span class="line">126</span><br><span class="line">127</span><br><span class="line">128</span><br><span class="line">129</span><br><span class="line">130</span><br><span class="line">131</span><br><span class="line">132</span><br><span class="line">133</span><br><span class="line">134</span><br><span class="line">135</span><br><span class="line">136</span><br><span class="line">137</span><br><span class="line">138</span><br><span class="line">139</span><br><span class="line">140</span><br><span class="line">141</span><br><span class="line">142</span><br><span class="line">143</span><br><span class="line">144</span><br><span class="line">145</span><br><span class="line">146</span><br><span class="line">147</span><br><span class="line">148</span><br><span class="line">149</span><br><span class="line">150</span><br><span class="line">151</span><br><span class="line">152</span><br><span class="line">153</span><br></pre></td><td class="code"><pre><span class="line"><span class="meta">#<span class="meta-keyword">include</span> <span class="meta-string">"bits/stdc++.h"</span></span></span><br><span class="line"><span class="keyword">using</span> <span class="keyword">namespace</span> <span class="built_in">std</span>;</span><br><span class="line"><span class="keyword">typedef</span> <span class="keyword">long</span> <span class="keyword">long</span> ll;</span><br><span class="line"><span class="keyword">const</span> ll N=<span class="number">1000010</span>;</span><br><span class="line">ll m,n,k,p;</span><br><span class="line"><span class="class"><span class="keyword">struct</span> <span class="title">matrix</span>&#123;</span></span><br><span class="line">    ll v[<span class="number">5</span>][<span class="number">5</span>];</span><br><span class="line">    ll x,y;</span><br><span class="line">&#125;f,A,B,I,ans;</span><br><span class="line">ll e[N],g[N],inv[N],vis[N];</span><br><span class="line"><span class="function"><span class="keyword">void</span> <span class="title">exgcd</span><span class="params">(ll a,ll b,ll&amp; d,ll&amp; x,ll&amp; y)</span></span>&#123;<span class="comment">//这些是求逆元</span></span><br><span class="line">    <span class="keyword">if</span>(!b)&#123;</span><br><span class="line">        d=a;</span><br><span class="line">        x=<span class="number">1</span>;y=<span class="number">0</span>;</span><br><span class="line">        <span class="keyword">return</span>;</span><br><span class="line">    &#125;</span><br><span class="line">    exgcd(b,a%b,d,y,x);</span><br><span class="line">    y-=x*(a/b);</span><br><span class="line">&#125;</span><br><span class="line"><span class="function">ll <span class="title">getinv</span><span class="params">(ll a)</span></span>&#123;</span><br><span class="line">    ll d,x,y;</span><br><span class="line">    exgcd(a,k,d,x,y);</span><br><span class="line">    <span class="keyword">return</span> d==<span class="number">1</span>?(x+k)%k:<span class="number">-1</span>;</span><br><span class="line">&#125;</span><br><span class="line">ostream&amp;<span class="keyword">operator</span>&lt;&lt;(ostream&amp;ous,matrix a)&#123;<span class="comment">//矩阵运算</span></span><br><span class="line">    <span class="keyword">for</span>(ll i=<span class="number">0</span>;i&lt;a.x;i++)&#123;</span><br><span class="line">        <span class="keyword">for</span>(ll j=<span class="number">0</span>;j&lt;a.y;j++)&#123;</span><br><span class="line">            ous&lt;&lt;a.v[i][j]&lt;&lt;<span class="string">' '</span>;</span><br><span class="line">        &#125;</span><br><span class="line">        ous&lt;&lt;<span class="built_in">endl</span>;</span><br><span class="line">    &#125;</span><br><span class="line">    <span class="keyword">return</span> ous;</span><br><span class="line">&#125;</span><br><span class="line">matrix <span class="keyword">operator</span>+(matrix a,matrix b)&#123;</span><br><span class="line">    matrix c;</span><br><span class="line">    <span class="keyword">if</span>(a.x!=b.x||a.y!=b.y)<span class="keyword">return</span> c;</span><br><span class="line">    ll x=a.x,y=a.y;</span><br><span class="line">    c.x=x;</span><br><span class="line">    c.y=y;</span><br><span class="line">    <span class="keyword">for</span>(ll i=<span class="number">0</span>;i&lt;x;i++)&#123;</span><br><span class="line">        <span class="keyword">for</span>(ll j=<span class="number">0</span>;j&lt;y;j++)&#123;</span><br><span class="line">            c.v[i][j]=(a.v[i][j]+b.v[i][j])%p;</span><br><span class="line">        &#125;</span><br><span class="line">    &#125;</span><br><span class="line">    <span class="keyword">return</span> c;</span><br><span class="line">&#125;</span><br><span class="line">matrix <span class="keyword">operator</span>*(matrix a,matrix b)&#123;</span><br><span class="line">    matrix c;</span><br><span class="line">    <span class="keyword">if</span>(a.y!=b.x)<span class="keyword">return</span> c;</span><br><span class="line">    ll x=a.x,y=b.y,z=a.y;</span><br><span class="line">    c.x=x;</span><br><span class="line">    c.y=y;</span><br><span class="line">    <span class="keyword">for</span>(ll i=<span class="number">0</span>;i&lt;x;i++)&#123;</span><br><span class="line">        <span class="keyword">for</span>(ll j=<span class="number">0</span>;j&lt;y;j++)&#123;</span><br><span class="line">            c.v[i][j]=<span class="number">0</span>;</span><br><span class="line">            <span class="keyword">for</span>(ll k=<span class="number">0</span>;k&lt;z;k++)&#123;</span><br><span class="line">                c.v[i][j]=(c.v[i][j]+a.v[i][k]*b.v[k][j])%p;</span><br><span class="line">            &#125;</span><br><span class="line">        &#125;</span><br><span class="line">    &#125;</span><br><span class="line">    <span class="keyword">return</span> c;</span><br><span class="line">&#125;</span><br><span class="line">matrix <span class="keyword">operator</span>^(matrix a,ll b)&#123;<span class="comment">//我把快速幂也重载了</span></span><br><span class="line">    matrix c=I;</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>)c=c*a;</span><br><span class="line">        a=a*a;</span><br><span class="line">        b&gt;&gt;=<span class="number">1</span>;</span><br><span class="line">    &#125;</span><br><span class="line">    <span class="keyword">return</span> c;</span><br><span class="line">&#125;</span><br><span class="line"><span class="function"><span class="keyword">void</span> <span class="title">init</span><span class="params">()</span></span>&#123;</span><br><span class="line">    A.x=A.y=B.x=B.y=<span class="number">3</span>;</span><br><span class="line">    I.x=I.y=<span class="number">3</span>;</span><br><span class="line">    f.x=<span class="number">3</span>;f.y=<span class="number">1</span>;</span><br><span class="line">    f.v[<span class="number">1</span>][<span class="number">0</span>]=f.v[<span class="number">2</span>][<span class="number">0</span>]=<span class="number">1</span>;</span><br><span class="line">    A.v[<span class="number">0</span>][<span class="number">0</span>]=A.v[<span class="number">0</span>][<span class="number">1</span>]=A.v[<span class="number">1</span>][<span class="number">0</span>]=A.v[<span class="number">2</span>][<span class="number">2</span>]=<span class="number">1</span>;</span><br><span class="line">    B.v[<span class="number">0</span>][<span class="number">0</span>]=B.v[<span class="number">0</span>][<span class="number">1</span>]=B.v[<span class="number">1</span>][<span class="number">0</span>]=B.v[<span class="number">2</span>][<span class="number">2</span>]=<span class="number">1</span>;</span><br><span class="line">    B.v[<span class="number">0</span>][<span class="number">2</span>]=<span class="number">-1</span>;</span><br><span class="line">    I.v[<span class="number">0</span>][<span class="number">0</span>]=I.v[<span class="number">1</span>][<span class="number">1</span>]=I.v[<span class="number">2</span>][<span class="number">2</span>]=<span class="number">1</span>;</span><br><span class="line">&#125;</span><br><span class="line"><span class="function"><span class="keyword">int</span> <span class="title">main</span><span class="params">()</span></span>&#123;</span><br><span class="line">    <span class="comment">// freopen("rabbit.in","r",stdin);</span></span><br><span class="line">    <span class="comment">// freopen("rabbit.out","w",stdout);</span></span><br><span class="line">    init();</span><br><span class="line">    <span class="built_in">cin</span>&gt;&gt;n&gt;&gt;k&gt;&gt;p;</span><br><span class="line">    <span class="keyword">int</span> a=<span class="number">1</span>,b=<span class="number">2</span>;</span><br><span class="line">    <span class="built_in">memset</span>(e,<span class="number">-1</span>,<span class="keyword">sizeof</span>(e));</span><br><span class="line">    <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">1</span>;i&lt;=k*k;i++)&#123;<span class="comment">//先预处理e和g</span></span><br><span class="line">        <span class="keyword">if</span>(a==<span class="number">1</span>&amp;&amp;b==<span class="number">1</span>)<span class="keyword">break</span>;</span><br><span class="line">        <span class="keyword">if</span>(!inv[b])inv[b]=getinv(b);</span><br><span class="line">        <span class="keyword">if</span>(inv[b]==<span class="number">-1</span>)&#123;</span><br><span class="line">            swap(a,b);</span><br><span class="line">            b=(a+b)%k;</span><br><span class="line">            <span class="keyword">continue</span>;</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">if</span>(e[inv[b]]==<span class="number">-1</span>)&#123;</span><br><span class="line">            e[inv[b]]=i;</span><br><span class="line">            g[inv[b]]=a*inv[b]%k;</span><br><span class="line">        &#125;</span><br><span class="line">        swap(a,b);</span><br><span class="line">        b=(a+b)%k;</span><br><span class="line">    &#125;</span><br><span class="line">    ll flag=<span class="number">0</span>;</span><br><span class="line">    ll cur=<span class="number">1</span>,ptr=<span class="number">0</span>;</span><br><span class="line">    <span class="keyword">while</span>(<span class="literal">true</span>)&#123;<span class="comment">//沿着路线走一走，碰到循环就退出</span></span><br><span class="line">        <span class="keyword">if</span>(vis[cur])&#123;</span><br><span class="line">            flag=cur;</span><br><span class="line">            <span class="keyword">break</span>;</span><br><span class="line">        &#125;</span><br><span class="line">        vis[cur]=<span class="number">1</span>;</span><br><span class="line">        <span class="keyword">if</span>(ptr+e[cur]+<span class="number">1</span>&gt;n||e[cur]==<span class="number">-1</span>)&#123;</span><br><span class="line">            f=(A^(n-ptr))*f;</span><br><span class="line">            <span class="keyword">break</span>;</span><br><span class="line">        &#125;</span><br><span class="line">        f=(A^(e[cur]+<span class="number">1</span>))*f;</span><br><span class="line">        ptr+=e[cur]+<span class="number">1</span>;</span><br><span class="line">        <span class="keyword">if</span>(ptr&gt;=n)<span class="keyword">break</span>;</span><br><span class="line">        f=B*f;</span><br><span class="line">        ptr++;</span><br><span class="line">        cur=g[cur];</span><br><span class="line">        <span class="keyword">if</span>(ptr&gt;=n)<span class="keyword">break</span>;</span><br><span class="line">    &#125;</span><br><span class="line">    matrix C;</span><br><span class="line">    ll siz;</span><br><span class="line">    <span class="keyword">if</span>(flag)&#123;<span class="comment">//如果在循环里，那就求一下C</span></span><br><span class="line">        siz=e[flag]+<span class="number">2</span>;</span><br><span class="line">        C=B*(A^(e[flag]+<span class="number">1</span>));</span><br><span class="line">        <span class="keyword">for</span>(ll i=g[flag];i!=flag;i=g[i])&#123;</span><br><span class="line">            C=B*(A^(e[i]+<span class="number">1</span>))*C;</span><br><span class="line">            siz+=e[i]+<span class="number">2</span>;</span><br><span class="line">        &#125;</span><br><span class="line">        f=(C^((n-ptr)/siz))*f;<span class="comment">//这一行做了大部分运算</span></span><br><span class="line">        <span class="keyword">if</span>((n-ptr)%siz)&#123;<span class="comment">//还有剩下的部分，那就接着算完</span></span><br><span class="line">            ptr=n-((n-ptr)%siz);</span><br><span class="line">            <span class="keyword">while</span>(<span class="literal">true</span>)&#123;</span><br><span class="line">                <span class="keyword">if</span>(ptr+e[cur]+<span class="number">1</span>&gt;n)&#123;</span><br><span class="line">                    f=(A^(n-ptr))*f;</span><br><span class="line">                    <span class="keyword">break</span>;</span><br><span class="line">                &#125;</span><br><span class="line">                f=(A^(e[cur]+<span class="number">1</span>))*f;</span><br><span class="line">                ptr+=e[cur]+<span class="number">1</span>;</span><br><span class="line">                <span class="keyword">if</span>(ptr&gt;=n)<span class="keyword">break</span>;</span><br><span class="line">                f=B*f;</span><br><span class="line">                ptr++;</span><br><span class="line">                cur=g[cur];</span><br><span class="line">                <span class="keyword">if</span>(ptr&gt;=n)<span class="keyword">break</span>;</span><br><span class="line">            &#125;</span><br><span class="line">        &#125;</span><br><span class="line">    &#125; </span><br><span class="line">    <span class="built_in">cout</span>&lt;&lt;(f.v[<span class="number">0</span>][<span class="number">0</span>]+p)%p&lt;&lt;<span class="built_in">endl</span>;<span class="comment">//完结</span></span><br><span class="line">    <span class="keyword">return</span> <span class="number">0</span>;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure></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/rabbit/" title="兔农">https://schtonn.github.io/blog/posts/rabbit/</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/gen-func/" rel="prev" title="生成函数"><i class="fa fa-chevron-left"></i> 生成函数</a></div><div class="post-nav-item"> <a href="/blog/posts/matrix-pow-complex/" rel="next" title="复杂矩阵快速幂">复杂矩阵快速幂<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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