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| <h1>辗转相除法和裴蜀定理</h1> | ||
| <time datetime=2024-06-05T09:04:35+0800 class="post-date">Wed, Jun 5, 2024</time> | ||
| <p>数论基础第二弹！</p> | ||
| <h2 id="辗转相除法">辗转相除法</h2> | ||
| <p>又称欧几里得算法，算法关键在于 | ||
| $$\gcd(a,b)=\gcd(b,a\bmod b).$$</p> | ||
| <blockquote> | ||
| <p>证：不妨设 $a>b,a=qb+r$，则 $r=a\bmod b$，设 $d\mid a,d\mid b$，则 $\dfrac{r}{d}=\dfrac{a}{d}-\dfrac{bk}{d}$。</p> | ||
| <p>由上式，$\dfrac{r}{d}$ 为整数，即 $d\mid r$，所以对 $a,b$ 的公约数，也是 $b,r$ 的公约数。</p> | ||
| <p>反过来，设 $d\mid b,d\mid r$，则 $\dfrac{a}{d}=\dfrac{r}{d}+\dfrac{bk}{d}$。</p> | ||
| <p>显然 $\dfrac{a}{d}$ 是整数，即 $d\mid d$，故 $b,r$ 的公约数也是 $a,b$ 的公约数。</p> | ||
| <p>既然公约数相同，那么最大公约数也相同。</p> | ||
| </blockquote> | ||
| <p>容易写出递归版：</p> | ||
| <div class="highlight"><pre tabindex="0" style="color:#f8f8f2;background-color:#272822;-moz-tab-size:4;-o-tab-size:4;tab-size:4;"><code class="language-cpp" data-lang="cpp"><span style="display:flex;"><span><span style="color:#66d9ef">int</span> <span style="color:#a6e22e">gcd</span>(<span style="color:#66d9ef">int</span> a, <span style="color:#66d9ef">int</span> b) { <span style="color:#66d9ef">return</span> b<span style="color:#f92672">?</span> gcd(b, a <span style="color:#f92672">%</span> b)<span style="color:#f92672">:</span> a; } | ||
| </span></span></code></pre></div><h3 id="时间复杂度">时间复杂度</h3> | ||
| <p>欧几里得算法的最坏时间复杂度为 $O(n)$，这里 $n$ 是输入数据在二进制下的长度。换句话说，复杂度是 $O(\log a)$（默认 $a,b$ 同阶）。</p> | ||
| <blockquote> | ||
| <p>证：</p> | ||
| <ul> | ||
| <li>$a\lt b$ 时，下一次迭代到 $\gcd(b,a);$</li> | ||
| <li>$a\ge b$ 时，下一次迭代到 $\gcd(b,a\bmod b)$，而 $a\bmod b$ 至少让 $a$ 折半，即该过程最多发生 $O(\log a)=O(n)$ 次。</li> | ||
| </ul> | ||
| <p>情况一发生后一定会发生情况二，故前者发生次数不多于后者发生次数。</p> | ||
| <p>因此最多递归 $O(n)$ 次。</p> | ||
| </blockquote> | ||
| <p>欧几里得算法达到最坏时间复杂度仅当输入数据是斐波那契数列<sup id="fnref:1"><a href="#fn:1" class="footnote-ref" role="doc-noteref">1</a></sup>的相邻两项。</p> | ||
| <h3 id="更相减损术">更相减损术</h3> | ||
| <p>由于取模的时间常数高（尤其在高精运算中不可忽视），我们可以利用加减法代替乘除。</p> | ||
| <p>更相减损术的核心是： | ||
| $$\gcd(a,b)=\gcd(b,a-b).$$</p> | ||
| <p>注意到若 $a\gg b$，时间开销将不可接受，若能进行高效除以 $2$ 的操作（如位运算），可以进行优化。</p> | ||
| <p>若 $2\mid a,2\mid b$，则 $\gcd(a,b)=2\gcd\left(\dfrac{a}{2},\dfrac{b}{2}\right);$</p> | ||
| <p>不然，若 $2\mid a$（$2\mid b$ 同理），$\gcd(a,b)=\gcd\left(\dfrac{a}{2},b\right).$</p> | ||
| <p>由于两次操作内 $a,b$ 之一必将减半，时间复杂度为 $O(\log a).$</p> | ||
| <h2 id="裴蜀定理和扩展欧几里得算法">裴蜀定理和扩展欧几里得算法</h2> | ||
| <blockquote> | ||
| <p>设 $a,b$ 是不全为零的整数，对任意整数 $x,y$，满足 $\gcd(a,b)\mid ax+by$，且存在整数 $x,y$，使得 $ax+by=\gcd(a,b).$</p> | ||
| </blockquote> | ||
| <p>前者显然，后者构造即可，下面进行构造。</p> | ||
| <p>不妨设 $a\ge b,\gcd(a,b)=1$，使用归纳法，若 $b=0$，此时 $a=1$，显然有一组解 $x=1,y=0;$</p> | ||
| <p>不然，假设我们已知 $(a\bmod b,b)$ 为系数时的一组解 $(x_0,y_0)$，尝试推出 $(a,b)$ 为系数的一组解： | ||
| $$\begin{aligned}(a\bmod b)x_0+by_0&=1;\\(a-\left\lfloor\dfrac{a}{b}\right\rfloor b)x_0+by_0&=1;\\ax_0+b(y_0-\left\lfloor\dfrac{a}{b}\right\rfloor x_0)&=1.\end{aligned}$$</p> | ||
| <p>故解为 $(x_0,y_0-\left\lfloor\dfrac{a}{b}\right\rfloor x_0)$，由数学归纳法，对任意 $\gcd(a,b)=1$，均能构造 $(x,y)$ 使 $ax+by=1$，进而原命题得证。</p> | ||
| <p>容易写出代码：</p> | ||
| <div class="highlight"><pre tabindex="0" style="color:#f8f8f2;background-color:#272822;-moz-tab-size:4;-o-tab-size:4;tab-size:4;"><code class="language-cpp" data-lang="cpp"><span style="display:flex;"><span><span style="color:#66d9ef">int</span> <span style="color:#a6e22e">exgcd</span>(<span style="color:#66d9ef">int</span> a, <span style="color:#66d9ef">int</span> b, <span style="color:#66d9ef">int</span> <span style="color:#f92672">&</span>x, <span style="color:#66d9ef">int</span> <span style="color:#f92672">&</span>y) { | ||
| </span></span><span style="display:flex;"><span> <span style="color:#66d9ef">if</span> (<span style="color:#f92672">!</span>b) { | ||
| </span></span><span style="display:flex;"><span> x <span style="color:#f92672">=</span> <span style="color:#ae81ff">1</span>, y <span style="color:#f92672">=</span> <span style="color:#ae81ff">0</span>; | ||
| </span></span><span style="display:flex;"><span> <span style="color:#66d9ef">return</span> a; | ||
| </span></span><span style="display:flex;"><span> } | ||
| </span></span><span style="display:flex;"><span> <span style="color:#66d9ef">int</span> d <span style="color:#f92672">=</span> exgcd(b, a <span style="color:#f92672">%</span> b, y, x); | ||
| </span></span><span style="display:flex;"><span> y <span style="color:#f92672">-=</span> a <span style="color:#f92672">/</span> b <span style="color:#f92672">*</span> x; | ||
| </span></span><span style="display:flex;"><span> <span style="color:#66d9ef">return</span> d; | ||
| </span></span><span style="display:flex;"><span>} | ||
| </span></span></code></pre></div><p>由于计算可行解在求最大公因数的过程中，这种算法被称作扩展欧几里得算法。</p> | ||
| <p>有时我们担心，扩展欧几里得算法给出的可行解过大，使得发生溢出；幸运的是，可以证明，上述算法给出的解满足 $|x|<b,|y|<a.$</p> | ||
| <p>归纳法易证。</p> | ||
| <div class="footnotes" role="doc-endnotes"> | ||
| <hr> | ||
| <ol> | ||
| <li id="fn:1"> | ||
| <p>斐波那契数列，又称兔子数列，定义如下：$f_0=0,f_1=1,f_n=f_{n-1}+f_{n-2}(n>1).$ <a href="#fnref:1" class="footnote-backref" role="doc-backlink">↩︎</a></p> | ||
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| </ol> | ||
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