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<!--[if lt IE 7]> <html class="no-js lt-ie9 lt-ie8 lt-ie7"> <![endif]--><!--[if IE 7]> <html class="no-js lt-ie9 lt-ie8"> <![endif]--><!--[if IE 8]> <html class="no-js lt-ie9"> <![endif]--><!--[if gt IE 8]><!--><html class="no-js"> <head><meta charset="utf-8"><meta http-equiv="X-UA-Compatible" content="IE=edge,chrome=1"><title>Моё решение задачи 134</title><meta name="description" content><meta name="viewport" content="width=device-width"><link rel="stylesheet" href="../../theme/css/normalize.css"><link href="http://fonts.googleapis.com/css?family=Philosopher&subset=latin,cyrillic" rel="stylesheet" type="text/css"><link href="http://fonts.googleapis.com/css?family=Forum&subset=cyrillic" rel="stylesheet" type="text/css"><link href="//fonts.googleapis.com/css?family=Oswald" rel="stylesheet" type="text/css"><link href="http://fonts.googleapis.com/css?family=Ubuntu+Mono" rel="stylesheet" type="text/css"><link href="http://fonts.googleapis.com/css?family=PT+Sans" rel="stylesheet" type="text/css"><link rel="stylesheet" href="../../theme/css/font-awesome.min.css"><link rel="stylesheet" href="../../theme/css/main.css"><link rel="stylesheet" href="../../theme/css/blog.css"><link rel="stylesheet" href="../../theme/css/github.css"><link href="http://likemath.ru/feeds/all.atom.xml" type="application/atom+xml" rel="alternate" title="Блог 529 Atom Feed"><link href="http://likemath.ru/feeds/all.rss.xml" type="application/rss+xml" rel="alternate" title="Блог 529 RSS Feed"><script src="../../theme/js/vendor/modernizr-2.6.2.min.js"></script></head><body><!--[if lt IE 7]>
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<![endif]--><div id="wrapper"><header id="sidebar" class="side-shadow"><hgroup id="site-header"><a id="site-title" href="../.."><h2><i class="icon-pencil"></i> Блог 529</h2></a><p id="site-desc"> Project Euler и остальное </p></hgroup><nav><ul id="nav-links"><li><a href="../../">Главная</a></li><li><a href="../../pages/projects.html">Мои проекты</a></li><li><a href="../../pages/about.html">Об авторе</a></li><li><a href="../../feeds/feed.atom.xml">Atom feed</a></li></ul></nav><footer id="site-info"><p> Powered by Pelican. </p></footer></header><div id="post-container"><ol id="post-list"><li><article class="post-entry"><header class="entry-header"><time class="post-time" datetime="2016-10-30T17:40:00+03:00" pubdate> Вс 30 Октябрь 2016 </time><a href="../../posts/moio-reshenie-zadachi-134/" rel="bookmark"><h1>Моё решение задачи&nbsp;134</h1></a></header><section class="post-content"><p>Назовём <em>порождающим</em> для двух последовательных простых <span class="math">\(p_1 &lt; p_2\)</span> наименьшее натуральное число, что оно закачивается на <span class="math">\(p_1\)</span> и при этом делится на <span class="math">\(p_2\)</span>. Необходимо найти сумму порождающих для всех <span class="math">\(p_1 \in \left[ 5; 10^6&nbsp;\right]\)</span></p><p>Например, если <span class="math">\(p_1 = 19\)</span>, то следующее простое <span class="math">\(p_2 = 23\)</span>. Тогда порождающим будет число <span class="math">\(1219\)</span>, при этом <span class="math">\(1219 \: \vdots \: 23\)</span>.</p><p>Полное условие можно найти <a href="https://projecteuler.net/problem=134">тут</a></p><p>Несмотря на то, что сложность задачи 45%, для её решения достаточно выписать&nbsp;условие.</p><p>Пусть <span class="math">\(p_1\)</span> содержит в себе <span class="math">\(k\)</span> цифр, т.е. <span class="math">\(n = r \cdot 10^k + p_1\)</span>, где <span class="math">\(r\)</span> &#8212; какое-то натуральное число с отрезка <span class="math">\(\left[ 1; p_2-1&nbsp;\right]\)</span></p><p>Давайте посчитаем остатки по модулю <span class="math">\(p_2\)</span>: <span class="math">\(n \equiv r \cdot 10^k + p_1 \equiv 0\)</span>. Отсюда получим явную формулу для <span class="math">\(r\)</span>: </p><div class="math">$$ r \equiv -p_1 \cdot 10^{-k} \equiv -p_1 \cdot 10^{p_2 -1-k} $$</div><p>Комментарии:</p><ol><li>Так как <span class="math">\(a^p \equiv a \mod p\)</span>, то верно что <span class="math">\(a^{-k} \equiv a^{p -1-k} \mod&nbsp;p\)</span></li><li>Это всё бессмысленно, если не знать про <a href="https://ru.wikipedia.org/wiki/Алгоритмы_быстрого_возведения_в_степень">алгоритм быстрого возведения в степень</a>, который делает асимптотическую сложность возведения в степень&nbsp;логарифмической.</li></ol><p>У нас есть явная формула для порождающего, и мы знаем как её быстро посчитать. Ниже приведён код на Python с использованием <a href="http://www.sympy.org/ru/">sympy</a>.</p><div class="highlight"><pre><span class="code-line"><span></span><span class="kn">from</span> <span class="nn">sympy</span> <span class="kn">import</span> <span class="n">primerange</span> <span class="c1"># для получения простых чисел</span></span>
<span class="code-line"></span>
<span class="code-line"><span class="c1"># быстрое возведение в степень по модулю</span></span>
<span class="code-line"><span class="k">def</span> <span class="nf">fast_pow</span><span class="p">(</span><span class="n">x</span><span class="p">,</span> <span class="n">y</span><span class="p">,</span> <span class="n">modulo</span><span class="p">):</span></span>
<span class="code-line"> <span class="k">if</span> <span class="n">y</span> <span class="o">==</span> <span class="mi">0</span><span class="p">:</span></span>
<span class="code-line"> <span class="k">return</span> <span class="mi">1</span></span>
<span class="code-line"> <span class="n">p</span> <span class="o">=</span> <span class="n">fast_pow</span><span class="p">(</span><span class="n">x</span><span class="p">,</span> <span class="n">y</span> <span class="o">//</span> <span class="mi">2</span><span class="p">,</span> <span class="n">modulo</span><span class="p">)</span></span>
<span class="code-line"> <span class="n">p</span> <span class="o">=</span> <span class="p">(</span><span class="n">p</span> <span class="o">*</span> <span class="n">p</span><span class="p">)</span> <span class="o">%</span> <span class="n">modulo</span></span>
<span class="code-line"> <span class="k">if</span> <span class="n">y</span> <span class="o">%</span> <span class="mi">2</span><span class="p">:</span></span>
<span class="code-line"> <span class="n">p</span> <span class="o">=</span> <span class="p">(</span><span class="n">p</span> <span class="o">*</span> <span class="n">x</span><span class="p">)</span> <span class="o">%</span> <span class="n">modulo</span></span>
<span class="code-line"> <span class="k">return</span> <span class="n">p</span></span>
<span class="code-line"></span>
<span class="code-line"><span class="c1"># нам нужно первое простое, которое больше 10^6 -- 10^6+3</span></span>
<span class="code-line"><span class="n">primes</span> <span class="o">=</span> <span class="nb">list</span><span class="p">(</span><span class="n">primerange</span><span class="p">(</span><span class="mi">5</span><span class="p">,</span><span class="mi">10</span><span class="o">**</span><span class="mi">6</span><span class="o">+</span><span class="mi">4</span><span class="p">))</span> </span>
<span class="code-line"></span>
<span class="code-line"><span class="n">sm</span> <span class="o">=</span> <span class="mi">0</span></span>
<span class="code-line"></span>
<span class="code-line"><span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="nb">len</span><span class="p">(</span><span class="n">primes</span><span class="p">)</span> <span class="o">-</span> <span class="mi">1</span><span class="p">):</span></span>
<span class="code-line"> <span class="n">digs</span> <span class="o">=</span> <span class="nb">len</span><span class="p">(</span><span class="nb">str</span><span class="p">(</span><span class="n">primes</span><span class="p">[</span><span class="n">i</span><span class="p">]))</span> <span class="c1"># количество цифр</span></span>
<span class="code-line"> <span class="n">r</span> <span class="o">=</span> <span class="p">(</span><span class="n">primes</span><span class="p">[</span><span class="n">i</span><span class="o">+</span><span class="mi">1</span><span class="p">]</span><span class="o">**</span><span class="mi">2</span> <span class="o">-</span> <span class="n">primes</span><span class="p">[</span><span class="n">i</span><span class="p">]</span> <span class="o">*</span> <span class="n">fast_pow</span><span class="p">(</span><span class="mi">10</span><span class="p">,</span> <span class="n">primes</span><span class="p">[</span><span class="n">i</span><span class="o">+</span><span class="mi">1</span><span class="p">]</span> <span class="o">-</span> <span class="mi">1</span> <span class="o">-</span> <span class="n">digs</span><span class="p">,</span> <span class="n">primes</span><span class="p">[</span><span class="n">i</span><span class="o">+</span><span class="mi">1</span><span class="p">]))</span> <span class="o">%</span> <span class="n">primes</span><span class="p">[</span><span class="n">i</span><span class="o">+</span><span class="mi">1</span><span class="p">]</span></span>
<span class="code-line"> <span class="n">sm</span> <span class="o">+=</span> <span class="n">r</span> <span class="o">*</span> <span class="mi">10</span><span class="o">**</span><span class="n">digs</span> <span class="o">+</span> <span class="n">primes</span><span class="p">[</span><span class="n">i</span><span class="p">]</span></span>
<span class="code-line"></span>
<span class="code-line"><span class="k">print</span><span class="p">(</span><span class="s1">&#39;Result is {}&#39;</span><span class="o">.</span><span class="n">format</span><span class="p">(</span><span class="n">sm</span><span class="p">))</span></span>
</pre></div><p>Ответ: <strong>18613426663617118</strong></p><script type="text/javascript">if (!document.getElementById('mathjaxscript_pelican_#%@#$@#')) {
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