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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>Блог 529</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 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"><h1>Archives for Блог 529</h1><dl><dt>Пт 22 Июль 2016</dt><dd><a href="./posts/nakhozhdenie-summy-k-ykh-stepenei/">Нахождение суммы k-ых степеней</a></dd><dt>Чт 17 Март 2016</dt><dd><a href="./posts/wallabag-i-realnaia-zhizn/">Wallabag и реальная жизнь</a></dd><dt>Вс 10 Январь 2016</dt><dd><a href="./posts/kak-ia-shakhmatnogo-bota-pisal/">Как я шахматного бота писал</a></dd><dt>Вс 02 Август 2015</dt><dd><a href="./posts/crossgen-v10/">CrossGen v1.0</a></dd><dt>Пт 17 Июль 2015</dt><dd><a href="./posts/moio-reshenie-zadachi-60/">Моё решение задачи 60</a></dd><dt>Пт 03 Июль 2015</dt><dd><a href="./posts/eshchio-odno-vychislenie-vyrazhenii/">Ещё одно вычисление выражений</a></dd><dt>Пт 17 Апрель 2015</dt><dd><a href="./posts/moi-pervyi-post/">Мой первый пост</a></dd></dl></div><script>
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<![endif]--><div id="wrapper"><h1>Archives for Блог 529</h1><dl><dt>Пт 21 Октябрь 2016</dt><dd><a href="./posts/moio-reshenie-zadachi-146/">Моё решение задачи 146</a></dd><dt>Пт 22 Июль 2016</dt><dd><a href="./posts/nakhozhdenie-summy-k-ykh-stepenei/">Нахождение суммы k-ых степеней</a></dd><dt>Чт 17 Март 2016</dt><dd><a href="./posts/wallabag-i-realnaia-zhizn/">Wallabag и реальная жизнь</a></dd><dt>Вс 10 Январь 2016</dt><dd><a href="./posts/kak-ia-shakhmatnogo-bota-pisal/">Как я шахматного бота писал</a></dd><dt>Вс 02 Август 2015</dt><dd><a href="./posts/crossgen-v10/">CrossGen v1.0</a></dd><dt>Пт 17 Июль 2015</dt><dd><a href="./posts/moio-reshenie-zadachi-60/">Моё решение задачи 60</a></dd><dt>Пт 03 Июль 2015</dt><dd><a href="./posts/eshchio-odno-vychislenie-vyrazhenii/">Ещё одно вычисление выражений</a></dd><dt>Пт 17 Апрель 2015</dt><dd><a href="./posts/moi-pervyi-post/">Мой первый пост</a></dd></dl></div><script>
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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>Блог 529 - Алексей Лобанов</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-coffee"></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="2016-07-22T13:35:00+03:00" pubdate> Пт 22 Июль 2016 </time><a href="../posts/nakhozhdenie-summy-k-ykh-stepenei/" rel="bookmark"><h1>Нахождение суммы k-ых степеней</h1></a></header><section class="post-content"><p>Как придумать формулу для суммы <span class="math">\(1^5 + 2^5 + 3^5 + \ldots + n^5\)</span> и есть ли она вообще?</p><script type="text/javascript">if (!document.getElementById('mathjaxscript_pelican_#%@#$@#')) {
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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-coffee"></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="2016-10-21T17:40:00+03:00" pubdate> Пт 21 Октябрь 2016 </time><a href="../posts/moio-reshenie-zadachi-146/" rel="bookmark"><h1>Моё решение задачи 146</h1></a></header><section class="post-content"><p>Краткое условие: необходимо найти сумму всех натуральных <span class="math">\(n\)</span>, что <span class="math">\(n^2+1\)</span>, <span class="math">\(n^2+3\)</span>, <span class="math">\(n^2+7\)</span>, <span class="math">\(n^2+9\)</span>, <span class="math">\(n^2+13\)</span>, и <span class="math">\(n^2+27\)</span> будут последовательными простыми числами.</p><script type="text/javascript">if (!document.getElementById('mathjaxscript_pelican_#%@#$@#')) {
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</script></section></article></li><hr><li><article class="post-entry"><header class="entry-header"><time class="post-time" datetime="2016-07-22T13:35:00+03:00" pubdate> Пт 22 Июль 2016 </time><a href="../posts/nakhozhdenie-summy-k-ykh-stepenei/" rel="bookmark"><h1>Нахождение суммы k-ых степеней</h1></a></header><section class="post-content"><p>Как придумать формулу для суммы <span class="math">\(1^5 + 2^5 + 3^5 + \ldots + n^5\)</span> и есть ли она вообще?</p><script type="text/javascript">if (!document.getElementById('mathjaxscript_pelican_#%@#$@#')) {
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<![endif]--><div id="wrapper"><h1>Authors on Блог 529</h1><ul> <li><a href="./author/aleksei-lobanov.html">Алексей Лобанов</a> (7)</li></ul></div><script>
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<![endif]--><div id="wrapper"><h1>Authors on Блог 529</h1><ul> <li><a href="./author/aleksei-lobanov.html">Алексей Лобанов</a> (8)</li></ul></div><script>
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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-coffee"></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="2015-07-17T17:40:00+03:00" pubdate> Пт 17 Июль 2015 </time><a href="../posts/moio-reshenie-zadachi-60/" rel="bookmark"><h1>Моё решение задачи 60</h1></a></header><section class="post-content"><p>Краткое условие: необходимо найти множество из пяти простых чисел с минимальной суммой такое, что после “склеивания” в любом порядке любых двух чисел из него тоже будет простое число.</p></section></article></li><hr></ol><div class="paginator"> Page 1 / 1 </div></div></div><script>
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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-coffee"></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="2016-10-21T17:40:00+03:00" pubdate> Пт 21 Октябрь 2016 </time><a href="../posts/moio-reshenie-zadachi-146/" rel="bookmark"><h1>Моё решение задачи 146</h1></a></header><section class="post-content"><p>Краткое условие: необходимо найти сумму всех натуральных <span class="math">\(n\)</span>, что <span class="math">\(n^2+1\)</span>, <span class="math">\(n^2+3\)</span>, <span class="math">\(n^2+7\)</span>, <span class="math">\(n^2+9\)</span>, <span class="math">\(n^2+13\)</span>, и <span class="math">\(n^2+27\)</span> будут последовательными простыми числами.</p><script type="text/javascript">if (!document.getElementById('mathjaxscript_pelican_#%@#$@#')) {
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</script></section></article></li><hr><li><article class="post-entry"><header class="entry-header"><time class="post-time" datetime="2015-07-17T17:40:00+03:00" pubdate> Пт 17 Июль 2015 </time><a href="../posts/moio-reshenie-zadachi-60/" rel="bookmark"><h1>Моё решение задачи 60</h1></a></header><section class="post-content"><p>Краткое условие: необходимо найти множество из пяти простых чисел с минимальной суммой такое, что после “склеивания” в любом порядке любых двух чисел из него тоже будет простое число.</p></section></article></li><hr></ol><div class="paginator"> Page 1 / 1 </div></div></div><script>
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<feed xmlns="http://www.w3.org/2005/Atom"><title>Блог 529</title><link href="http://likemath.ru/" rel="alternate"></link><link href="http://likemath.ru/feeds/all.atom.xml" rel="self"></link><id>http://likemath.ru/</id><updated>2016-10-21T17:40:00+03:00</updated><entry><title>Моё решение задачи 146</title><link href="http://likemath.ru/posts/moio-reshenie-zadachi-146/" rel="alternate"></link><published>2016-10-21T17:40:00+03:00</published><author><name>Алексей Лобанов</name></author><id>tag:likemath.ru,2016-10-21:posts/moio-reshenie-zadachi-146/</id><summary type="html"><p>Краткое условие: необходимо найти сумму всех натуральных <span class="math">\(n\)</span>, что <span class="math">\(n^2+1\)</span>, <span class="math">\(n^2+3\)</span>, <span class="math">\(n^2+7\)</span>, <span class="math">\(n^2+9\)</span>, <span class="math">\(n^2+13\)</span>, и <span class="math">\(n^2+27\)</span> будут последовательными простыми&nbsp;числами.</p>
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</script></summary><category term="Project Euler"></category><category term="c++"></category><category term="FLINT"></category></entry><entry><title>Нахождение суммы k-ых степеней</title><link href="http://likemath.ru/posts/nakhozhdenie-summy-k-ykh-stepenei/" rel="alternate"></link><published>2016-07-22T13:35:00+03:00</published><author><name>Алексей Лобанов</name></author><id>tag:likemath.ru,2016-07-22:posts/nakhozhdenie-summy-k-ykh-stepenei/</id><summary type="html"><p>Как придумать формулу для суммы <span class="math">\(1^5 + 2^5 + 3^5 + \ldots + n^5\)</span> и есть ли она&nbsp;вообще?</p>
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</script></description><dc:creator xmlns:dc="http://purl.org/dc/elements/1.1/">Алексей Лобанов</dc:creator><pubDate>Fri, 21 Oct 2016 17:40:00 +0300</pubDate><guid isPermaLink="false">tag:likemath.ru,2016-10-21:posts/moio-reshenie-zadachi-146/</guid><category>Project Euler</category><category>c++</category><category>FLINT</category></item><item><title>Нахождение суммы k-ых степеней</title><link>http://likemath.ru/posts/nakhozhdenie-summy-k-ykh-stepenei/</link><description><p>Как придумать формулу для суммы <span class="math">\(1^5 + 2^5 + 3^5 + \ldots + n^5\)</span> и есть ли она&nbsp;вообще?</p>
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<feed xmlns="http://www.w3.org/2005/Atom"><title>Блог 529</title><link href="http://likemath.ru/" rel="alternate"></link><link href="http://likemath.ru/feeds/feed.atom.xml" rel="self"></link><id>http://likemath.ru/</id><updated>2016-10-21T17:40:00+03:00</updated><entry><title>Моё решение задачи 146</title><link href="http://likemath.ru/posts/moio-reshenie-zadachi-146/" rel="alternate"></link><published>2016-10-21T17:40:00+03:00</published><author><name>Алексей Лобанов</name></author><id>tag:likemath.ru,2016-10-21:posts/moio-reshenie-zadachi-146/</id><summary type="html"><p>Краткое условие: необходимо найти сумму всех натуральных <span class="math">\(n\)</span>, что <span class="math">\(n^2+1\)</span>, <span class="math">\(n^2+3\)</span>, <span class="math">\(n^2+7\)</span>, <span class="math">\(n^2+9\)</span>, <span class="math">\(n^2+13\)</span>, и <span class="math">\(n^2+27\)</span> будут последовательными простыми&nbsp;числами.</p>
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</script></summary><category term="Project Euler"></category><category term="c++"></category><category term="FLINT"></category></entry><entry><title>CrossGen v1.0</title><link href="http://likemath.ru/posts/crossgen-v10/" rel="alternate"></link><published>2015-08-04T17:40:00+03:00</published><author><name>Алексей Лобанов</name></author><id>tag:likemath.ru,2015-08-02:posts/crossgen-v10/</id><summary type="html"><p>Читая хабр, случайно натолкнулся на идею сделать программу, которая по заданной кроссвордной сетке находит способ её заполнить. В этом посте вкратце напишу про моё решение и первую версию&nbsp;приложения.</p></summary><category term="проект"></category><category term="c++"></category><category term="wxWidgets"></category></entry><entry><title>Моё решение задачи 60</title><link href="http://likemath.ru/posts/moio-reshenie-zadachi-60/" rel="alternate"></link><published>2015-11-22T23:41:00+03:00</published><author><name>Алексей Лобанов</name></author><id>tag:likemath.ru,2015-07-17:posts/moio-reshenie-zadachi-60/</id><summary type="html"><p>Краткое условие: необходимо найти множество из пяти простых чисел с минимальной суммой такое, что после &#8220;склеивания&#8221; в любом порядке любых двух чисел из него тоже будет простое&nbsp;число.</p></summary><category term="Project Euler"></category><category term="c++"></category><category term="BGL"></category></entry></feed>
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<feed xmlns="http://www.w3.org/2005/Atom"><title>Блог 529</title><link href="http://likemath.ru/" rel="alternate"></link><link href="http://likemath.ru/feeds/tag-project-euler.atom.xml" rel="self"></link><id>http://likemath.ru/</id><updated>2016-10-21T17:40:00+03:00</updated><entry><title>Моё решение задачи 146</title><link href="http://likemath.ru/posts/moio-reshenie-zadachi-146/" rel="alternate"></link><published>2016-10-21T17:40:00+03:00</published><author><name>Алексей Лобанов</name></author><id>tag:likemath.ru,2016-10-21:posts/moio-reshenie-zadachi-146/</id><summary type="html"><p>Краткое условие: необходимо найти сумму всех натуральных <span class="math">\(n\)</span>, что <span class="math">\(n^2+1\)</span>, <span class="math">\(n^2+3\)</span>, <span class="math">\(n^2+7\)</span>, <span class="math">\(n^2+9\)</span>, <span class="math">\(n^2+13\)</span>, и <span class="math">\(n^2+27\)</span> будут последовательными простыми&nbsp;числами.</p>
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</script></summary><category term="Project Euler"></category><category term="c++"></category><category term="FLINT"></category></entry><entry><title>Моё решение задачи 60</title><link href="http://likemath.ru/posts/moio-reshenie-zadachi-60/" rel="alternate"></link><published>2015-11-22T23:41:00+03:00</published><author><name>Алексей Лобанов</name></author><id>tag:likemath.ru,2015-07-17:posts/moio-reshenie-zadachi-60/</id><summary type="html"><p>Краткое условие: необходимо найти множество из пяти простых чисел с минимальной суммой такое, что после &#8220;склеивания&#8221; в любом порядке любых двух чисел из него тоже будет простое&nbsp;число.</p></summary><category term="Project Euler"></category><category term="c++"></category><category term="BGL"></category></entry></feed>
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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-coffee"></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="2016-07-22T13:35:00+03:00" pubdate> Пт 22 Июль 2016 </time><a href="./posts/nakhozhdenie-summy-k-ykh-stepenei/" rel="bookmark"><h1>Нахождение суммы k-ых степеней</h1></a></header><section class="post-content"><p>Как придумать формулу для суммы <span class="math">\(1^5 + 2^5 + 3^5 + \ldots + n^5\)</span> и есть ли она вообще?</p><script type="text/javascript">if (!document.getElementById('mathjaxscript_pelican_#%@#$@#')) {
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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-coffee"></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="2016-10-21T17:40:00+03:00" pubdate> Пт 21 Октябрь 2016 </time><a href="./posts/moio-reshenie-zadachi-146/" rel="bookmark"><h1>Моё решение задачи 146</h1></a></header><section class="post-content"><p>Краткое условие: необходимо найти сумму всех натуральных <span class="math">\(n\)</span>, что <span class="math">\(n^2+1\)</span>, <span class="math">\(n^2+3\)</span>, <span class="math">\(n^2+7\)</span>, <span class="math">\(n^2+9\)</span>, <span class="math">\(n^2+13\)</span>, и <span class="math">\(n^2+27\)</span> будут последовательными простыми числами.</p><script type="text/javascript">if (!document.getElementById('mathjaxscript_pelican_#%@#$@#')) {
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</script></section></article></li><hr><li><article class="post-entry"><header class="entry-header"><time class="post-time" datetime="2016-07-22T13:35:00+03:00" pubdate> Пт 22 Июль 2016 </time><a href="./posts/nakhozhdenie-summy-k-ykh-stepenei/" rel="bookmark"><h1>Нахождение суммы k-ых степеней</h1></a></header><section class="post-content"><p>Как придумать формулу для суммы <span class="math">\(1^5 + 2^5 + 3^5 + \ldots + n^5\)</span> и есть ли она вообще?</p><script type="text/javascript">if (!document.getElementById('mathjaxscript_pelican_#%@#$@#')) {
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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 146</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-coffee"></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="2016-10-21T17:40:00+03:00" pubdate> Пт 21 Октябрь 2016 </time><a href="../../posts/moio-reshenie-zadachi-146/" rel="bookmark"><h1>Моё решение задачи 146</h1></a></header><section class="post-content"><p>Необходимо найти сумму всех натуральных <span class="math">\(n\)</span>, что <span class="math">\(n^2+1\)</span>, <span class="math">\(n^2+3\)</span>, <span class="math">\(n^2+7\)</span>, <span class="math">\(n^2+9\)</span>, <span class="math">\(n^2+13\)</span>, и <span class="math">\(n^2+27\)</span> будут <em>последовательными</em> простыми числами.</p><p>Полное условие можно найти <a href="https://projecteuler.net/problem=146">тут</a></p><p>Хочется отметить, что сложность у задачи 50%, а на текущий момент её решило меньше 4000 человек. Тем не менее, мне она показалось простой. Простейшее решение отработало очень быстро.</p><p>Для начала, можно отметить, что в лоб проверять условие очень долго. Проверять на простоту числа порядка <span class="math">\(10^{15}\)</span> достаточно сложно, поэтому их нужно как-то отсеять.</p><p>Самое простое — не рассматривать те <span class="math">\(n\)</span>, что хотя бы одно из <span class="math">\(n^2+1\)</span>, <span class="math">\(n^2+3\)</span>, <span class="math">\(n^2+7\)</span>, <span class="math">\(n^2+9\)</span>, <span class="math">\(n^2+13\)</span>, и <span class="math">\(n^2+27\)</span> будет заведомо делиться на какое-то маленькое простое число. Это даёт достаточно хорошие результаты: из 150 миллионов чисел, после отсеивания по простым числам <span class="math">\(< 3000\)</span> (этот параметр я подбирал уже после решения задач: если он слишком маленький, то будет слишком много проверок на простоту, если же слишком большой, то мы делаем слишком много работы, чтобы отсеять несколько чисел), останется меньше <span class="math">\(2000\)</span> чисел. Их уже можно проверить непосредственно. </p><p>Тогда алгоритм может быть таким:</p><ol><li>Находим простые числа меньше <span class="math">\(3000\)</span>.</li><li>Для каждого из них находим допустимые остатки.</li><li>Для каждого из чисел от <span class="math">\(1\)</span> до <span class="math">\(n\)</span> проверяем, что остатки по всем простым хорошие.</li><li>Непосредственно проверяем условие. Важно не забыть проверить <strong>не</strong>простоту оставшихся нечётных чисел из диапазона <span class="math">\(n^2 + 1 \ldots n^2 + 27\)</span> там могут быть (и будут!) другие простые числа.</li></ol><p>Непосредственно сам поиск такой клики можно реализовать тривиально. Ниже мой код на C++11 с использованием библиотек Flint и primesieve. Распараллеливание хоть и просится, но смысла не имеет, т.к. я получил ответ менее, чем за 5 секунд.</p><div class="highlight"><pre><span class="code-line"><span class="cm">/*</span></span>
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<span class="code-line"><span class="cm"> * Problem 146 on Project Euler</span></span>
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<span class="code-line"><span class="cm"> * Aleksey Lobanov (c) 2016</span></span>
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<span class="code-line"><span class="cm"> */</span></span>
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<span class="code-line"></span>
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<span class="code-line"><span class="cp">#include</span> <span class="cpf"><iostream></span><span class="cp"></span></span>
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<span class="code-line"><span class="cp">#include</span> <span class="cpf"><vector></span><span class="cp"></span></span>
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<span class="code-line"><span class="cp">#include</span> <span class="cpf"><cstdint></span><span class="cp"></span></span>
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<span class="code-line"><span class="cp">#include</span> <span class="cpf"><set></span><span class="cp"></span></span>
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<span class="code-line"><span class="cp">#include</span> <span class="cpf"><iomanip></span><span class="cp"></span></span>
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<span class="code-line"><span class="cp">#include</span> <span class="cpf"><algorithm></span><span class="cp"></span></span>
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<span class="code-line"></span>
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<span class="code-line"><span class="cp">#include</span> <span class="cpf">"fmpzxx.h"</span><span class="cp"></span></span>
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<span class="code-line"><span class="cp">#include</span> <span class="cpf">"arithxx.h"</span><span class="cp"></span></span>
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<span class="code-line"></span>
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<span class="code-line"><span class="cp">#include</span> <span class="cpf">"primesieve.hpp"</span><span class="cp"></span></span>
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<span class="code-line"></span>
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<span class="code-line"><span class="k">using</span> <span class="k">namespace</span> <span class="n">std</span><span class="p">;</span></span>
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<span class="code-line"><span class="k">using</span> <span class="k">namespace</span> <span class="n">flint</span><span class="p">;</span></span>
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<span class="code-line"></span>
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<span class="code-line"><span class="kt">bool</span> <span class="nf">is_prime</span><span class="p">(</span><span class="kt">int64_t</span> <span class="n">num</span><span class="p">)</span> <span class="p">{</span></span>
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<span class="code-line"> <span class="n">fmpz_factorxx</span> <span class="n">fact</span><span class="p">;</span></span>
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<span class="code-line"> <span class="n">fact</span><span class="p">.</span><span class="n">set_factor</span><span class="p">(</span><span class="n">num</span><span class="p">);</span></span>
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<span class="code-line"> <span class="k">if</span> <span class="p">(</span> <span class="n">fact</span><span class="p">.</span><span class="n">size</span><span class="p">()</span> <span class="o">!=</span> <span class="mi">1</span> <span class="p">)</span></span>
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<span class="code-line"> <span class="k">return</span> <span class="nb">false</span><span class="p">;</span></span>
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<span class="code-line"> <span class="k">if</span> <span class="p">(</span> <span class="n">fact</span><span class="p">.</span><span class="n">exp</span><span class="p">(</span><span class="mi">0</span><span class="p">)</span> <span class="o">!=</span> <span class="mi">1</span><span class="p">)</span> </span>
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<span class="code-line"> <span class="k">return</span> <span class="nb">false</span><span class="p">;</span></span>
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<span class="code-line"> <span class="k">return</span> <span class="nb">true</span><span class="p">;</span></span>
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<span class="code-line"><span class="p">}</span></span>
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<span class="code-line"></span>
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<span class="code-line"><span class="kt">bool</span> <span class="nf">is_possible</span><span class="p">(</span><span class="kt">int64_t</span> <span class="n">num</span><span class="p">,</span> <span class="k">const</span> <span class="n">vector</span><span class="o"><</span><span class="kt">int64_t</span><span class="o">></span> <span class="o">&</span><span class="n">to_add</span><span class="p">)</span> <span class="p">{</span></span>
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<span class="code-line"> <span class="k">for</span> <span class="p">(</span><span class="k">auto</span> <span class="o">&&</span><span class="nl">add</span><span class="p">:</span> <span class="n">to_add</span><span class="p">)</span></span>
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<span class="code-line"> <span class="k">if</span> <span class="p">(</span> <span class="o">!</span><span class="n">is_prime</span><span class="p">(</span><span class="n">num</span><span class="o">*</span><span class="n">num</span> <span class="o">+</span> <span class="n">add</span><span class="p">)</span> <span class="p">)</span></span>
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<span class="code-line"> <span class="k">return</span> <span class="nb">false</span><span class="p">;</span></span>
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|
<span class="code-line"></span>
|
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<span class="code-line"> <span class="c1">// primes must be consecutive</span></span>
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<span class="code-line"> <span class="c1">// so we need check, that other numbers like n^2 + i is not primes</span></span>
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<span class="code-line"> <span class="n">vector</span><span class="o"><</span><span class="kt">int64_t</span><span class="o">></span> <span class="n">other_adds</span><span class="p">;</span></span>
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<span class="code-line"> <span class="k">for</span> <span class="p">(</span><span class="kt">size_t</span> <span class="n">i</span> <span class="o">=</span> <span class="n">to_add</span><span class="p">[</span><span class="mi">0</span><span class="p">]</span> <span class="o">+</span> <span class="mi">2</span><span class="p">;</span> <span class="n">i</span> <span class="o"><</span> <span class="n">to_add</span><span class="p">[</span><span class="n">to_add</span><span class="p">.</span><span class="n">size</span><span class="p">()</span> <span class="o">-</span> <span class="mi">1</span><span class="p">];</span> <span class="n">i</span> <span class="o">+=</span> <span class="mi">2</span><span class="p">)</span> </span>
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<span class="code-line"> <span class="k">if</span> <span class="p">(</span> <span class="o">!</span><span class="n">binary_search</span><span class="p">(</span><span class="n">to_add</span><span class="p">.</span><span class="n">begin</span><span class="p">(),</span> <span class="n">to_add</span><span class="p">.</span><span class="n">end</span><span class="p">(),</span> <span class="n">i</span><span class="p">)</span> <span class="p">)</span></span>
|
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<span class="code-line"> <span class="n">other_adds</span><span class="p">.</span><span class="n">push_back</span><span class="p">(</span><span class="n">i</span><span class="p">);</span></span>
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<span class="code-line"></span>
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|
<span class="code-line"> <span class="k">for</span> <span class="p">(</span><span class="k">auto</span> <span class="o">&&</span><span class="nl">add</span><span class="p">:</span> <span class="n">other_adds</span><span class="p">)</span></span>
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<span class="code-line"> <span class="k">if</span> <span class="p">(</span> <span class="n">is_prime</span><span class="p">(</span><span class="n">num</span><span class="o">*</span><span class="n">num</span> <span class="o">+</span> <span class="n">add</span><span class="p">)</span> <span class="p">)</span></span>
|
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|
<span class="code-line"> <span class="k">return</span> <span class="nb">false</span><span class="p">;</span> </span>
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|
<span class="code-line"></span>
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|
<span class="code-line"> <span class="k">return</span> <span class="nb">true</span><span class="p">;</span></span>
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|
<span class="code-line"><span class="p">}</span></span>
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|
<span class="code-line"></span>
|
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|
<span class="code-line"><span class="kt">int</span> <span class="nf">main</span><span class="p">()</span> <span class="p">{</span></span>
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<span class="code-line"> <span class="k">const</span> <span class="n">vector</span><span class="o"><</span><span class="kt">int64_t</span><span class="o">></span> <span class="n">to_add</span> <span class="o">=</span> <span class="p">{</span><span class="mi">1</span><span class="p">,</span> <span class="mi">3</span><span class="p">,</span> <span class="mi">7</span><span class="p">,</span> <span class="mi">9</span><span class="p">,</span> <span class="mi">13</span><span class="p">,</span> <span class="mi">27</span><span class="p">};</span></span>
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|
<span class="code-line"></span>
|
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<span class="code-line"> <span class="k">const</span> <span class="kt">int64_t</span> <span class="n">MAX_N</span> <span class="o">=</span> <span class="mi">1l</span><span class="o">*</span><span class="mi">150</span><span class="o">*</span><span class="mi">1000</span><span class="o">*</span><span class="mi">1000</span><span class="p">;</span></span>
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<span class="code-line"> <span class="k">const</span> <span class="kt">int64_t</span> <span class="n">MAX_PRIME</span> <span class="o">=</span> <span class="mi">3000</span><span class="p">;</span></span>
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<span class="code-line"></span>
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<span class="code-line"> <span class="n">vector</span><span class="o"><</span><span class="kt">int64_t</span><span class="o">></span> <span class="n">sieve_primes</span><span class="p">;</span></span>
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<span class="code-line"> <span class="n">primesieve</span><span class="o">::</span><span class="n">generate_primes</span><span class="p">(</span><span class="n">MAX_PRIME</span><span class="p">,</span> <span class="o">&</span><span class="n">sieve_primes</span><span class="p">);</span></span>
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<span class="code-line"></span>
|
||||||
|
<span class="code-line"> <span class="n">vector</span><span class="o"><</span> <span class="n">vector</span> <span class="o"><</span><span class="kt">int64_t</span><span class="o">></span> <span class="o">></span> <span class="n">good_remainders</span><span class="p">;</span> </span>
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|
<span class="code-line"> <span class="k">for</span> <span class="p">(</span><span class="k">auto</span> <span class="o">&&</span><span class="nl">prime</span><span class="p">:</span> <span class="n">sieve_primes</span><span class="p">)</span> <span class="p">{</span></span>
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<span class="code-line"> <span class="n">set</span><span class="o"><</span><span class="kt">int64_t</span><span class="o">></span> <span class="n">remainders</span><span class="p">;</span></span>
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<span class="code-line"> <span class="k">for</span><span class="p">(</span><span class="kt">int64_t</span> <span class="n">rem</span> <span class="o">=</span> <span class="mi">0</span><span class="p">;</span> <span class="n">rem</span> <span class="o"><</span> <span class="n">prime</span><span class="p">;</span> <span class="o">++</span><span class="n">rem</span><span class="p">)</span></span>
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<span class="code-line"> <span class="n">remainders</span><span class="p">.</span><span class="n">insert</span><span class="p">(</span><span class="n">rem</span><span class="p">);</span></span>
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<span class="code-line"></span>
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<span class="code-line"> <span class="n">set</span><span class="o"><</span><span class="kt">int64_t</span><span class="o">></span> <span class="n">base_remainders</span><span class="p">;</span></span>
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|
<span class="code-line"> <span class="k">for</span> <span class="p">(</span><span class="k">auto</span> <span class="o">&&</span><span class="nl">base</span><span class="p">:</span> <span class="n">to_add</span><span class="p">)</span></span>
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<span class="code-line"> <span class="n">base_remainders</span><span class="p">.</span><span class="n">insert</span><span class="p">(</span><span class="n">base</span> <span class="o">%</span> <span class="n">prime</span><span class="p">);</span></span>
|
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|
<span class="code-line"></span>
|
||||||
|
<span class="code-line"> <span class="k">for</span> <span class="p">(</span><span class="kt">int64_t</span> <span class="n">rem</span> <span class="o">=</span> <span class="mi">0</span><span class="p">;</span> <span class="n">rem</span> <span class="o"><</span> <span class="n">prime</span><span class="p">;</span> <span class="o">++</span><span class="n">rem</span><span class="p">)</span></span>
|
||||||
|
<span class="code-line"> <span class="k">for</span> <span class="p">(</span><span class="k">auto</span> <span class="o">&&</span><span class="nl">base_rem</span><span class="p">:</span> <span class="n">base_remainders</span><span class="p">)</span></span>
|
||||||
|
<span class="code-line"> <span class="k">if</span> <span class="p">(</span> <span class="p">(</span><span class="n">rem</span><span class="o">*</span><span class="n">rem</span> <span class="o">+</span> <span class="n">base_rem</span><span class="p">)</span> <span class="o">%</span> <span class="n">prime</span> <span class="o">==</span> <span class="mi">0</span> <span class="p">)</span></span>
|
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|
<span class="code-line"> <span class="n">remainders</span><span class="p">.</span><span class="n">erase</span><span class="p">(</span><span class="n">rem</span><span class="p">);</span></span>
|
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|
<span class="code-line"></span>
|
||||||
|
<span class="code-line"> <span class="n">good_remainders</span><span class="p">.</span><span class="n">push_back</span><span class="p">(</span><span class="n">vector</span><span class="o"><</span><span class="kt">int64_t</span><span class="o">></span><span class="p">(</span><span class="n">remainders</span><span class="p">.</span><span class="n">begin</span><span class="p">(),</span> <span class="n">remainders</span><span class="p">.</span><span class="n">end</span><span class="p">()));</span></span>
|
||||||
|
<span class="code-line"> <span class="p">}</span></span>
|
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|
<span class="code-line"></span>
|
||||||
|
<span class="code-line"> <span class="kt">size_t</span> <span class="n">cnt</span> <span class="o">=</span> <span class="mi">0</span><span class="p">;</span></span>
|
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|
<span class="code-line"> <span class="kt">size_t</span> <span class="n">sum</span> <span class="o">=</span> <span class="mi">0</span><span class="p">;</span></span>
|
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|
<span class="code-line"></span>
|
||||||
|
<span class="code-line"> <span class="cm">/* WARNING</span></span>
|
||||||
|
<span class="code-line"><span class="cm"> * for small n can be that</span></span>
|
||||||
|
<span class="code-line"><span class="cm"> * n^2 + [1 or 3 or .. ] is prime in sieve primes</span></span>
|
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|
<span class="code-line"><span class="cm"> * but there is only one n < 315410 is 10,</span></span>
|
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|
<span class="code-line"><span class="cm"> * so we need add 10 to n</span></span>
|
||||||
|
<span class="code-line"><span class="cm"> */</span></span>
|
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|
<span class="code-line"> <span class="n">sum</span> <span class="o">+=</span> <span class="mi">10</span><span class="p">;</span></span>
|
||||||
|
<span class="code-line"></span>
|
||||||
|
<span class="code-line"> <span class="k">for</span> <span class="p">(</span><span class="kt">int64_t</span> <span class="n">i</span> <span class="o">=</span> <span class="mi">1</span><span class="p">;</span> <span class="n">i</span> <span class="o"><</span> <span class="n">MAX_N</span><span class="p">;</span> <span class="o">++</span><span class="n">i</span><span class="p">)</span> <span class="p">{</span></span>
|
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|
<span class="code-line"> <span class="kt">bool</span> <span class="n">is_good</span> <span class="o">=</span> <span class="nb">true</span><span class="p">;</span></span>
|
||||||
|
<span class="code-line"> <span class="k">for</span> <span class="p">(</span><span class="kt">size_t</span> <span class="n">prime_ind</span> <span class="o">=</span> <span class="mi">0</span><span class="p">;</span> <span class="n">prime_ind</span> <span class="o"><</span> <span class="n">sieve_primes</span><span class="p">.</span><span class="n">size</span><span class="p">();</span> <span class="o">++</span><span class="n">prime_ind</span><span class="p">)</span> <span class="p">{</span></span>
|
||||||
|
<span class="code-line"> <span class="k">if</span> <span class="p">(</span> <span class="o">!</span><span class="n">binary_search</span><span class="p">(</span><span class="n">good_remainders</span><span class="p">[</span><span class="n">prime_ind</span><span class="p">].</span><span class="n">begin</span><span class="p">(),</span> <span class="n">good_remainders</span><span class="p">[</span><span class="n">prime_ind</span><span class="p">].</span><span class="n">end</span><span class="p">(),</span> <span class="n">i</span> <span class="o">%</span> <span class="n">sieve_primes</span><span class="p">[</span><span class="n">prime_ind</span><span class="p">])</span> <span class="p">)</span> <span class="p">{</span></span>
|
||||||
|
<span class="code-line"> <span class="n">is_good</span> <span class="o">=</span> <span class="nb">false</span><span class="p">;</span></span>
|
||||||
|
<span class="code-line"> <span class="k">break</span><span class="p">;</span></span>
|
||||||
|
<span class="code-line"> <span class="p">}</span></span>
|
||||||
|
<span class="code-line"> <span class="p">}</span></span>
|
||||||
|
<span class="code-line"> <span class="k">if</span> <span class="p">(</span> <span class="n">is_good</span> <span class="p">)</span> <span class="p">{</span></span>
|
||||||
|
<span class="code-line"> <span class="n">cnt</span><span class="o">++</span><span class="p">;</span></span>
|
||||||
|
<span class="code-line"> <span class="k">if</span> <span class="p">(</span> <span class="n">is_possible</span><span class="p">(</span><span class="n">i</span><span class="p">,</span> <span class="n">to_add</span><span class="p">)</span> <span class="p">)</span> <span class="p">{</span></span>
|
||||||
|
<span class="code-line"> <span class="n">sum</span> <span class="o">+=</span> <span class="n">i</span><span class="p">;</span></span>
|
||||||
|
<span class="code-line"> <span class="p">}</span> </span>
|
||||||
|
<span class="code-line"> <span class="p">}</span></span>
|
||||||
|
<span class="code-line"> <span class="p">}</span></span>
|
||||||
|
<span class="code-line"> <span class="n">cout</span> <span class="o"><<</span> <span class="s">"count = "</span> <span class="o"><<</span> <span class="n">cnt</span> <span class="o"><<</span> <span class="n">endl</span><span class="p">;</span></span>
|
||||||
|
<span class="code-line"> <span class="n">cout</span> <span class="o"><<</span> <span class="s">"Result is: "</span> <span class="o"><<</span> <span class="n">sum</span> <span class="o"><<</span> <span class="n">endl</span><span class="p">;</span></span>
|
||||||
|
<span class="code-line"> <span class="k">return</span> <span class="mi">0</span><span class="p">;</span></span>
|
||||||
|
<span class="code-line"><span class="p">}</span></span>
|
||||||
|
</pre></div><p>Ответ: <strong>676333270</strong></p><script type="text/javascript">if (!document.getElementById('mathjaxscript_pelican_#%@#$@#')) {
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</script></section><hr><aside class="post-meta"><p>Категория: <a href="../../category/project-euler.html">Project Euler</a></p><p>Теги: <a href="../../tag/project-euler.html">Project Euler</a>, <a href="../../tag/c.html">c++</a>, <a href="../../tag/flint.html">FLINT</a>, </p></aside><hr><div class="comments"><div id="disqus_thread"></div><script type="text/javascript">
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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-coffee"></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="2015-08-02T17:40:00+03:00" pubdate> Вс 02 Август 2015 </time><a href="../posts/crossgen-v10/" rel="bookmark"><h1>CrossGen v1.0</h1></a></header><section class="post-content"><p>Читая хабр, случайно натолкнулся на идею сделать программу, которая по заданной кроссвордной сетке находит способ её заполнить. В этом посте вкратце напишу про моё решение и первую версию приложения.</p></section></article></li><hr><li><article class="post-entry"><header class="entry-header"><time class="post-time" datetime="2015-07-17T17:40:00+03:00" pubdate> Пт 17 Июль 2015 </time><a href="../posts/moio-reshenie-zadachi-60/" rel="bookmark"><h1>Моё решение задачи 60</h1></a></header><section class="post-content"><p>Краткое условие: необходимо найти множество из пяти простых чисел с минимальной суммой такое, что после “склеивания” в любом порядке любых двух чисел из него тоже будет простое число.</p></section></article></li><hr></ol><div class="paginator"> Page 1 / 1 </div></div></div><script>
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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-coffee"></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="2016-10-21T17:40:00+03:00" pubdate> Пт 21 Октябрь 2016 </time><a href="../posts/moio-reshenie-zadachi-146/" rel="bookmark"><h1>Моё решение задачи 146</h1></a></header><section class="post-content"><p>Краткое условие: необходимо найти сумму всех натуральных <span class="math">\(n\)</span>, что <span class="math">\(n^2+1\)</span>, <span class="math">\(n^2+3\)</span>, <span class="math">\(n^2+7\)</span>, <span class="math">\(n^2+9\)</span>, <span class="math">\(n^2+13\)</span>, и <span class="math">\(n^2+27\)</span> будут последовательными простыми числами.</p><script type="text/javascript">if (!document.getElementById('mathjaxscript_pelican_#%@#$@#')) {
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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-coffee"></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="2016-10-21T17:40:00+03:00" pubdate> Пт 21 Октябрь 2016 </time><a href="../posts/moio-reshenie-zadachi-146/" rel="bookmark"><h1>Моё решение задачи 146</h1></a></header><section class="post-content"><p>Краткое условие: необходимо найти сумму всех натуральных <span class="math">\(n\)</span>, что <span class="math">\(n^2+1\)</span>, <span class="math">\(n^2+3\)</span>, <span class="math">\(n^2+7\)</span>, <span class="math">\(n^2+9\)</span>, <span class="math">\(n^2+13\)</span>, и <span class="math">\(n^2+27\)</span> будут последовательными простыми числами.</p><script type="text/javascript">if (!document.getElementById('mathjaxscript_pelican_#%@#$@#')) {
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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-coffee"></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="2015-07-17T17:40:00+03:00" pubdate> Пт 17 Июль 2015 </time><a href="../posts/moio-reshenie-zadachi-60/" rel="bookmark"><h1>Моё решение задачи 60</h1></a></header><section class="post-content"><p>Краткое условие: необходимо найти множество из пяти простых чисел с минимальной суммой такое, что после “склеивания” в любом порядке любых двух чисел из него тоже будет простое число.</p></section></article></li><hr></ol><div class="paginator"> Page 1 / 1 </div></div></div><script>
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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-coffee"></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="2016-10-21T17:40:00+03:00" pubdate> Пт 21 Октябрь 2016 </time><a href="../posts/moio-reshenie-zadachi-146/" rel="bookmark"><h1>Моё решение задачи 146</h1></a></header><section class="post-content"><p>Краткое условие: необходимо найти сумму всех натуральных <span class="math">\(n\)</span>, что <span class="math">\(n^2+1\)</span>, <span class="math">\(n^2+3\)</span>, <span class="math">\(n^2+7\)</span>, <span class="math">\(n^2+9\)</span>, <span class="math">\(n^2+13\)</span>, и <span class="math">\(n^2+27\)</span> будут последовательными простыми числами.</p><script type="text/javascript">if (!document.getElementById('mathjaxscript_pelican_#%@#$@#')) {
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</script></section></article></li><hr><li><article class="post-entry"><header class="entry-header"><time class="post-time" datetime="2015-07-17T17:40:00+03:00" pubdate> Пт 17 Июль 2015 </time><a href="../posts/moio-reshenie-zadachi-60/" rel="bookmark"><h1>Моё решение задачи 60</h1></a></header><section class="post-content"><p>Краткое условие: необходимо найти множество из пяти простых чисел с минимальной суммой такое, что после “склеивания” в любом порядке любых двух чисел из него тоже будет простое число.</p></section></article></li><hr></ol><div class="paginator"> Page 1 / 1 </div></div></div><script>
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