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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-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 134</h1></a></header><section class="post-content"><p>Краткое условие: назовём <em>порождающим</em> для двух последовательных простых <span class="math">\(p_1 < p_2\)</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">\(p_1 \in \left[ 5; 10^6 \right]\)</span></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="2016-10-21T17:40:00+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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</script></section></article></li><hr><li><article class="post-entry"><header class="entry-header"><time class="post-time" datetime="2016-03-17T13:35:00+03:00" pubdate> Чт 17 Март 2016 </time><a href="../posts/wallabag-i-realnaia-zhizn/" rel="bookmark"><h1>Wallabag и реальная жизнь</h1></a></header><section class="post-content"><p>Как я устанавливал известнейшее свободное решение для отложенного чтения, и что из этого получилось.</p></section></article></li><hr><li><article class="post-entry"><header class="entry-header"><time class="post-time" datetime="2016-01-10T13:35:00+03:00" pubdate> Вс 10 Январь 2016 </time><a href="../posts/kak-ia-shakhmatnogo-bota-pisal/" rel="bookmark"><h1>Как я шахматного бота писал</h1></a></header><section class="post-content"><p>Как я проверял шахматное приложение на “ботоустойчивость”.</p></section></article></li><hr><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><li><article class="post-entry"><header class="entry-header"><time class="post-time" datetime="2015-07-03T17:40:00+03:00" pubdate> Пт 03 Июль 2015 </time><a href="../posts/eshchio-odno-vychislenie-vyrazhenii/" rel="bookmark"><h1>Ещё одно вычисление выражений</h1></a></header><section class="post-content"><p>На хабре когда-то увидел статью про то, что в Яндексе двум сотрудникам дали задачу на написание приложения, для вычисления выражений. Менеджер справился за 4 часа, а программист за два. Я решил попробовать свои силы.</p></section></article></li><hr><li><article class="post-entry"><header class="entry-header"><time class="post-time" datetime="2015-04-17T13:35:00+03:00" pubdate> Пт 17 Апрель 2015 </time><a href="../posts/moi-pervyi-post/" rel="bookmark"><h1>Мой первый пост</h1></a></header><section class="post-content"><p>Кратко о том, почему я сделал блог.</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-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="2016-07-22T13:35:00+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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</script></section></article></li><hr><li><article class="post-entry"><header class="entry-header"><time class="post-time" datetime="2016-03-17T13:35:00+03:00" pubdate> Чт 17 Март 2016 </time><a href="../posts/wallabag-i-realnaia-zhizn/" rel="bookmark"><h1>Wallabag и реальная жизнь</h1></a></header><section class="post-content"><p>Как я устанавливал известнейшее свободное решение для отложенного чтения, и что из этого получилось.</p></section></article></li><hr><li><article class="post-entry"><header class="entry-header"><time class="post-time" datetime="2015-04-17T13:35:00+03:00" pubdate> Пт 17 Апрель 2015 </time><a href="../posts/moi-pervyi-post/" rel="bookmark"><h1>Мой первый пост</h1></a></header><section class="post-content"><p>Кратко о том, почему я сделал блог.</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-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 134</h1></a></header><section class="post-content"><p>Краткое условие: назовём <em>порождающим</em> для двух последовательных простых <span class="math">\(p_1 < p_2\)</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">\(p_1 \in \left[ 5; 10^6 \right]\)</span></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="2016-10-21T17:40:00+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>2016-10-30T17:40:00+03:00</updated><entry><title>Моё решение задачи 134</title><link href="http://likemath.ru/posts/moio-reshenie-zadachi-134/" rel="alternate"></link><published>2016-10-30T17:40:00+03:00</published><updated>2016-10-30T17:40:00+03:00</updated><author><name>Алексей Лобанов</name></author><id>tag:likemath.ru,2016-10-30:posts/moio-reshenie-zadachi-134/</id><summary type="html"><p>Краткое условие: 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>
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</script></summary><category term="Project Euler"></category><category term="Python"></category><category term="sympy"></category></entry><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></summary><category term="Project Euler"></category><category term="Python"></category><category term="sympy"></category></entry><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><updated>2016-10-21T17:40:00+03:00</updated><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="математика"></category></entry><entry><title>Wallabag и реальная жизнь</title><link href="http://likemath.ru/posts/wallabag-i-realnaia-zhizn/" rel="alternate"></link><published>2016-03-17T13:35:00+03:00</published><author><name>Алексей Лобанов</name></author><id>tag:likemath.ru,2016-03-17:posts/wallabag-i-realnaia-zhizn/</id><summary type="html"><p>Как я устанавливал известнейшее свободное решение для отложенного чтения, и что из этого&nbsp;получилось.</p></summary><category term="wallabag"></category><category term="open source"></category><category term="Go"></category></entry><entry><title>Как я шахматного бота писал</title><link href="http://likemath.ru/posts/kak-ia-shakhmatnogo-bota-pisal/" rel="alternate"></link><published>2016-01-10T13:35:00+03:00</published><author><name>Алексей Лобанов</name></author><id>tag:likemath.ru,2016-01-10:posts/kak-ia-shakhmatnogo-bota-pisal/</id><summary type="html"><p>Как я проверял шахматное приложение на&nbsp;&#8220;ботоустойчивость&#8221;.</p></summary><category term="проект"></category><category term="шахматы"></category><category term="бот"></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><entry><title>Ещё одно вычисление выражений</title><link href="http://likemath.ru/posts/eshchio-odno-vychislenie-vyrazhenii/" rel="alternate"></link><published>2015-07-03T17:40:00+03:00</published><author><name>Алексей Лобанов</name></author><id>tag:likemath.ru,2015-07-03:posts/eshchio-odno-vychislenie-vyrazhenii/</id><summary type="html"><p>На хабре когда-то увидел статью про то, что в Яндексе двум сотрудникам дали задачу на написание приложения, для вычисления выражений. Менеджер справился за 4 часа, а программист за два. Я решил попробовать свои&nbsp;силы.</p></summary><category term="проект"></category></entry><entry><title>Мой первый пост</title><link href="http://likemath.ru/posts/moi-pervyi-post/" rel="alternate"></link><published>2015-04-17T13:35:00+03:00</published><author><name>Алексей Лобанов</name></author><id>tag:likemath.ru,2015-04-17:posts/moi-pervyi-post/</id><summary type="html"><p>Кратко о том, почему я сделал&nbsp;блог.</p></summary><category term="блог"></category></entry></feed>
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</script></summary><category term="математика"></category></entry><entry><title>Wallabag и реальная жизнь</title><link href="http://likemath.ru/posts/wallabag-i-realnaia-zhizn/" rel="alternate"></link><published>2016-03-17T13:35:00+03:00</published><updated>2016-03-17T13:35:00+03:00</updated><author><name>Алексей Лобанов</name></author><id>tag:likemath.ru,2016-03-17:posts/wallabag-i-realnaia-zhizn/</id><summary type="html"><p>Как я устанавливал известнейшее свободное решение для отложенного чтения, и что из этого&nbsp;получилось.</p></summary><category term="wallabag"></category><category term="open source"></category><category term="Go"></category></entry><entry><title>Как я шахматного бота писал</title><link href="http://likemath.ru/posts/kak-ia-shakhmatnogo-bota-pisal/" rel="alternate"></link><published>2016-01-10T13:35:00+03:00</published><updated>2016-01-10T13:35:00+03:00</updated><author><name>Алексей Лобанов</name></author><id>tag:likemath.ru,2016-01-10:posts/kak-ia-shakhmatnogo-bota-pisal/</id><summary type="html"><p>Как я проверял шахматное приложение на&nbsp;&#8220;ботоустойчивость&#8221;.</p></summary><category term="проект"></category><category term="шахматы"></category><category term="бот"></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><updated>2015-08-04T17:40:00+03:00</updated><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><updated>2015-11-22T23:41:00+03:00</updated><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><entry><title>Ещё одно вычисление выражений</title><link href="http://likemath.ru/posts/eshchio-odno-vychislenie-vyrazhenii/" rel="alternate"></link><published>2015-07-03T17:40:00+03:00</published><updated>2015-07-03T17:40:00+03:00</updated><author><name>Алексей Лобанов</name></author><id>tag:likemath.ru,2015-07-03:posts/eshchio-odno-vychislenie-vyrazhenii/</id><summary type="html"><p>На хабре когда-то увидел статью про то, что в Яндексе двум сотрудникам дали задачу на написание приложения, для вычисления выражений. Менеджер справился за 4 часа, а программист за два. Я решил попробовать свои&nbsp;силы.</p></summary><category term="проект"></category></entry><entry><title>Мой первый пост</title><link href="http://likemath.ru/posts/moi-pervyi-post/" rel="alternate"></link><published>2015-04-17T13:35:00+03:00</published><updated>2015-04-17T13:35:00+03:00</updated><author><name>Алексей Лобанов</name></author><id>tag:likemath.ru,2015-04-17:posts/moi-pervyi-post/</id><summary type="html"><p>Кратко о том, почему я сделал&nbsp;блог.</p></summary><category term="блог"></category></entry></feed>
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</script></summary><category term="Project Euler"></category><category term="Python"></category><category term="sympy"></category></entry><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></summary><category term="Project Euler"></category><category term="Python"></category><category term="sympy"></category></entry><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><updated>2016-10-21T17:40:00+03:00</updated><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="математика"></category></entry><entry><title>Wallabag и реальная жизнь</title><link href="http://likemath.ru/posts/wallabag-i-realnaia-zhizn/" rel="alternate"></link><published>2016-03-17T13:35:00+03:00</published><author><name>Алексей Лобанов</name></author><id>tag:likemath.ru,2016-03-17:posts/wallabag-i-realnaia-zhizn/</id><summary type="html"><p>Как я устанавливал известнейшее свободное решение для отложенного чтения, и что из этого&nbsp;получилось.</p></summary><category term="wallabag"></category><category term="open source"></category><category term="Go"></category></entry><entry><title>Как я шахматного бота писал</title><link href="http://likemath.ru/posts/kak-ia-shakhmatnogo-bota-pisal/" rel="alternate"></link><published>2016-01-10T13:35:00+03:00</published><author><name>Алексей Лобанов</name></author><id>tag:likemath.ru,2016-01-10:posts/kak-ia-shakhmatnogo-bota-pisal/</id><summary type="html"><p>Как я проверял шахматное приложение на&nbsp;&#8220;ботоустойчивость&#8221;.</p></summary><category term="проект"></category><category term="шахматы"></category><category term="бот"></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><entry><title>Ещё одно вычисление выражений</title><link href="http://likemath.ru/posts/eshchio-odno-vychislenie-vyrazhenii/" rel="alternate"></link><published>2015-07-03T17:40:00+03:00</published><author><name>Алексей Лобанов</name></author><id>tag:likemath.ru,2015-07-03:posts/eshchio-odno-vychislenie-vyrazhenii/</id><summary type="html"><p>На хабре когда-то увидел статью про то, что в Яндексе двум сотрудникам дали задачу на написание приложения, для вычисления выражений. Менеджер справился за 4 часа, а программист за два. Я решил попробовать свои&nbsp;силы.</p></summary><category term="проект"></category></entry><entry><title>Мой первый пост</title><link href="http://likemath.ru/posts/moi-pervyi-post/" rel="alternate"></link><published>2015-04-17T13:35:00+03:00</published><author><name>Алексей Лобанов</name></author><id>tag:likemath.ru,2015-04-17:posts/moi-pervyi-post/</id><summary type="html"><p>Кратко о том, почему я сделал&nbsp;блог.</p></summary><category term="блог"></category></entry></feed>
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</script></summary><category term="математика"></category></entry><entry><title>Wallabag и реальная жизнь</title><link href="http://likemath.ru/posts/wallabag-i-realnaia-zhizn/" rel="alternate"></link><published>2016-03-17T13:35:00+03:00</published><updated>2016-03-17T13:35:00+03:00</updated><author><name>Алексей Лобанов</name></author><id>tag:likemath.ru,2016-03-17:posts/wallabag-i-realnaia-zhizn/</id><summary type="html"><p>Как я устанавливал известнейшее свободное решение для отложенного чтения, и что из этого&nbsp;получилось.</p></summary><category term="wallabag"></category><category term="open source"></category><category term="Go"></category></entry><entry><title>Как я шахматного бота писал</title><link href="http://likemath.ru/posts/kak-ia-shakhmatnogo-bota-pisal/" rel="alternate"></link><published>2016-01-10T13:35:00+03:00</published><updated>2016-01-10T13:35:00+03:00</updated><author><name>Алексей Лобанов</name></author><id>tag:likemath.ru,2016-01-10:posts/kak-ia-shakhmatnogo-bota-pisal/</id><summary type="html"><p>Как я проверял шахматное приложение на&nbsp;&#8220;ботоустойчивость&#8221;.</p></summary><category term="проект"></category><category term="шахматы"></category><category term="бот"></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><updated>2015-08-04T17:40:00+03:00</updated><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><updated>2015-11-22T23:41:00+03:00</updated><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><entry><title>Ещё одно вычисление выражений</title><link href="http://likemath.ru/posts/eshchio-odno-vychislenie-vyrazhenii/" rel="alternate"></link><published>2015-07-03T17:40:00+03:00</published><updated>2015-07-03T17:40:00+03:00</updated><author><name>Алексей Лобанов</name></author><id>tag:likemath.ru,2015-07-03:posts/eshchio-odno-vychislenie-vyrazhenii/</id><summary type="html"><p>На хабре когда-то увидел статью про то, что в Яндексе двум сотрудникам дали задачу на написание приложения, для вычисления выражений. Менеджер справился за 4 часа, а программист за два. Я решил попробовать свои&nbsp;силы.</p></summary><category term="проект"></category></entry><entry><title>Мой первый пост</title><link href="http://likemath.ru/posts/moi-pervyi-post/" rel="alternate"></link><published>2015-04-17T13:35:00+03:00</published><updated>2015-04-17T13:35:00+03:00</updated><author><name>Алексей Лобанов</name></author><id>tag:likemath.ru,2015-04-17:posts/moi-pervyi-post/</id><summary type="html"><p>Кратко о том, почему я сделал&nbsp;блог.</p></summary><category term="блог"></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/proekty.atom.xml" rel="self"></link><id>http://likemath.ru/</id><updated>2016-01-10T13:35:00+03:00</updated><entry><title>Как я шахматного бота писал</title><link href="http://likemath.ru/posts/kak-ia-shakhmatnogo-bota-pisal/" rel="alternate"></link><published>2016-01-10T13:35:00+03:00</published><updated>2016-01-10T13:35:00+03:00</updated><author><name>Алексей Лобанов</name></author><id>tag:likemath.ru,2016-01-10:posts/kak-ia-shakhmatnogo-bota-pisal/</id><summary type="html"><p>Как я проверял шахматное приложение на&nbsp;&#8220;ботоустойчивость&#8221;.</p></summary><category term="проект"></category><category term="шахматы"></category><category term="бот"></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><updated>2015-08-04T17:40:00+03:00</updated><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>Ещё одно вычисление выражений</title><link href="http://likemath.ru/posts/eshchio-odno-vychislenie-vyrazhenii/" rel="alternate"></link><published>2015-07-03T17:40:00+03:00</published><updated>2015-07-03T17:40:00+03:00</updated><author><name>Алексей Лобанов</name></author><id>tag:likemath.ru,2015-07-03:posts/eshchio-odno-vychislenie-vyrazhenii/</id><summary type="html"><p>На хабре когда-то увидел статью про то, что в Яндексе двум сотрудникам дали задачу на написание приложения, для вычисления выражений. Менеджер справился за 4 часа, а программист за два. Я решил попробовать свои&nbsp;силы.</p></summary><category term="проект"></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/project-euler.atom.xml" rel="self"></link><id>http://likemath.ru/</id><updated>2016-10-30T17:40:00+03:00</updated><entry><title>Моё решение задачи 134</title><link href="http://likemath.ru/posts/moio-reshenie-zadachi-134/" rel="alternate"></link><published>2016-10-30T17:40:00+03:00</published><author><name>Алексей Лобанов</name></author><id>tag:likemath.ru,2016-10-30:posts/moio-reshenie-zadachi-134/</id><summary type="html"><p>Краткое условие: 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>
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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/project-euler.atom.xml" rel="self"></link><id>http://likemath.ru/</id><updated>2016-10-30T17:40:00+03:00</updated><entry><title>Моё решение задачи 134</title><link href="http://likemath.ru/posts/moio-reshenie-zadachi-134/" rel="alternate"></link><published>2016-10-30T17:40:00+03:00</published><updated>2016-10-30T17:40:00+03:00</updated><author><name>Алексей Лобанов</name></author><id>tag:likemath.ru,2016-10-30:posts/moio-reshenie-zadachi-134/</id><summary type="html"><p>Краткое условие: 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>
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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-flint.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><updated>2016-10-21T17:40:00+03:00</updated><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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<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-go.atom.xml" rel="self"></link><id>http://likemath.ru/</id><updated>2016-03-17T13:35:00+03:00</updated><entry><title>Wallabag и реальная жизнь</title><link href="http://likemath.ru/posts/wallabag-i-realnaia-zhizn/" rel="alternate"></link><published>2016-03-17T13:35:00+03:00</published><author><name>Алексей Лобанов</name></author><id>tag:likemath.ru,2016-03-17:posts/wallabag-i-realnaia-zhizn/</id><summary type="html"><p>Как я устанавливал известнейшее свободное решение для отложенного чтения, и что из этого&nbsp;получилось.</p></summary><category term="wallabag"></category><category term="open source"></category><category term="Go"></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-go.atom.xml" rel="self"></link><id>http://likemath.ru/</id><updated>2016-03-17T13:35:00+03:00</updated><entry><title>Wallabag и реальная жизнь</title><link href="http://likemath.ru/posts/wallabag-i-realnaia-zhizn/" rel="alternate"></link><published>2016-03-17T13:35:00+03:00</published><updated>2016-03-17T13:35:00+03:00</updated><author><name>Алексей Лобанов</name></author><id>tag:likemath.ru,2016-03-17:posts/wallabag-i-realnaia-zhizn/</id><summary type="html"><p>Как я устанавливал известнейшее свободное решение для отложенного чтения, и что из этого&nbsp;получилось.</p></summary><category term="wallabag"></category><category term="open source"></category><category term="Go"></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-matematika.atom.xml" rel="self"></link><id>http://likemath.ru/</id><updated>2016-07-22T13:35:00+03:00</updated><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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<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-matematika.atom.xml" rel="self"></link><id>http://likemath.ru/</id><updated>2016-07-22T13:35:00+03:00</updated><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><updated>2016-07-22T13:35:00+03:00</updated><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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<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-open-source.atom.xml" rel="self"></link><id>http://likemath.ru/</id><updated>2016-03-17T13:35:00+03:00</updated><entry><title>Wallabag и реальная жизнь</title><link href="http://likemath.ru/posts/wallabag-i-realnaia-zhizn/" rel="alternate"></link><published>2016-03-17T13:35:00+03:00</published><updated>2016-03-17T13:35:00+03:00</updated><author><name>Алексей Лобанов</name></author><id>tag:likemath.ru,2016-03-17:posts/wallabag-i-realnaia-zhizn/</id><summary type="html"><p>Как я устанавливал известнейшее свободное решение для отложенного чтения, и что из этого&nbsp;получилось.</p></summary><category term="wallabag"></category><category term="open source"></category><category term="Go"></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-proekt.atom.xml" rel="self"></link><id>http://likemath.ru/</id><updated>2016-01-10T13:35:00+03:00</updated><entry><title>Как я шахматного бота писал</title><link href="http://likemath.ru/posts/kak-ia-shakhmatnogo-bota-pisal/" rel="alternate"></link><published>2016-01-10T13:35:00+03:00</published><author><name>Алексей Лобанов</name></author><id>tag:likemath.ru,2016-01-10:posts/kak-ia-shakhmatnogo-bota-pisal/</id><summary type="html"><p>Как я проверял шахматное приложение на&nbsp;&#8220;ботоустойчивость&#8221;.</p></summary><category term="проект"></category><category term="шахматы"></category><category term="бот"></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>Ещё одно вычисление выражений</title><link href="http://likemath.ru/posts/eshchio-odno-vychislenie-vyrazhenii/" rel="alternate"></link><published>2015-07-03T17:40:00+03:00</published><author><name>Алексей Лобанов</name></author><id>tag:likemath.ru,2015-07-03:posts/eshchio-odno-vychislenie-vyrazhenii/</id><summary type="html"><p>На хабре когда-то увидел статью про то, что в Яндексе двум сотрудникам дали задачу на написание приложения, для вычисления выражений. Менеджер справился за 4 часа, а программист за два. Я решил попробовать свои&nbsp;силы.</p></summary><category term="проект"></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>2016-10-30T17:40:00+03:00</updated><entry><title>Моё решение задачи 134</title><link href="http://likemath.ru/posts/moio-reshenie-zadachi-134/" rel="alternate"></link><published>2016-10-30T17:40:00+03:00</published><author><name>Алексей Лобанов</name></author><id>tag:likemath.ru,2016-10-30:posts/moio-reshenie-zadachi-134/</id><summary type="html"><p>Краткое условие: 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>
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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>2016-10-30T17:40:00+03:00</updated><entry><title>Моё решение задачи 134</title><link href="http://likemath.ru/posts/moio-reshenie-zadachi-134/" rel="alternate"></link><published>2016-10-30T17:40:00+03:00</published><updated>2016-10-30T17:40:00+03:00</updated><author><name>Алексей Лобанов</name></author><id>tag:likemath.ru,2016-10-30:posts/moio-reshenie-zadachi-134/</id><summary type="html"><p>Краткое условие: 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>
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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-python.atom.xml" rel="self"></link><id>http://likemath.ru/</id><updated>2016-10-30T17:40:00+03:00</updated><entry><title>Моё решение задачи 134</title><link href="http://likemath.ru/posts/moio-reshenie-zadachi-134/" rel="alternate"></link><published>2016-10-30T17:40:00+03:00</published><updated>2016-10-30T17:40:00+03:00</updated><author><name>Алексей Лобанов</name></author><id>tag:likemath.ru,2016-10-30:posts/moio-reshenie-zadachi-134/</id><summary type="html"><p>Краткое условие: 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>
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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 134</h1></a></header><section class="post-content"><p>Краткое условие: назовём <em>порождающим</em> для двух последовательных простых <span class="math">\(p_1 < p_2\)</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">\(p_1 \in \left[ 5; 10^6 \right]\)</span></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="2016-10-21T17:40:00+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"><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="2015-07-03T17:40:00+03:00" pubdate> Пт 03 Июль 2015 </time><a href="../../posts/eshchio-odno-vychislenie-vyrazhenii/" rel="bookmark"><h1>Ещё одно вычисление выражений</h1></a></header><section class="post-content"><p>Задачка кажется не очень сложной, даже, если не знать как её делать (я не знал). Целью является быстрое вычисление чего-то типа <code>4 * ( 5 + 7 ^ 4)</code>. Для это я парсил исходную строку в список токенов, а затем непосредственно вычислял, что получится.</p><p>Я решил, что проще всего будет реализовать (а мне потом и понять) алгоритм, когда после каждого действия будет выполняться некий “хороший” инвариант. Первое что приходит в голову — это то, что истинность выражение после выполнения операции не меняется (<span class="caps">TITO</span> соблюдается). То есть выражение <code>3 + 5</code> можно заменить на <code>8</code> или хотя бы на <code>4 * 2</code>.</p><p>Непосредственно сама обработка является несколькими проходами, так что в каждом проходе мы избавляемся от операций одного приоритета. <code>4 + 5 * 3</code> заменяется на <code>4 + 15, 7 - 5 * 2^3</code> заменяется на <code>7 - 5*8</code>. Таким образом, каждый цикл тривиален, и легко задавать приоритеты операций.</p><p>Если использовать один список как контейнер для токенов и при работе изменять непосредственно его, сохраняя указанные инварианты, то сложность получается <span class="math">\(O\left( N \right)\)</span>, где <span class="math">\(N\)</span> — число токенов.</p><p>Времени на непосредственно кодирование ушло часа три-четыре, но в это время не входит продумывание мелких деталей.</p><p>Всё написано на C++11. Исходники лежат на <a href="https://github.com/AlekseyLobanov/ExprEvaler">GitHub</a> и <a href="https://bitbucket.org/Begemot23/exprevaler">BitBucket</a>.</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-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="2015-04-17T13:35:00+03:00" pubdate> Пт 17 Апрель 2015 </time><a href="../../posts/moi-pervyi-post/" rel="bookmark"><h1>Мой первый пост</h1></a></header><section class="post-content"><p>Главная причина появления заключается в том, что мне захотелось использовать хоть как-то купленный домен (я купил его только ради почты). Плюс мне бы хотелось проще давать контактные данные, а адрес сайта достаточно простой.</p><p>Пока я планирую публиковать свои, возможно не лучшие, но рабочие решения для задач из Project Euler (<a href="https://projecteuler.net/">projecteuler.net</a>). На самом деле, решения значительного числа задач уже есть в Сети, но на английском. Хотя, это и не есть большая проблема.</p></section><hr><aside class="post-meta"><p>Категория: <a href="../../category/misc.html">misc</a></p><p>Теги: <a href="../../tag/blog.html">блог</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-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="2015-04-17T13:35:00+03:00" pubdate> Пт 17 Апрель 2015 </time><a href="../../posts/moi-pervyi-post/" rel="bookmark"><h1>Мой первый пост</h1></a></header><section class="post-content"><p>Главная причина появления заключается в том, что мне захотелось использовать хоть как-то купленный домен (я купил его только ради почты). Плюс мне бы хотелось проще давать контактные данные, а адрес сайта достаточно простой.</p><p>Пока я планирую публиковать свои, возможно не лучшие, но рабочие решения для задач из Project Euler (<a href="https://projecteuler.net/">projecteuler.net</a>). На самом деле, решения значительного числа задач уже есть в Сети, но на английском. Хотя, это и не есть большая проблема.</p></section><hr><aside class="post-meta"><p>Категория: <a href="../../category/misc.html">misc</a></p><p>Теги: <a href="../../tag/blog.html">блог</a>, </p></aside><hr></article></li></ol></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-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 134</h1></a></header><section class="post-content"><p>Назовём <em>порождающим</em> для двух последовательных простых <span class="math">\(p_1 < p_2\)</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">\(p_1 \in \left[ 5; 10^6 \right]\)</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 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> — какое-то натуральное число с отрезка <span class="math">\(\left[ 1; p_2-1 \right]\)</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>: <div class="math">$$ r \equiv -p_1 \cdot 10^{-k} \equiv -p_1 \cdot 10^{p_2 -1-k} $$</div></p><p>Комментарии:</p><ol><li>Так как <span class="math">\(a^p \equiv a \mod p\)</span>, то верно что <span class="math">\(a^{-k} \equiv a^{p -1-k} \mod p\)</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 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 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>
|
||||
<![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 134</h1></a></header><section class="post-content"><p>Назовём <em>порождающим</em> для двух последовательных простых <span class="math">\(p_1 < p_2\)</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">\(p_1 \in \left[ 5; 10^6 \right]\)</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 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> — какое-то натуральное число с отрезка <span class="math">\(\left[ 1; p_2-1 \right]\)</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">\(a^{-k} \equiv a^{p -1-k} \mod p\)</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 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>
|
||||
@@ -25,19 +25,30 @@
|
||||
<span class="code-line"></span>
|
||||
<span class="code-line"><span class="k">print</span><span class="p">(</span><span class="s1">'Result is {}'</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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<![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="2016-10-21T17:40:00+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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<![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="2016-10-21T17:40:00+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></span><span class="cm">/*</span></span>
|
||||
<span class="code-line"><span class="cm"> * Problem 146 on Project Euler</span></span>
|
||||
<span class="code-line"><span class="cm"> * Aleksey Lobanov (c) 2016</span></span>
|
||||
<span class="code-line"><span class="cm"> */</span></span>
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@@ -108,19 +108,30 @@
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||||
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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="2016-10-21T17:40:00+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="2015-08-02T17:40:00+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-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="2016-10-21T17:40:00+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-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="2016-07-22T13:35:00+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-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 134</h1></a></header><section class="post-content"><p>Краткое условие: назовём <em>порождающим</em> для двух последовательных простых <span class="math">\(p_1 < p_2\)</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">\(p_1 \in \left[ 5; 10^6 \right]\)</span></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>Блог 529 - Python</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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<p class="chromeframe">You are using an <strong>outdated</strong> browser. Please <a href="http://browsehappy.com/">upgrade your browser</a> or <a href="http://www.google.com/chromeframe/?redirect=true">activate Google Chrome Frame</a> to improve your experience.</p>
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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 134</h1></a></header><section class="post-content"><p>Краткое условие: назовём <em>порождающим</em> для двух последовательных простых <span class="math">\(p_1 < p_2\)</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">\(p_1 \in \left[ 5; 10^6 \right]\)</span></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>Блог 529 - sympy</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 134</h1></a></header><section class="post-content"><p>Краткое условие: назовём <em>порождающим</em> для двух последовательных простых <span class="math">\(p_1 < p_2\)</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">\(p_1 \in \left[ 5; 10^6 \right]\)</span></p><script type="text/javascript">if (!document.getElementById('mathjaxscript_pelican_#%@#$@#')) {
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