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<feed xmlns="http://www.w3.org/2005/Atom"><title>Блог 529 - c++</title><link href="https://likemath.ru/" rel="alternate"></link><link href="https://likemath.ru/feeds/tag-c.atom.xml" rel="self"></link><id>https://likemath.ru/</id><updated>2016-10-21T17:40:00+03:00</updated><subtitle>Project Euler и остальное</subtitle><entry><title>Моё решение задачи 146</title><link href="https://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">&lt;p&gt;Краткое условие: необходимо найти сумму всех натуральных &lt;span class="math"&gt;\(n\)&lt;/span&gt;, что &lt;span class="math"&gt;\(n^2+1\)&lt;/span&gt;, &lt;span class="math"&gt;\(n^2+3\)&lt;/span&gt;, &lt;span class="math"&gt;\(n^2+7\)&lt;/span&gt;, &lt;span class="math"&gt;\(n^2+9\)&lt;/span&gt;, &lt;span class="math"&gt;\(n^2+13\)&lt;/span&gt;, и &lt;span class="math"&gt;\(n^2+27\)&lt;/span&gt; будут последовательными простыми&amp;nbsp;числами.&lt;/p&gt;
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&lt;/script&gt;</summary><category term="Project Euler"></category><category term="c++"></category><category term="FLINT"></category></entry><entry><title>CrossGen v1.0</title><link href="https://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">&lt;p&gt;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&amp;nbsp;приложения.&lt;/p&gt;</summary><category term="проект"></category><category term="c++"></category><category term="wxWidgets"></category></entry><entry><title>Моё решение задачи 60</title><link href="https://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">&lt;p&gt;Краткое условие: необходимо найти множество из пяти простых чисел с минимальной суммой такое, что после &amp;#8220;склеивания&amp;#8221; в любом порядке любых двух чисел из него тоже будет простое&amp;nbsp;число.&lt;/p&gt;</summary><category term="Project Euler"></category><category term="c++"></category><category term="BGL"></category></entry></feed>