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<article>
<header>
<h1 class="entry-title"><a href="/blog/2014/04/24/project-euler-simple/">Project Euler Without Descriptions</a></h1>
<p class="meta">
<time datetime="2014-04-24T15:22:00+08:00" pubdate data-updated="true">Apr 24<span>th</span>, 2014</time>
| <a href="/blog/2014/04/24/project-euler-simple/#disqus_thread">Comments</a>
</p>
</header>
<div class="entry-content"></p>
<div id="table-of-contents">
<h2>Table of Contents</h2>
<div id="text-table-of-contents">
<ul>
<li><a href="#sec-1">1 Straight Forward</a></li>
<li><a href="#sec-2">2 Answer</a></li>
</ul>
</div>
</div>
<div id="outline-container-1" class="outline-2">
<h2 id="sec-1">Straight Forward</h2>
<div class="outline-text-2" id="text-1">
<table border="2" cellspacing="0" cellpadding="6" rules="groups" frame="hsides">
<colgroup><col class="right" /></colgroup>
<colgroup><col class="right" /></colgroup><colgroup><col class="right" /></colgroup><colgroup><col class="right" /></colgroup><colgroup><col class="right" /></colgroup>
<thead>
</thead>
<tbody>
<tr><td class="right">32</td><td class="right">52</td><td class="right">41</td><td class="right">53</td><td class="right">38</td></tr>
</tbody>
<tbody>
<tr><td class="right">45228</td><td class="right">142857</td><td class="right">7652413</td><td class="right">4075</td><td class="right">932718654</td></tr>
</tbody>
<tbody>
<tr><td class="right">50</td><td class="right">56</td><td class="right">46</td><td class="right">43</td><td class="right">47</td></tr>
</tbody>
<tbody>
<tr><td class="right">997651</td><td class="right">972</td><td class="right">5777</td><td class="right">16695334890</td><td class="right">134043</td></tr>
</tbody>
<tbody>
<tr><td class="right">49</td><td class="right">55</td><td class="right">44</td><td class="right">97</td><td class="right"></td></tr>
</tbody>
<tbody>
<tr><td class="right">296962999629</td><td class="right">249</td><td class="right">5482660</td><td class="right">8739992577</td><td class="right"></td></tr>
</tbody>
</table>
</div>
</div>
<div id="outline-container-2" class="outline-2">
<h2 id="sec-2">Answer</h2>
<div class="outline-text-2" id="text-2">
<p>Source:<a href="https://github.com/kingFighter/project-euler">project-euler</a>
</p></div>
</div>
</div>
</article>
<article>
<header>
<h1 class="entry-title"><a href="/blog/2014/04/10/project-euler-33/">Project Euler Problem 33</a></h1>
<p class="meta">
<time datetime="2014-04-10T22:38:00+08:00" pubdate data-updated="true">Apr 10<span>th</span>, 2014</time>
| <a href="/blog/2014/04/10/project-euler-33/#disqus_thread">Comments</a>
</p>
</header>
<div class="entry-content"></p>
<div id="table-of-contents">
<h2>Table of Contents</h2>
<div id="text-table-of-contents">
<ul>
<li><a href="#sec-1">1 Problem</a></li>
<li><a href="#sec-2">2 Solution</a></li>
<li><a href="#sec-3">3 Answer</a></li>
</ul>
</div>
</div>
<div id="outline-container-1" class="outline-2">
<h2 id="sec-1">Problem</h2>
<div class="outline-text-2" id="text-1">
<blockquote>
<p>The fraction 49/98 is a curious fraction, as an inexperienced mathematician in attempting to simplify it may incorrectly believe that 49/98 = 4/8, which is correct, is obtained by cancelling the 9s.
</p>
<p>
We shall consider fractions like, 30/50 = 3/5, to be trivial examples.
</p>
<p>
There are exactly four non-trivial examples of this type of fraction, less than one in value, and containing two digits in the numerator and denominator.
</p>
<p>
If the product of these four fractions is given in its lowest common terms, find the value of the denominator.
</p>
</blockquote>
</div>
</div>
<div id="outline-container-2" class="outline-2">
<h2 id="sec-2">Solution</h2>
<div class="outline-text-2" id="text-2">
<p>很简单,解空间很小,可直接Brute Force.
</p>
<p>
但我们可以进一步缩小解空间.其可能的四种形式:
</p><ol>
<li>\begin{eqnarray*}
\frac{10a+b}{10c+b} = \frac{a}{c}
\end{eqnarray*}
</li>
<li>\begin{eqnarray*}
\frac{10a+b}{10b+c} = \frac{a}{c}
\end{eqnarray*}
</li>
<li>\begin{eqnarray*}
\frac{10b+a}{10c+b} = \frac{a}{c}
\end{eqnarray*}
</li>
<li>\begin{eqnarray*}
\frac{10b+a}{10b+c} = \frac{a}{c}
\end{eqnarray*}
</li>
</ol>
<p>
对于1,4 化简得出a=c,与条件矛盾
</p>
<p>
对于3,可利用a,b,c <=9 及a<b可得到矛盾
</p>
<p>
对此只有2才是可能的形式
</p></div>
</div>
<div id="outline-container-3" class="outline-2">
<h2 id="sec-3">Answer</h2>
<div class="outline-text-2" id="text-3">
<p>100
</p>
<p>
Source:<a href="https://github.com/kingFighter/project-euler">project-euler</a>
</p></div>
</div>
</div>
</article>
<article>
<header>
<h1 class="entry-title"><a href="/blog/2014/04/05/project-euler-39/">Project Euler Problem 39</a></h1>
<p class="meta">
<time datetime="2014-04-05T00:09:00+08:00" pubdate data-updated="true">Apr 5<span>th</span>, 2014</time>
| <a href="/blog/2014/04/05/project-euler-39/#disqus_thread">Comments</a>
</p>
</header>
<div class="entry-content"></p>
<div id="table-of-contents">
<h2>Table of Contents</h2>
<div id="text-table-of-contents">
<ul>
<li><a href="#sec-1">1 Problem</a></li>
<li><a href="#sec-2">2 Solution</a></li>
<li><a href="#sec-3">3 Answer</a></li>
</ul>
</div>
</div>
<div id="outline-container-1" class="outline-2">
<h2 id="sec-1">Problem</h2>
<div class="outline-text-2" id="text-1">
<blockquote>
<p>If p is the perimeter of a right angle triangle with integral length sides, {a,b,c}, there are exactly three solutions for p = 120.
</p>
<p>
{20,48,52}, {24,45,51}, {30,40,50}
</p>
<p>
For which value of p ≤ 1000, is the number of solutions maximised?
</p>
</blockquote>
</div>
</div>
<div id="outline-container-2" class="outline-2">
<h2 id="sec-2">Solution</h2>
<div class="outline-text-2" id="text-2">
<p>简单,Brute Force.
</p></div>
</div>
<div id="outline-container-3" class="outline-2">
<h2 id="sec-3">Answer</h2>
<div class="outline-text-2" id="text-3">
<p>840
</p>
<p>
Source:<a href="https://github.com/kingFighter/project-euler">project-euler</a>
</p></div>
</div>
</div>
</article>
<article>
<header>
<h1 class="entry-title"><a href="/blog/2014/04/04/project-euler-45/">Project Euler Problem 45</a></h1>
<p class="meta">
<time datetime="2014-04-04T23:28:00+08:00" pubdate data-updated="true">Apr 4<span>th</span>, 2014</time>
| <a href="/blog/2014/04/04/project-euler-45/#disqus_thread">Comments</a>
</p>
</header>
<div class="entry-content"></p>
<div id="table-of-contents">
<h2>Table of Contents</h2>
<div id="text-table-of-contents">
<ul>
<li><a href="#sec-1">1 Problem</a></li>
<li><a href="#sec-2">2 Solution</a></li>
<li><a href="#sec-3">3 Answer</a></li>
</ul>
</div>
</div>
<div id="outline-container-1" class="outline-2">
<h2 id="sec-1">Problem</h2>
<div class="outline-text-2" id="text-1">
<blockquote>
<p>Triangle, pentagonal, and hexagonal numbers are generated by the following formulae:
</p>
<p>
Triangle Tn=n(n+1)/2 1, 3, 6, 10, 15, …
Pentagonal Pn=n(3n−1)/2 1, 5, 12, 22, 35, …
Hexagonal Hn=n(2n−1) 1, 6, 15, 28, 45, …
It can be verified that T285 = P165 = H143 = 40755.
</p>
<p>
Find the next triangle number that is also pentagonal and hexagonal.
</p>
</blockquote>
</div>
</div>
<div id="outline-container-2" class="outline-2">
<h2 id="sec-2">Solution</h2>
<div class="outline-text-2" id="text-2">
<p>简单,一元二次方程组.
</p></div>
</div>
<div id="outline-container-3" class="outline-2">
<h2 id="sec-3">Answer</h2>
<div class="outline-text-2" id="text-3">
<p>1533776805
</p>
<p>
Source:<a href="https://github.com/kingFighter/project-euler">project-euler</a>
</p></div>
</div>
</div>
</article>
<article>
<header>
<h1 class="entry-title"><a href="/blog/2014/03/14/project-euler-42/">Project Euler Problem 42</a></h1>
<p class="meta">
<time datetime="2014-03-14T21:11:00+08:00" pubdate data-updated="true">Mar 14<span>th</span>, 2014</time>
| <a href="/blog/2014/03/14/project-euler-42/#disqus_thread">Comments</a>
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</header>
<div class="entry-content"></p>
<div id="table-of-contents">
<h2>Table of Contents</h2>
<div id="text-table-of-contents">
<ul>
<li><a href="#sec-1">1 Problem</a></li>
<li><a href="#sec-2">2 Solution</a></li>
<li><a href="#sec-3">3 Answer</a></li>
</ul>
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<h2 id="sec-1">Problem</h2>
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<blockquote>
<p>The nth term of the sequence of triangle numbers is given by, tn = ½n(n+1); so the first ten triangle numbers are:
</p>
<p>
1, 3, 6, 10, 15, 21, 28, 36, 45, 55, …
</p>
<p>
By converting each letter in a word to a number corresponding to its alphabetical position and adding these values we form a word value. For example, the word value for SKY is 19 + 11 + 25 = 55 = t10. If the word value is a triangle number then we shall call the word a triangle word.
</p>
<p>
Using <a href="https://projecteuler.net/project/words.txt">words.txt</a> (right click and ‘Save Link/Target As…’), a 16K text file containing nearly two-thousand common English words, how many are triangle words?
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<h2 id="sec-2">Solution</h2>
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<p>太简单
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<h2 id="sec-3">Answer</h2>
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<p>162
</p>
<p>
Source:<a href="https://github.com/kingFighter/project-euler">project-euler</a>
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</article>
<article>
<header>
<h1 class="entry-title"><a href="/blog/2014/03/03/project-euler-31/">Project Euler Problem 31</a></h1>
<p class="meta">
<time datetime="2014-03-03T20:45:00+08:00" pubdate data-updated="true">Mar 3<span>rd</span>, 2014</time>
| <a href="/blog/2014/03/03/project-euler-31/#disqus_thread">Comments</a>
</p>
</header>
<div class="entry-content"></p>
<div id="table-of-contents">
<h2>Table of Contents</h2>
<div id="text-table-of-contents">
<ul>
<li><a href="#sec-1">1 Problem</a></li>
<li><a href="#sec-2">2 Solution</a>
<ul>
<li><a href="#sec-2-1">2.1 Brute Force</a></li>
<li><a href="#sec-2-2">2.2 Recursion</a></li>
<li><a href="#sec-2-3">2.3 DP</a></li>
</ul>
</li>
<li><a href="#sec-3">3 Answer</a></li>
</ul>
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<h2 id="sec-1">Problem</h2>
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<blockquote>
<p>In England the currency is made up of pound, £, and pence, p, and there are eight coins in general circulation:
</p>
<p>
1p, 2p, 5p, 10p, 20p, 50p, £1 (100p) and £2 (200p).
</p>
<p>
It is possible to make £2 in the following way:
</p>
<p>
1×£1 + 1×50p + 2×20p + 1×5p + 1×2p + 3×1p
</p>
<p>
How many different ways can £2 be made using any number of coins?
</p>
</blockquote>
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<h2 id="sec-2">Solution</h2>
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<h3 id="sec-2-1">Brute Force</h3>
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<p>可以用穷举法:
</p>
<p>
200a+100b+50c+20d+10e+5f+2g+h = 200
</p>
<p>
穷尽a,b,c,d,e,f,g,h所有的可能.a:0~1, b:0~2…., 不过知道a = 1时, b:0,
可缩小a,b,c….大小范围, 见 C++Brute1, 类似的见C++Brute2
</p></div>
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<h3 id="sec-2-2">Recursion</h3>
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<p>很直接的方法:
</p><ol>
<li>用一个200的coin, 没用其它的
</li>
<li>用零个200的coin, 由最大coin为100的来换零钱200
</li>
<li>用两个100的coin, 没用其它的
</li>
<li>用一个100的coin, 由最大coin为50的来换零钱100
</li>
<li>用零个100的coin, 由最大coin为50的来换零钱200
</li>
</ol>
<p> …..
</p>
<p>
这里大家应该能看出规律, 用公式表示:
</p>
<p>
\begin{eqnarray*}
f(t, c) = 1 & & if & c =1 &or& t=0 & \newline
f(t,c) = \sum_{i=0}^{\left \lfloor \frac{t}{c} \right \rfloor}f(t-ic, s(c)) & & if & c > 1 & and & t>0 &
\end{eqnarray*}
其中t是要换的钱, c是当前最大的coin,s(c)是下一个比c小的coin
</p>
<p>
见 C++Formula, 稍有不同的解法见 C++Formula1
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<h3 id="sec-2-3">DP</h3>
<div class="outline-text-3" id="text-2-3">
<p>见<a href="http://www.mathblog.dk/project-euler-31-combinations-english-currency-denominations/">分析</a>
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<h2 id="sec-3">Answer</h2>
<div class="outline-text-2" id="text-3">
<p>73682
</p>
<p>
Source:<a href="http://1drv.ms/Nn1zWO">C++Formula</a>, <a href="http://1drv.ms/1eVztrC">C++Formula1,</a> <a href="http://1drv.ms/1cnATAU">C++Brute1</a>, <a href="http://1drv.ms/1gMSteo">C++Brute2</a>, <a href="http://1drv.ms/1eVzMCU">dp</a>
</p></div>
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</article>
<article>
<header>
<h1 class="entry-title"><a href="/blog/2014/03/02/project-euler-26/">Project Euler Problem 26</a></h1>
<p class="meta">
<time datetime="2014-03-02T14:13:00+08:00" pubdate data-updated="true">Mar 2<span>nd</span>, 2014</time>
| <a href="/blog/2014/03/02/project-euler-26/#disqus_thread">Comments</a>
</p>
</header>
<div class="entry-content"></p>
<div id="table-of-contents">
<h2>Table of Contents</h2>
<div id="text-table-of-contents">
<ul>
<li><a href="#sec-1">1 Problem</a></li>
<li><a href="#sec-2">2 Solution</a></li>
<li><a href="#sec-3">3 Answer</a></li>
</ul>
</div>
</div>
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<h2 id="sec-1">Problem</h2>
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<blockquote>
<p>
A unit fraction contains 1 in the numerator. The decimal representation of the unit fractions with denominators 2 to 10 are given:
</p>
<p>
1/2 = 0.5
</p>
<p>
1/3 = 0.(3)
</p>
<p>
1/4 = 0.25
</p>
<p>
1/5 = 0.2
</p>
<p>
1/6 = 0.1(6)
</p>
<p>
1/7 = 0.(142857)
</p>
<p>
1/8 = 0.125
</p>
<p>
1/9 = 0.(1)
</p>
<p>
1/10 = 0.1
</p>
<p>
Where 0.1(6) means 0.166666…, and has a 1-digit recurring cycle. It can be seen that 1/7 has a 6-digit recurring cycle.
</p>
<p>
Find the value of d < 1000 for which 1/d contains the longest recurring cycle in its decimal fraction part.
</p>
</blockquote>
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<h2 id="sec-2">Solution</h2>
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<p> Brute Force,将小数部分保存vector中, 然后从头遍历寻找是否有相同的
pattern, 重复找到三次即认可.
</p>
<p>
更优的做法需要数学知识<sup><a class="footref" name="fnr-.1" href="#fn-.1">1</a></sup>
</p></div>
</div>
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<h2 id="sec-3">Answer</h2>
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<p>983
</p>
<p>
Source:<a href="http://1drv.ms/1eOxC7U">C++</a>
</p>
<div id="footnotes">
<h2 class="footnotes">Footnotes: </h2>
<div id="text-footnotes">
<p class="footnote"><sup><a class="footnum" name="fn-.1" href="#fnr-.1">1</a></sup> <a href="http://mathworld.wolfram.com/DecimalExpansion.html">http://mathworld.wolfram.com/DecimalExpansion.html</a>
</p>
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</div>
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</article>
<article>
<header>
<h1 class="entry-title"><a href="/blog/2013/08/28/project-euler-40/">Project Euler Problem 40</a></h1>
<p class="meta">
<time datetime="2013-08-28T21:51:00+08:00" pubdate data-updated="true">Aug 28<span>th</span>, 2013</time>
| <a href="/blog/2013/08/28/project-euler-40/#disqus_thread">Comments</a>
</p>
</header>
<div class="entry-content"></p>
<div id="table-of-contents">
<h2>Table of Contents</h2>
<div id="text-table-of-contents">
<ul>
<li><a href="#sec-1">1 Problem</a></li>
<li><a href="#sec-2">2 Solution</a></li>
<li><a href="#sec-3">3 Answer</a></li>
</ul>
</div>
</div>
<div id="outline-container-1" class="outline-2">
<h2 id="sec-1"><span class="section-number-2">1</span> Problem</h2>
<div class="outline-text-2" id="text-1">
<blockquote>
<p>An irrational decimal fraction is created by concatenating the positive integers:
</p>
<p>
0.123456789101112131415161718192021…
</p>
<p>
It can be seen that the \(12^{th}\) digit of the fractional part is 1.
</p>
<p>
If dn represents the nth digit of the fractional part, find the value of the following expression.
</p>
\begin{eqnarray*}
d_{1} × d_{10} × d_{100} × d_{1000} × d_{10000} × d_{100000} × d_{1000000}
\end{eqnarray*}
</blockquote>
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<h2 id="sec-2"><span class="section-number-2">2</span> Solution</h2>
<div class="outline-text-2" id="text-2">
<p>二分法,我们无法直接得知第ith的digit,但是容易知道number n所在的位置.要找ith的digit,其必夹在1~i number之间的某个位置.如：
</p>
<p>
找1000th的digit
</p><ol>
<li>其必夹在1~1000 number之间,易知 (1 + 1000) / 2 = 500 的最后一位0在序列中的位置为1392th(number 11的最后一位1的位置为13),比1000th大,则必在1~499间
</li>
<li>如此往复,我们可以找到最接近的number 370,其最后一位0为1002th,则1000th为digit 3
</li>
</ol>