atom.xml

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  <title><![CDATA[eBash]]></title>
  <link href="http://kingfighter.github.io//atom.xml" rel="self"/>
  <link href="http://kingfighter.github.io//"/>
  <updated>2017-04-01T11:27:41+08:00</updated>
  <id>http://kingfighter.github.io//</id>
  <author>
    <name><![CDATA[Lv Kaiyang(Kevin Lui)]]></name>
    
  </author>
  <generator uri="http://octopress.org/">Octopress</generator>

  
  <entry>
    <title type="html"><![CDATA[project euler without descriptions]]></title>
    <link href="http://kingfighter.github.io//blog/2014/04/24/project-euler-simple/"/>
    <updated>2014-04-24T15:22:00+08:00</updated>
    <id>http://kingfighter.github.io//blog/2014/04/24/project-euler-simple</id>
    <content type="html"><![CDATA[</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>
]]></content>
  </entry>
  
  <entry>
    <title type="html"><![CDATA[project euler problem 33]]></title>
    <link href="http://kingfighter.github.io//blog/2014/04/10/project-euler-33/"/>
    <updated>2014-04-10T22:38:00+08:00</updated>
    <id>http://kingfighter.github.io//blog/2014/04/10/project-euler-33</id>
    <content type="html"><![CDATA[</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 &lt;=9 及a&lt;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>
]]></content>
  </entry>
  
  <entry>
    <title type="html"><![CDATA[project euler problem 39]]></title>
    <link href="http://kingfighter.github.io//blog/2014/04/05/project-euler-39/"/>
    <updated>2014-04-05T00:09:00+08:00</updated>
    <id>http://kingfighter.github.io//blog/2014/04/05/project-euler-39</id>
    <content type="html"><![CDATA[</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>
]]></content>
  </entry>
  
  <entry>
    <title type="html"><![CDATA[project euler problem 45]]></title>
    <link href="http://kingfighter.github.io//blog/2014/04/04/project-euler-45/"/>
    <updated>2014-04-04T23:28:00+08:00</updated>
    <id>http://kingfighter.github.io//blog/2014/04/04/project-euler-45</id>
    <content type="html"><![CDATA[</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, &hellip;
Pentagonal              Pn=n(3n−1)/2            1, 5, 12, 22, 35, &hellip;
Hexagonal               Hn=n(2n−1)              1, 6, 15, 28, 45, &hellip;
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>
]]></content>
  </entry>
  
  <entry>
    <title type="html"><![CDATA[project euler problem 42]]></title>
    <link href="http://kingfighter.github.io//blog/2014/03/14/project-euler-42/"/>
    <updated>2014-03-14T21:11:00+08:00</updated>
    <id>http://kingfighter.github.io//blog/2014/03/14/project-euler-42</id>
    <content type="html"><![CDATA[</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 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, &hellip;
</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 &#8216;Save Link/Target As&hellip;&#8217;), a 16K text file containing nearly two-thousand common English words, how many are triangle words?
</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>太简单
</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>162
</p>
<p>
Source:<a href="https://github.com/kingFighter/project-euler">project-euler</a>
</p></div>
</div>
]]></content>
  </entry>
  
  <entry>
    <title type="html"><![CDATA[project euler problem 31]]></title>
    <link href="http://kingfighter.github.io//blog/2014/03/03/project-euler-31/"/>
    <updated>2014-03-03T20:45:00+08:00</updated>
    <id>http://kingfighter.github.io//blog/2014/03/03/project-euler-31</id>
    <content type="html"><![CDATA[</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>
</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>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>


</div>

</div>

<div id="outline-container-2" class="outline-2">
<h2 id="sec-2">Solution</h2>
<div class="outline-text-2" id="text-2">


</div>

<div id="outline-container-2-1" class="outline-3">
<h3 id="sec-2-1">Brute Force</h3>
<div class="outline-text-3" id="text-2-1">

<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&hellip;., 不过知道a = 1时, b:0,
可缩小a,b,c&hellip;.大小范围, 见 C++Brute1, 类似的见C++Brute2
</p></div>

</div>

<div id="outline-container-2-2" class="outline-3">
<h3 id="sec-2-2">Recursion</h3>
<div class="outline-text-3" id="text-2-2">

<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>   &hellip;..
</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
</p>
</div>

</div>

<div id="outline-container-2-3" class="outline-3">
<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>
</p></div>
</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>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>
</div>
]]></content>
  </entry>
  
  <entry>
    <title type="html"><![CDATA[project euler problem 26]]></title>
    <link href="http://kingfighter.github.io//blog/2014/03/02/project-euler-26/"/>
    <updated>2014-03-02T14:13:00+08:00</updated>
    <id>http://kingfighter.github.io//blog/2014/03/02/project-euler-26</id>
    <content type="html"><![CDATA[</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>
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&hellip;, 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 &lt; 1000 for which 1/d contains the longest recurring cycle in its decimal fraction part.
</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>  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>

<div id="outline-container-3" class="outline-2">
<h2 id="sec-3">Answer</h2>
<div class="outline-text-2" id="text-3">

<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>


</div>
</div>

</div>
</div>
]]></content>
  </entry>
  
  <entry>
    <title type="html"><![CDATA[project euler problem 40]]></title>
    <link href="http://kingfighter.github.io//blog/2013/08/28/project-euler-40/"/>
    <updated>2013-08-28T21:51:00+08:00</updated>
    <id>http://kingfighter.github.io//blog/2013/08/28/project-euler-40</id>
    <content type="html"><![CDATA[</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&hellip;
</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>


</div>

</div>

<div id="outline-container-2" class="outline-2">
<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>


<p>
依此法可以找到任意ith的digit
</p></div>

</div>

<div id="outline-container-3" class="outline-2">
<h2 id="sec-3"><span class="section-number-2">3</span> Answer</h2>
<div class="outline-text-2" id="text-3">

<p>  210
</p>
<p>
Source:<a href="http://sdrv.ms/1cjMvR3">C++</a>
</p>





</div>
</div>
]]></content>
  </entry>
  
  <entry>
    <title type="html"><![CDATA[project euler problem 37]]></title>
    <link href="http://kingfighter.github.io//blog/2013/08/26/project-euler-37/"/>
    <updated>2013-08-26T23:38:00+08:00</updated>
    <id>http://kingfighter.github.io//blog/2013/08/26/project-euler-37</id>
    <content type="html"><![CDATA[</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>The number 3797 has an interesting property. Being prime itself, it is possible to continuously remove digits from left to right, and remain prime at each stage: 3797, 797, 97, and 7. Similarly we can work from right to left: 3797, 379, 37, and 3.
</p>
<p>
Find the sum of the only eleven primes that are both truncatable from left to right and right to left.
</p>
<p>
NOTE: 2, 3, 5, and 7 are not considered to be truncatable primes.
</p>
</blockquote>


</div>

</div>

<div id="outline-container-2" class="outline-2">
<h2 id="sec-2"><span class="section-number-2">2</span> Solution</h2>
<div class="outline-text-2" id="text-2">

<p>brute force
</p></div>

</div>

<div id="outline-container-3" class="outline-2">
<h2 id="sec-3"><span class="section-number-2">3</span> Answer</h2>
<div class="outline-text-2" id="text-3">

<p>  748317
</p>
<p>
Source:<a href="http://sdrv.ms/17eVSQE">C++</a>
</p>




</div>
</div>
]]></content>
  </entry>
  
  <entry>
    <title type="html"><![CDATA[丰乳肥臀]]></title>
    <link href="http://kingfighter.github.io//blog/2013/06/15/fengrufeitun/"/>
    <updated>2013-06-15T16:35:00+08:00</updated>
    <id>http://kingfighter.github.io//blog/2013/06/15/fengrufeitun</id>
    <content type="html"><![CDATA[以前从来没有听说过莫言,直到他得奖了,然后我在图书馆找书时,就顺手借了本<a href="http://zh.wikipedia.org/wiki/丰乳肥臀">丰乳肥臀</a>。
</p>
<p>
我承认我很少看&#8221;正经&#8221;的小说,觉得无趣,抑或是因为无法沉下心来.
</p>
<p>
这本书讲述了从抗日到改革开放的上官家庭的命运。
</p>
<p>
上官鲁氏当真是十分的了不得,丈夫无能,饱受婆家欺凌,受人凌辱,她没有死去,她又十分的睿智,早已看出上官家女人的悲惨的命运.
</p>
<blockquote>

<p>我变了，也没变。这十几年里，上官家的人，像韭菜一样，一茬茬的死，一茬茬的发，有生就有死，死容易，活难，越难越要活。越不怕死越要挣扎着活。我要看到我的后代儿孙浮上水来那一天，你们都要给我争气！
</p>
</blockquote>


<p>
可惜,她的宝贝,混血儿上官金童却纯粹是个毫无出息的人,虽然一表人才,却是个一辈子吊在女人乳头上的窝囊废.窝囊归窝囊,却也是好人一个.
</p>
<p>
最讨厌的是六亲不认的上官盼弟与其女鲁胜利,都只为自己的前途,最终也没有个好下场.
</p>
<p>
八姐上官玉女最是让人怜惜
</p><blockquote>

<p>上官玉女二十多岁时，心理状态还像个小姑娘，胆怯的小姑娘，畏缩的小姑娘。她终生
都像蛹一样缩在茧里，生怕给家里人增添麻烦。
</p>
</blockquote>


]]></content>
  </entry>
  
  <entry>
    <title type="html"><![CDATA[把妹达人]]></title>
    <link href="http://kingfighter.github.io//blog/2013/06/15/The-Game/"/>
    <updated>2013-06-15T14:09:00+08:00</updated>
    <id>http://kingfighter.github.io//blog/2013/06/15/The-Game</id>
    <content type="html"><![CDATA[<a href="http://book.douban.com/subject/3374558/">这本书</a> 看上去像是泡妞攻略手册,其实不如说是挫男的脱变史.
</p>
<p>
一群挫男用技巧在夜店把妹,自己变成了机器,最终逃离了这个圈子.
</p>
<p>
但其中的一些观点却也不是没有道理.
</p>
<blockquote>

<p>除了自信和微笑，我们学到，雄性领袖的其他特质是注重仪表、保持幽默感、与他人搏感
情、扮演一个空间里的社交中心
</p>
</blockquote>


<p>
自信和微笑,这两样总是富有感染力的
</p>
<blockquote>

<p>三秒法则.
</p>
</blockquote>


<p>
很多时候瞻前顾后与周密慎重只是一个度的问题.
</p>

<p>
方法是死的,用好用坏就是由人了.我们大可以用其中的技巧来与周围的人进行交际,来追求自己的另一半.但也切记不要忘记自己原来的初衷.
</p>]]></content>
  </entry>
  
  <entry>
    <title type="html"><![CDATA[project euler problem 27]]></title>
    <link href="http://kingfighter.github.io//blog/2013/06/14/project-euler-27/"/>
    <updated>2013-06-14T23:31:00+08:00</updated>
    <id>http://kingfighter.github.io//blog/2013/06/14/project-euler-27</id>
    <content type="html"><![CDATA[</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>Euler discovered the remarkable quadratic formula:
</p>
<p>
n² + n + 41
</p>
<p>
It turns out that the formula will produce 40 primes for the consecutive values n = 0 to 39. However, when n = 40, 402 + 40 + 41 = 40(40 + 1) + 41 is divisible by 41, and certainly when n = 41, 41² + 41 + 41 is clearly divisible by 41.
</p>
<p>
The incredible formula  n² - 79n + 1601 was discovered, which produces 80 primes for the consecutive values n = 0 to 79. The product of the coefficients, -79 and 1601, is -126479.
</p>
<p>
Considering quadratics of the form:
</p>
<p>
n² + an + b, where |a| &lt; 1000 and |b| &lt; 1000
</p>
<p>
where |n| is the modulus/absolute value of n <br/>
e.g. |11| = 11 and |-4| = 4 
</p>
<p>
Find the product of the coefficients, a and b, for the quadratic expression that produces the maximum number of primes for consecutive values of n, starting with n = 0.
</p>
</blockquote>


</div>

</div>

<div id="outline-container-2" class="outline-2">
<h2 id="sec-2"><span class="section-number-2">2</span> Solution</h2>
<div class="outline-text-2" id="text-2">

<p> Brute Force,当然如果n为0时,可知b必为素数.还可以试图代入n = 1等,得出a,b相应性质,减小复杂度
</p></div>

</div>

<div id="outline-container-3" class="outline-2">
<h2 id="sec-3"><span class="section-number-2">3</span> Answer</h2>
<div class="outline-text-2" id="text-3">

<p>  -59231
</p>
<p>
Source:<a href="http://sdrv.ms/12MpHJ9">C++</a>
</p></div>
</div>
]]></content>
  </entry>
  
  <entry>
    <title type="html"><![CDATA[project euler problem 35]]></title>
    <link href="http://kingfighter.github.io//blog/2013/06/14/project-euler-35/"/>
    <updated>2013-06-14T22:44:00+08:00</updated>
    <id>http://kingfighter.github.io//blog/2013/06/14/project-euler-35</id>
    <content type="html"><![CDATA[</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>  The number, 197, is called a circular prime because all rotations of the digits: 197, 971, and 719, are themselves prime.
</p>
<p>
There are thirteen such primes below 100: 2, 3, 5, 7, 11, 13, 17, 31, 37, 71, 73, 79, and 97.
</p>
<p>
How many circular primes are there below one million?
</p>
</blockquote>


</div>

</div>

<div id="outline-container-2" class="outline-2">
<h2 id="sec-2"><span class="section-number-2">2</span> Solution</h2>
<div class="outline-text-2" id="text-2">

<p>  brute force,没什么好说的
</p></div>

</div>

<div id="outline-container-3" class="outline-2">
<h2 id="sec-3"><span class="section-number-2">3</span> Answer</h2>
<div class="outline-text-2" id="text-3">

<p>  55
</p>
<p>
Source:<a href="http://sdrv.ms/12MpN3F">C++</a>
</p></div>
</div>
]]></content>
  </entry>
  
  <entry>
    <title type="html"><![CDATA[project euler problem 36]]></title>
    <link href="http://kingfighter.github.io//blog/2013/06/10/project-euler-36/"/>
    <updated>2013-06-10T20:24:00+08:00</updated>
    <id>http://kingfighter.github.io//blog/2013/06/10/project-euler-36</id>
    <content type="html"><![CDATA[</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 Generate</a></li>
</ul>
</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>  The decimal number, 585 = \(1001001001_2\) (binary), is palindromic in both bases.
</p>
<p>
Find the sum of all numbers, less than one million, which are palindromic in base 10 and base 2.
</p>
<p>
(Please note that the palindromic number, in either base, may not include leading zeros.)
</p>
</blockquote>


</div>

</div>

<div id="outline-container-2" class="outline-2">
<h2 id="sec-2"><span class="section-number-2">2</span> Solution</h2>
<div class="outline-text-2" id="text-2">


</div>

<div id="outline-container-2-1" class="outline-3">
<h3 id="sec-2-1"><span class="section-number-3">2.1</span> Brute Force</h3>
<div class="outline-text-3" id="text-2-1">

<p>   因为任何进制数字开头不能为0,则二进制表示时最低位必为1,故数字必为奇数.<br/>
   接下来只要对所有范围内的奇数进行回文数检查即可.<br/>
   检查方法:
</p><ol>
<li>将数字每位分解,相应位对比是否相同(121,第一位和第三位比较)
</li>
<li>将数字倒过来写,比较与原数是否相等(123,反过来即321不等)

<p>
      主要利用整除与模,类似于进制间的转换
</p></li>
</ol>

</div>

</div>

<div id="outline-container-2-2" class="outline-3">
<h3 id="sec-2-2"><span class="section-number-3">2.2</span> Generate</h3>
<div class="outline-text-3" id="text-2-2">

<p>   直接在范围内产生要求的二进制回文数,再检查是否为十进制的回文数.检查方法同 Brute Force<br/>
   每个数可生成两个回文数: 01 -&gt; 0110 or 010 <br/>
   具体操作主要用整除与模,对于二进制来说则有更快捷的shift等位操作.
</p></div>
</div>

</div>

<div id="outline-container-3" class="outline-2">
<h2 id="sec-3"><span class="section-number-2">3</span> Answer</h2>
<div class="outline-text-2" id="text-3">

<p>  872187
</p>
<p>  
Source:<a href="http://sdrv.ms/17DumxU">C++</a>
</p>



</div>
</div>
]]></content>
  </entry>
  
  <entry>
    <title type="html"><![CDATA[project euler problem 34]]></title>
    <link href="http://kingfighter.github.io//blog/2013/06/09/project-euler-34/"/>
    <updated>2013-06-09T16:52:00+08:00</updated>
    <id>http://kingfighter.github.io//blog/2013/06/09/project-euler-34</id>
    <content type="html"><![CDATA[</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>145 is a curious number, as 1! + 4! + 5! = 1 + 24 + 120 = 145.
</p>
<p>
Find the sum of all numbers which are equal to the sum of the factorial of their digits.
</p>
<p>
Note: as 1! = 1 and 2! = 2 are not sums they are not included.
</p>
</blockquote>


</div>

</div>

<div id="outline-container-2" class="outline-2">
<h2 id="sec-2"><span class="section-number-2">2</span> Solution</h2>
<div class="outline-text-2" id="text-2">

<p>  问题的关键只要找出上界就行.跟<a href="http://kingfighter.github.io/blog/2013/05/24/project-euler-30/">Project Euler Problem 30</a> 类似,设上界为n位数字,则
</p>
<p>  
  \(10^n\) &lt; 9! * n 
</p>
<p>
  易知n最大为7,更确切的说其上界为 9! * n = 2540160; <br/>
  剩下的就是遍历查找了
</p></div>

</div>

<div id="outline-container-3" class="outline-2">
<h2 id="sec-3"><span class="section-number-2">3</span> Answer</h2>
<div class="outline-text-2" id="text-3">

<p>  40730
</p>
<p>
Source:<a href="http://sdrv.ms/17Dui18">C++</a>
</p></div>
</div>
]]></content>
  </entry>
  
  <entry>
    <title type="html"><![CDATA[project euler problem 23]]></title>
    <link href="http://kingfighter.github.io//blog/2013/06/09/project-euler-23/"/>
    <updated>2013-06-09T16:21:00+08:00</updated>
    <id>http://kingfighter.github.io//blog/2013/06/09/project-euler-23</id>
    <content type="html"><![CDATA[</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>  A perfect number is a number for which the sum of its proper divisors is exactly equal to the number. For example, the sum of the proper divisors of 28 would be 1 + 2 + 4 + 7 + 14 = 28, which means that 28 is a perfect number.
</p>
<p>
A number n is called deficient if the sum of its proper divisors is less than n and it is called abundant if this sum exceeds n.
</p>
<p>
As 12 is the smallest abundant number, 1 + 2 + 3 + 4 + 6 = 16, the smallest number that can be written as the sum of two abundant numbers is 24. By mathematical analysis, it can be shown that all integers greater than 28123 can be written as the sum of two abundant numbers. However, this upper limit cannot be reduced any further by analysis even though it is known that the greatest number that cannot be expressed as the sum of two abundant numbers is less than this limit.
</p>
<p>
Find the sum of all the positive integers which cannot be written as the sum of two abundant numbers.
</p>
</blockquote>


</div>

</div>

<div id="outline-container-2" class="outline-2">
<h2 id="sec-2"><span class="section-number-2">2</span> Solution</h2>
<div class="outline-text-2" id="text-2">

<p>  上界很明显为28123
</p>
<ol>
<li>计算到上界为止所有数的和,记为sum,并将abundant数保存
</li>
<li>将abundant数两两相加,如不大于上界,则将这两数的和记为可以表示为abundant数相加,即不满足题意要求的数
</li>
<li>所有可表示为abundant数相加的数其和记为abundantsSum
</li>
<li>所求答案即为 sum - abundantsSum
</li>
</ol>


</div>

</div>

<div id="outline-container-3" class="outline-2">
<h2 id="sec-3"><span class="section-number-2">3</span> Answer</h2>
<div class="outline-text-2" id="text-3">

<p>  4179871
</p>
<p>
Source:<a href="http://sdrv.ms/18lliPp">C++</a>
</p></div>
</div>
]]></content>
  </entry>
  
  <entry>
    <title type="html"><![CDATA[project euler problem 29]]></title>
    <link href="http://kingfighter.github.io//blog/2013/06/08/project-euler-29/"/>
    <updated>2013-06-08T22:12:00+08:00</updated>
    <id>http://kingfighter.github.io//blog/2013/06/08/project-euler-29</id>
    <content type="html"><![CDATA[</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>Consider all integer combinations of \(a^b\) for 2 \(\leq\) a \(\leq\) 5 and 2 \(\leq\) b \(\leq\) 5:
</p>
<p>
\(2^2\) =4, \(2^3\) =8, \(2^4\) =16, \(2^5\) =32
\(3^2\) =9, \(3^3\) =27, \(3^4\) =81, \(3^5\) =243
\(4^2\) =16, \(4^3\) =64, \(4^4\) =256, \(4^5\) =1024
\(5^2\) =25, \(5^3\) =125, \(5^4\) =625, \(5^5\) =3125
</p>
<p>
If they are then placed in numerical order, with any repeats removed, we get the following sequence of 15 distinct terms:
</p>
<p>
4, 8, 9, 16, 25, 27, 32, 64, 81, 125, 243, 256, 625, 1024, 3125
</p>
<p>
How many distinct terms are in the sequence generated by \(a^b\)  for 2 \(\leq\) a \(\leq\) 100 and 2 \(\leq\) b \(\leq\) 100?
</p>
</blockquote>


</div>

</div>

<div id="outline-container-2" class="outline-2">
<h2 id="sec-2"><span class="section-number-2">2</span> Solution</h2>
<div class="outline-text-2" id="text-2">

<p>  Set集合
</p></div>

</div>

<div id="outline-container-3" class="outline-2">
<h2 id="sec-3"><span class="section-number-2">3</span> Answer</h2>
<div class="outline-text-2" id="text-3">

<p>  9183
</p>
<p>  
Source:<a href="http://sdrv.ms/14L7x4S">C++</a>
</p>



</div>
</div>
]]></content>
  </entry>
  
  <entry>
    <title type="html"><![CDATA[project euler problem 67]]></title>
    <link href="http://kingfighter.github.io//blog/2013/05/31/project-euler-67/"/>
    <updated>2013-05-31T20:47:00+08:00</updated>
    <id>http://kingfighter.github.io//blog/2013/05/31/project-euler-67</id>
    <content type="html"><![CDATA[</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>  By starting at the top of the triangle below and moving to adjacent numbers on the row below, the maximum total from top to bottom is 23.
</p>


  <p style="text-align:center;font-family:courier new;font-size:12pt;"><span style="color:#ff0000;"><b>3</b></span><br />
<span style="color:#ff0000;"><b>7</b></span> 4<br />
2 <span style="color:#ff0000;"><b>4</b></span> 6<br />
8 5 <span style="color:#ff0000;"><b>9</b></span> 3</p>


<p>
That is, 3 + 7 + 4 + 9 = 23.
</p>
<p>
Find the maximum total from top to bottom in <a href="http://projecteuler.net/project/triangle.txt">triangle.txt</a> (right click and &#8216;Save Link/Target As&hellip;&#8217;), a 15K text file containing a triangle with one-hundred rows.
</p>
<p>
NOTE: This is a much more difficult version of <a href="http://kingfighter.github.io/blog/2013/05/06/project-euler-18/">Problem 18</a>. It is not possible to try every route to solve this problem, as there are 2<sup>99</sup> altogether! If you could check one trillion (10<sup>12</sup>) routes every second it would take over twenty billion years to check them all. There is an efficient algorithm to solve it. ;o)
</p>

</blockquote>


</div>

</div>

<div id="outline-container-2" class="outline-2">
<h2 id="sec-2"><span class="section-number-2">2</span> Solution</h2>
<div class="outline-text-2" id="text-2">

<p>  请查看<a href="http://kingfighter.github.io/blog/2013/05/06/project-euler-18/">Problem 18</a>
</p></div>

</div>

<div id="outline-container-3" class="outline-2">
<h2 id="sec-3"><span class="section-number-2">3</span> Answer</h2>
<div class="outline-text-2" id="text-3">

<p>  7273
</p>
<p>
Source:<a href="http://sdrv.ms/18Fx2v3">C++</a>
</p>



</div>
</div>
]]></content>
  </entry>
  
  <entry>
    <title type="html"><![CDATA[project euler problem 30]]></title>
    <link href="http://kingfighter.github.io//blog/2013/05/24/project-euler-30/"/>
    <updated>2013-05-24T22:20:00+08:00</updated>
    <id>http://kingfighter.github.io//blog/2013/05/24/project-euler-30</id>
    <content type="html"><![CDATA[</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>Surprisingly there are only three numbers that can be written as the sum of fourth powers of their digits:
\begin{eqnarray*}
1634 = 1^4 +6^4 + 3^4 + 4^4 &#92;
\end{eqnarray*}
\begin{eqnarray*}
8208 = 8^4 + 2^4 + 0^4 + 8^4 &#92;
\end{eqnarray*}
\begin{eqnarray*}
9474 = 9^4 + 4^4 + 7^4 + 4^4 &#92;
\end{eqnarray*}
As 1 = \(1^4\) is not a sum it is not included.
</p>
<p>
The sum of these numbers is 1634 + 8208 + 9474 = 19316.
</p>
<p>
Find the sum of all the numbers that can be written as the sum of fifth powers of their digits.
</p>
</blockquote>


</div>

</div>

<div id="outline-container-2" class="outline-2">
<h2 id="sec-2"><span class="section-number-2">2</span> Solution</h2>
<div class="outline-text-2" id="text-2">

<p>  很直接,关键是找出上界.个位数是不符合题意的。我们检查多少位时其位数上的值的五次方的和不能表示相应的位数.我们可以看到,到7位数时,其最大就只能表示413343,更不用说8位,9位等了.故上界为354294
</p>
<p>  
  \begin{eqnarray*}
  9^5 + 9^5 = 118098 &#92;
  \end{eqnarray*}
  \begin{eqnarray*}
  9^5 + 9^5 + 9^5 = 177147 &#92;
  \end{eqnarray*}
  \begin{eqnarray*}
  9^5 + 9^5 + 9^5 + 9^5 = 236196 &#92;
  \end{eqnarray*}
  \begin{eqnarray*}
  9^5 + 9^5 + 9^5 + 9^5 + 9^5 = 295245 &#92;
  \end{eqnarray*}
  \begin{eqnarray*}
  9^5 + 9^5 + 9^5 + 9^5 + 9^5 + 9^5 = 354294 &#92;
  \end{eqnarray*}
  \begin{eqnarray*}
  9^5 + 9^5 + 9^5 + 9^5 + 9^5 + 9^5 + 9^5 = 413343 &#92;
  \end{eqnarray*}

</p></div>

</div>

<div id="outline-container-3" class="outline-2">
<h2 id="sec-3"><span class="section-number-2">3</span> Answer</h2>
<div class="outline-text-2" id="text-3">

<p>  443839
</p>
<p>
Source:<a href="http://sdrv.ms/10SnRyK">C++</a>
</p>



</div>
</div>
]]></content>
  </entry>
  
  <entry>
    <title type="html"><![CDATA[project euler problem 24]]></title>
    <link href="http://kingfighter.github.io//blog/2013/05/23/project-euler-24/"/>
    <updated>2013-05-23T23:10:00+08:00</updated>
    <id>http://kingfighter.github.io//blog/2013/05/23/project-euler-24</id>
    <content type="html"><![CDATA[</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>A permutation is an ordered arrangement of objects. For example, 3124 is one possible permutation of the digits 1, 2, 3 and 4. If all of the permutations are listed numerically or alphabetically, we call it lexicographic order. The lexicographic permutations of 0, 1 and 2 are:
</p>
<p>
012   021   102   120   201   210
</p>
<p>
What is the millionth lexicographic permutation of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9?
</p>
</blockquote>


</div>

</div>

<div id="outline-container-2" class="outline-2">
<h2 id="sec-2"><span class="section-number-2">2</span> Solution</h2>
<div class="outline-text-2" id="text-2">

<p>  看到排列组合的问题,立马就想到了用排列组合的数学方法来求解。<br/>
  只需要高中数学的知识即可,计算过程如下:<br/>
  9!*2 &lt;＝ 1000000 &lt; 9!*3 <br/>
  最高位为从小到大排列未被使用的第三位数字2<br/>
  1000000 – 9!*2 = 274240
</p>
<p>
  8!*6 &lt;＝ 27420 &lt; 8!*7<br/>
  现在数字2已经被使用,故从小到大排列未被使用的第七位数字为7<br/>
  27420 – 241920 = 32320
</p>
<p>
  &hellip;..
</p>
<p>
  按这个过程计算出来的为2783915604,但这并不是真正所求的,而是所求序列的下一个。很容易知道2783915604序列的前一个数为2783915460
</p></div>

</div>

<div id="outline-container-3" class="outline-2">
<h2 id="sec-3"><span class="section-number-2">3</span> Answer</h2>
<div class="outline-text-2" id="text-3">

<p>  2783915460  
</p>
<p>
Source:<a href="http://sdrv.ms/14Fp1Cl">C++</a>
</p>



</div>
</div>
]]></content>
  </entry>
  
</feed>