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	<title>ความเห็นบน: บรรทัดเดียวก็พอ #1 factorial</title>
	<atom:link href="http://www.kowito.com/2009/07/08/%e0%b8%9a%e0%b8%a3%e0%b8%a3%e0%b8%97%e0%b8%b1%e0%b8%94%e0%b9%80%e0%b8%94%e0%b8%b5%e0%b8%a2%e0%b8%a7%e0%b8%81%e0%b9%87%e0%b8%9e%e0%b8%ad-1-factorial/feed/" rel="self" type="application/rss+xml" />
	<link>http://www.kowito.com/2009/07/08/%e0%b8%9a%e0%b8%a3%e0%b8%a3%e0%b8%97%e0%b8%b1%e0%b8%94%e0%b9%80%e0%b8%94%e0%b8%b5%e0%b8%a2%e0%b8%a7%e0%b8%81%e0%b9%87%e0%b8%9e%e0%b8%ad-1-factorial/</link>
	<description>Try not to become a man of success but a man of value.</description>
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		<title>โดย: nuboat</title>
		<link>http://www.kowito.com/2009/07/08/%e0%b8%9a%e0%b8%a3%e0%b8%a3%e0%b8%97%e0%b8%b1%e0%b8%94%e0%b9%80%e0%b8%94%e0%b8%b5%e0%b8%a2%e0%b8%a7%e0%b8%81%e0%b9%87%e0%b8%9e%e0%b8%ad-1-factorial/#comment-6</link>
		<dc:creator>nuboat</dc:creator>
		<pubDate>Wed, 08 Jul 2009 12:12:59 +0000</pubDate>
		<guid isPermaLink="false">http://www.kowito.com/2009/07/08/%e0%b8%9a%e0%b8%a3%e0%b8%a3%e0%b8%97%e0%b8%b1%e0%b8%94%e0%b9%80%e0%b8%94%e0%b8%b5%e0%b8%a2%e0%b8%a7%e0%b8%81%e0%b9%87%e0%b8%9e%e0%b8%ad-1-factorial/#comment-6</guid>
		<description>def q220ss(n):
    return &#039;n&#039;.join([&#039;-&#039;*((n+1)/2 - (i &lt;= (n+1)/2 and i or 1 + n - i)) + &#039;*&#039;*((i  1) + &#039;-&#039;*((i  2 and (i &lt;= (n+1)/2 and i or 1 + n - i) - 2 or 0) + &#039;*&#039;*((i &lt;= (n+1)/2 and i or 1 + n - i) == 1) + &#039;-&#039;*((i  1) + &#039;-&#039;*((i  2 and (i &lt;= (n+1)/2 and i or 1 + n - i) - 2 or 0) + &#039;*&#039;*((i  1) + &#039;-&#039;*((n+1)/2 - (i &lt;= (n+1)/2 and i or 1 + n - i)) for i in range(1, n+1) ])

เล่นบ้างบรรทัดเดียวก็เกินพอ อยากรู้ว่าทำอะไรเอาไปรันเองนะครับ ^ ^</description>
		<content:encoded><![CDATA[<p>def q220ss(n):<br />
    return &#8216;n&#8217;.join(['-'*((n+1)/2 - (i &lt;= (n+1)/2 and i or 1 + n - i)) + &#039;*&#039;*((i  1) + '-'*((i  2 and (i &lt;= (n+1)/2 and i or 1 + n - i) - 2 or 0) + &#039;*&#039;*((i &lt;= (n+1)/2 and i or 1 + n - i) == 1) + &#039;-&#039;*((i  1) + '-'*((i  2 and (i &lt;= (n+1)/2 and i or 1 + n - i) - 2 or 0) + &#039;*&#039;*((i  1) + '-'*((n+1)/2 - (i &lt;= (n+1)/2 and i or 1 + n - i)) for i in range(1, n+1) ])</p>
<p>เล่นบ้างบรรทัดเดียวก็เกินพอ อยากรู้ว่าทำอะไรเอาไปรันเองนะครับ ^ ^</p>
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	<item>
		<title>โดย: bact'</title>
		<link>http://www.kowito.com/2009/07/08/%e0%b8%9a%e0%b8%a3%e0%b8%a3%e0%b8%97%e0%b8%b1%e0%b8%94%e0%b9%80%e0%b8%94%e0%b8%b5%e0%b8%a2%e0%b8%a7%e0%b8%81%e0%b9%87%e0%b8%9e%e0%b8%ad-1-factorial/#comment-7</link>
		<dc:creator>bact'</dc:creator>
		<pubDate>Wed, 08 Jul 2009 12:01:04 +0000</pubDate>
		<guid isPermaLink="false">http://www.kowito.com/2009/07/08/%e0%b8%9a%e0%b8%a3%e0%b8%a3%e0%b8%97%e0%b8%b1%e0%b8%94%e0%b9%80%e0%b8%94%e0%b8%b5%e0%b8%a2%e0%b8%a7%e0%b8%81%e0%b9%87%e0%b8%9e%e0%b8%ad-1-factorial/#comment-7</guid>
		<description>แหงะ บรรทัดเดียวจริง ๆ แต่หลาย statement นะ :p</description>
		<content:encoded><![CDATA[<p>แหงะ บรรทัดเดียวจริง ๆ แต่หลาย statement นะ :p</p>
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