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		<title><![CDATA[Gentoo中文社区 / C 练习实例23]]></title>
		<link>https://www.gentoo-zh.org/viewtopic.php?id=332</link>
		<description><![CDATA[C 练习实例23 最近发表的帖子。]]></description>
		<lastBuildDate>Mon, 29 Aug 2022 07:08:32 +0000</lastBuildDate>
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			<title><![CDATA[C 练习实例23]]></title>
			<link>https://www.gentoo-zh.org/viewtopic.php?pid=338#p338</link>
			<description><![CDATA[<p>题目：打印出如下图案（菱形）。</p><p>&#160; &#160;*<br />&#160; ***<br /> *****<br />*******<br /> *****<br />&#160; ***<br />&#160; &#160;*</p><p>程序分析：先把图形分成两部分来看待，前四行一个规律，后三行一个规律，利用双重for循环，第一层控制行，第二层控制列。</p><p>程序源代码：</p><p>#include &lt;stdio.h&gt;<br />int main()<br />{<br />&#160; &#160; int i,j,k;<br />&#160; &#160; for(i=0;i&lt;=3;i++) {<br />&#160; &#160; &#160; &#160; for(j=0;j&lt;=2-i;j++) {<br />&#160; &#160; &#160; &#160; &#160; &#160; printf(&quot; &quot;);<br />&#160; &#160; &#160; &#160; }<br />&#160; &#160; &#160; &#160; for(k=0;k&lt;=2*i;k++) {<br />&#160; &#160; &#160; &#160; &#160; &#160; printf(&quot;*&quot;);<br />&#160; &#160; &#160; &#160; }<br />&#160; &#160; &#160; &#160; printf(&quot;\n&quot;);<br />&#160; &#160; }<br />&#160; &#160; for(i=0;i&lt;=2;i++) {<br />&#160; &#160; &#160; &#160; for(j=0;j&lt;=i;j++) {<br />&#160; &#160; &#160; &#160; &#160; &#160; printf(&quot; &quot;);<br />&#160; &#160; &#160; &#160; }<br />&#160; &#160; &#160; &#160; for(k=0;k&lt;=4-2*i;k++) {<br />&#160; &#160; &#160; &#160; &#160; &#160; printf(&quot;*&quot;);<br />&#160; &#160; &#160; &#160; }<br />&#160; &#160; &#160; &#160; printf(&quot;\n&quot;);<br />&#160; &#160; }<br />&#160; <br />}</p>]]></description>
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			<pubDate>Mon, 29 Aug 2022 07:08:32 +0000</pubDate>
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