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	<updated>2026-05-14T17:48:10Z</updated>
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		<title>imported&gt;Unknown at 05:28, 7 February 2021</title>
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		<updated>2021-02-07T05:28:07Z</updated>

		<summary type="html">&lt;p&gt;&lt;/p&gt;
&lt;p&gt;&lt;b&gt;New page&lt;/b&gt;&lt;/p&gt;&lt;div&gt;== 소감 ==&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot;&lt;br /&gt;
|-&lt;br /&gt;
| 2006-01-08  Accepted 0.012 Minimum&lt;br /&gt;
|}&lt;br /&gt;
A, B 라는 입력이 들어왔을 때, n = B - A 로 본다면, n = 1 ~ 20 {1,2,3,3,4,4,5,5,5,6,6,6,7,7,7,7,8,8,8,8}. 간단한 일차 방정식 작성으로 풀 수 있다. &lt;br /&gt;
== 코드 ==&lt;br /&gt;
 // 846 - Steps&lt;br /&gt;
 #include &amp;amp;lt;iostream&amp;amp;gt;&lt;br /&gt;
 using namespace std;&lt;br /&gt;
 #include &amp;amp;lt;cmath&amp;amp;gt;&lt;br /&gt;
 &lt;br /&gt;
 inline void process(int n)&lt;br /&gt;
 {&lt;br /&gt;
 	int i = floor(sqrt(n));&lt;br /&gt;
 	for (; i * (i + 1) &amp;amp;lt;= n; i++) ;&lt;br /&gt;
 	if (n == i * (i - 1)) &lt;br /&gt;
 		cout &amp;amp;lt;&amp;amp;lt; 2 * (i - 1) &amp;amp;lt;&amp;amp;lt; endl;&lt;br /&gt;
 	else if (n &amp;amp;lt;= i * i) &lt;br /&gt;
 		cout &amp;amp;lt;&amp;amp;lt; 2 * i - 1 &amp;amp;lt;&amp;amp;lt; endl;&lt;br /&gt;
 	else &lt;br /&gt;
 		cout &amp;amp;lt;&amp;amp;lt; 2 * i &amp;amp;lt;&amp;amp;lt; endl;	&lt;br /&gt;
 }&lt;br /&gt;
 &lt;br /&gt;
 int main()&lt;br /&gt;
 {&lt;br /&gt;
 	int nCase, x, y;&lt;br /&gt;
 	cin &amp;amp;gt;&amp;amp;gt; nCase;&lt;br /&gt;
 	for (int i = 0; i &amp;amp;lt; nCase; i++)&lt;br /&gt;
 	{&lt;br /&gt;
 		cin &amp;amp;gt;&amp;amp;gt; x &amp;amp;gt;&amp;amp;gt; y;&lt;br /&gt;
 		process(y - x);&lt;br /&gt;
 	}&lt;br /&gt;
 	return 0;&lt;br /&gt;
 }&lt;br /&gt;
----&lt;br /&gt;
[[Steps]]&lt;/div&gt;</summary>
		<author><name>imported&gt;Unknown</name></author>
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