{"componentChunkName":"component---src-templates-blog-post-js","path":"/Problem-Solving/2020-01-19-9461번-파도반-수열/","result":{"data":{"site":{"siteMetadata":{"title":"Hun's Footsteps 🥷","author":"전여훈","siteUrl":"https://jeonyeohun.netlify.app","comment":{"disqusShortName":"","utterances":"jeonyeohun/jeonyeohun.github.io"},"sponsor":{"buyMeACoffeeId":"jeonyeohun"}}},"markdownRemark":{"id":"529600d5-52a8-56ca-b7bd-9d38594d4511","excerpt":"9461번: 파도반 수열 접근 방법: 파도반 수열을 나열해보면 P(N)은 P(N-2) 와 P(N-3)의 합으로 만들어지는 것을 알 수 있었다. 이것을 dp로 구현하면 끝. 통과 코드:","html":"<h3 id=\"9461번-파도반-수열\" style=\"position:relative;\"><a href=\"#9461%EB%B2%88-%ED%8C%8C%EB%8F%84%EB%B0%98-%EC%88%98%EC%97%B4\" aria-label=\"9461번 파도반 수열 permalink\" class=\"anchor before\"><svg aria-hidden=\"true\" focusable=\"false\" height=\"16\" version=\"1.1\" viewBox=\"0 0 16 16\" width=\"16\"><path fill-rule=\"evenodd\" d=\"M4 9h1v1H4c-1.5 0-3-1.69-3-3.5S2.55 3 4 3h4c1.45 0 3 1.69 3 3.5 0 1.41-.91 2.72-2 3.25V8.59c.58-.45 1-1.27 1-2.09C10 5.22 8.98 4 8 4H4c-.98 0-2 1.22-2 2.5S3 9 4 9zm9-3h-1v1h1c1 0 2 1.22 2 2.5S13.98 12 13 12H9c-.98 0-2-1.22-2-2.5 0-.83.42-1.64 1-2.09V6.25c-1.09.53-2 1.84-2 3.25C6 11.31 7.55 13 9 13h4c1.45 0 3-1.69 3-3.5S14.5 6 13 6z\"></path></svg></a><a href=\"https://www.acmicpc.net/problem/9461\">9461번: 파도반 수열</a></h3>\n<div class=\"gatsby-highlight\" data-language=\"text\"><pre class=\"language-text\"><code class=\"language-text\">문제\n오른쪽 그림과 같이 삼각형이 나선 모양으로 놓여져 있다. 첫 삼각형은 정삼각형으로 변의 길이는 1이다. 그 다음에는 다음과 같은 과정으로 정삼각형을 계속 추가한다. 나선에서 가장 긴 변의 길이를 k라 했을 때, 그 변에 길이가 k인 정삼각형을 추가한다.\n\n파도반 수열 P(N)은 나선에 있는 정삼각형의 변의 길이이다. P(1)부터 P(10)까지 첫 10개 숫자는 1, 1, 1, 2, 2, 3, 4, 5, 7, 9이다.\n\nN이 주어졌을 때, P(N)을 구하는 프로그램을 작성하시오.\n\n입력\n첫째 줄에 테스트 케이스의 개수 T가 주어진다. 각 테스트 케이스는 한 줄로 이루어져 있고, N이 주어진다. (1 ≤ N ≤ 100)\n\n출력\n각 테스트 케이스마다 P(N)을 출력한다.</code></pre></div>\n<h3 id=\"접근-방법\" style=\"position:relative;\"><a href=\"#%EC%A0%91%EA%B7%BC-%EB%B0%A9%EB%B2%95\" aria-label=\"접근 방법 permalink\" class=\"anchor before\"><svg aria-hidden=\"true\" focusable=\"false\" height=\"16\" version=\"1.1\" viewBox=\"0 0 16 16\" width=\"16\"><path fill-rule=\"evenodd\" d=\"M4 9h1v1H4c-1.5 0-3-1.69-3-3.5S2.55 3 4 3h4c1.45 0 3 1.69 3 3.5 0 1.41-.91 2.72-2 3.25V8.59c.58-.45 1-1.27 1-2.09C10 5.22 8.98 4 8 4H4c-.98 0-2 1.22-2 2.5S3 9 4 9zm9-3h-1v1h1c1 0 2 1.22 2 2.5S13.98 12 13 12H9c-.98 0-2-1.22-2-2.5 0-.83.42-1.64 1-2.09V6.25c-1.09.53-2 1.84-2 3.25C6 11.31 7.55 13 9 13h4c1.45 0 3-1.69 3-3.5S14.5 6 13 6z\"></path></svg></a>접근 방법:</h3>\n<p>파도반 수열을 나열해보면 P(N)은 P(N-2) 와 P(N-3)의 합으로 만들어지는 것을 알 수 있었다. 이것을 dp로 구현하면 끝.</p>\n<h3 id=\"통과-코드\" style=\"position:relative;\"><a href=\"#%ED%86%B5%EA%B3%BC-%EC%BD%94%EB%93%9C\" aria-label=\"통과 코드 permalink\" class=\"anchor before\"><svg aria-hidden=\"true\" focusable=\"false\" height=\"16\" version=\"1.1\" viewBox=\"0 0 16 16\" width=\"16\"><path fill-rule=\"evenodd\" d=\"M4 9h1v1H4c-1.5 0-3-1.69-3-3.5S2.55 3 4 3h4c1.45 0 3 1.69 3 3.5 0 1.41-.91 2.72-2 3.25V8.59c.58-.45 1-1.27 1-2.09C10 5.22 8.98 4 8 4H4c-.98 0-2 1.22-2 2.5S3 9 4 9zm9-3h-1v1h1c1 0 2 1.22 2 2.5S13.98 12 13 12H9c-.98 0-2-1.22-2-2.5 0-.83.42-1.64 1-2.09V6.25c-1.09.53-2 1.84-2 3.25C6 11.31 7.55 13 9 13h4c1.45 0 3-1.69 3-3.5S14.5 6 13 6z\"></path></svg></a>통과 코드:</h3>\n<div class=\"gatsby-highlight\" data-language=\"cpp\"><pre class=\"language-cpp\"><code class=\"language-cpp\"><span class=\"token macro property\"><span class=\"token directive-hash\">#</span><span class=\"token directive keyword\">include</span><span class=\"token string\">&lt;iostream></span></span>\n<span class=\"token macro property\"><span class=\"token directive-hash\">#</span><span class=\"token directive keyword\">include</span><span class=\"token string\">&lt;vector></span></span>\n\n<span class=\"token keyword\">using</span> <span class=\"token keyword\">namespace</span> std<span class=\"token punctuation\">;</span>\n\nvector<span class=\"token operator\">&lt;</span><span class=\"token keyword\">long</span> <span class=\"token keyword\">long</span><span class=\"token operator\">></span> <span class=\"token function\">dp</span><span class=\"token punctuation\">(</span><span class=\"token number\">102</span><span class=\"token punctuation\">,</span> <span class=\"token operator\">-</span><span class=\"token number\">1</span><span class=\"token punctuation\">)</span><span class=\"token punctuation\">;</span>\n\n<span class=\"token keyword\">long</span> <span class=\"token keyword\">long</span> <span class=\"token function\">p</span> <span class=\"token punctuation\">(</span><span class=\"token keyword\">int</span> n<span class=\"token punctuation\">)</span><span class=\"token punctuation\">{</span>\n    <span class=\"token keyword\">if</span> <span class=\"token punctuation\">(</span>dp<span class=\"token punctuation\">[</span>n<span class=\"token punctuation\">]</span> <span class=\"token operator\">!=</span> <span class=\"token operator\">-</span><span class=\"token number\">1</span><span class=\"token punctuation\">)</span> <span class=\"token keyword\">return</span> dp<span class=\"token punctuation\">[</span>n<span class=\"token punctuation\">]</span><span class=\"token punctuation\">;</span>\n\n    <span class=\"token keyword\">return</span> dp<span class=\"token punctuation\">[</span>n<span class=\"token punctuation\">]</span> <span class=\"token operator\">=</span> <span class=\"token function\">p</span><span class=\"token punctuation\">(</span>n<span class=\"token operator\">-</span><span class=\"token number\">2</span><span class=\"token punctuation\">)</span> <span class=\"token operator\">+</span> <span class=\"token function\">p</span><span class=\"token punctuation\">(</span>n<span class=\"token operator\">-</span><span class=\"token number\">3</span><span class=\"token punctuation\">)</span><span class=\"token punctuation\">;</span>\n<span class=\"token punctuation\">}</span>\n\n<span class=\"token keyword\">int</span> <span class=\"token function\">main</span> <span class=\"token punctuation\">(</span><span class=\"token punctuation\">)</span><span class=\"token punctuation\">{</span>\n    <span class=\"token keyword\">int</span> N<span class=\"token punctuation\">;</span>\n    cin <span class=\"token operator\">>></span> N<span class=\"token punctuation\">;</span>\n\n    <span class=\"token keyword\">for</span> <span class=\"token punctuation\">(</span><span class=\"token keyword\">int</span> i <span class=\"token operator\">=</span> <span class=\"token number\">1</span> <span class=\"token punctuation\">;</span> i <span class=\"token operator\">&lt;=</span> <span class=\"token number\">3</span> <span class=\"token punctuation\">;</span> i<span class=\"token operator\">++</span><span class=\"token punctuation\">)</span> dp<span class=\"token punctuation\">[</span>i<span class=\"token punctuation\">]</span> <span class=\"token operator\">=</span> <span class=\"token number\">1</span><span class=\"token punctuation\">;</span>\n\n    <span class=\"token keyword\">while</span><span class=\"token punctuation\">(</span>N<span class=\"token operator\">--</span><span class=\"token punctuation\">)</span><span class=\"token punctuation\">{</span>\n        <span class=\"token keyword\">int</span> t<span class=\"token punctuation\">;</span>\n        cin <span class=\"token operator\">>></span> t<span class=\"token punctuation\">;</span>\n        <span class=\"token function\">p</span><span class=\"token punctuation\">(</span>t<span class=\"token punctuation\">)</span><span class=\"token punctuation\">;</span>\n        cout <span class=\"token operator\">&lt;&lt;</span> dp<span class=\"token punctuation\">[</span>t<span class=\"token punctuation\">]</span> <span class=\"token operator\">&lt;&lt;</span> <span class=\"token string\">\"\\n\"</span><span class=\"token punctuation\">;</span>\n    <span class=\"token punctuation\">}</span>\n\n    <span class=\"token keyword\">return</span> <span class=\"token number\">0</span><span class=\"token punctuation\">;</span>\n<span class=\"token punctuation\">}</span></code></pre></div>","frontmatter":{"title":"[백준 알고리즘] 9461번: 파도반 수열","date":"May 05, 2020"}}},"pageContext":{"slug":"/Problem-Solving/2020-01-19-9461번-파도반-수열/","previous":{"fields":{"slug":"/Problem-Solving/2020-01-21-11047번-동전0/"},"frontmatter":{"title":"[백준 알고리즘] 11047번: 동전0","category":"Problem-Solving","draft":false}},"next":{"fields":{"slug":"/Problem-Solving/2020-01-18-1904번-01타일/"},"frontmatter":{"title":"[백준 알고리즘] 1904번: 01타일","category":"Problem-Solving","draft":false}}}},"staticQueryHashes":["2486386679","3128451518"]}