https://111qqz.com/en/tags/%E5%BE%AA%E7%8E%AF%E8%8A%82/Recent content in 循环节 on 111qqz's blogHugo -- gohugo.ioen© 2015-2026 111qqzWed, 01 Nov 2017 06:47:11 +0000https://111qqz.com/en/post/acm-icpc/2017/2017-11-01-hdu-6038/Wed, 01 Nov 2017 06:47:11 +0000https://111qqz.com/en/post/acm-icpc/2017/2017-11-01-hdu-6038/<p><a href="http://acm.hdu.edu.cn/showproblem.php?pid=6038" target="_blank" rel="noreferrer">http://acm.hdu.edu.cn/showproblem.php?pid=6038</a></p> <h1 class="relative group">题意： <div id="题意" class="anchor"></div> <span class="absolute top-0 w-6 transition-opacity opacity-0 -start-6 not-prose group-hover:opacity-100 select-none"> <a class="text-primary-300 dark:text-neutral-700 !no-underline" href="#%e9%a2%98%e6%84%8f" aria-label="Anchor">#</a> </span> </h1> <p>给出两个序列 a 和 b ，求满足  f[i]= b_{f[a[i]]} 的函数个数。</p>https://111qqz.com/en/post/acm-icpc/2017/2017-10-31-hdu-6043-2017-multi-university-training-contest-team-1-k-kazaqs-socks-efbc88e5beaae78eafe88a82efbc89/Tue, 31 Oct 2017 16:31:10 +0000https://111qqz.com/en/post/acm-icpc/2017/2017-10-31-hdu-6043-2017-multi-university-training-contest-team-1-k-kazaqs-socks-efbc88e5beaae78eafe88a82efbc89/<p><a href="http://acm.hdu.edu.cn/showproblem.php?pid=6043" target="_blank" rel="noreferrer">http://acm.hdu.edu.cn/showproblem.php?pid=6043</a></p> <h1 class="relative group">题意： <div id="题意" class="anchor"></div> <span class="absolute top-0 w-6 transition-opacity opacity-0 -start-6 not-prose group-hover:opacity-100 select-none"> <a class="text-primary-300 dark:text-neutral-700 !no-underline" href="#%e9%a2%98%e6%84%8f" aria-label="Anchor">#</a> </span> </h1> <p>n双袜子标号1到n，初始在抽屉里，每天早晨<strong>穿一双标号最小的袜子</strong>，晚上把脏袜子放到盆里，如果放完之后喷子里已经有了n-1双脏袜子，那么就要洗，然后在第二天晚上放回抽屉里。问第k天穿的是标号为几的袜子。</p>https://111qqz.com/en/post/acm-icpc/2017/2017-10-18-look-for-loop-section-in-generalized-fibonacci-sequence/Wed, 18 Oct 2017 05:38:43 +0000https://111qqz.com/en/post/acm-icpc/2017/2017-10-18-look-for-loop-section-in-generalized-fibonacci-sequence/<p>**问题：**给定<figure><img class="my-0 rounded-md" loading="lazy" decoding="async" fetchpriority="low" alt="" src="http://img.blog.csdn.net/20140512141202250" ></figure> ，满足<figure><img class="my-0 rounded-md" loading="lazy" decoding="async" fetchpriority="low" alt="" src="http://img.blog.csdn.net/20140512141344500" ></figure> ，求<figure><img class="my-0 rounded-md" loading="lazy" decoding="async" fetchpriority="low" alt="" src="http://img.blog.csdn.net/20140512141547343" ></figure> 的循</p> <p>环节长度。</p> <p>原理见<a href="http://blog.csdn.net/acdreamers/article/details/25616461" target="_blank" rel="noreferrer">广义Fibonacci数列找循环节</a></p> <p>这里只说做法</p>https://111qqz.com/en/post/acm-icpc/2016/2016-11-03-acdream-oj-1124-e596b5e596b5e79a84e98197e686be-e69690e6b3a2e982a3e5a591e695b0e58897e5beaae78eafe88a82/Thu, 03 Nov 2016 00:39:54 +0000https://111qqz.com/en/post/acm-icpc/2016/2016-11-03-acdream-oj-1124-e596b5e596b5e79a84e98197e686be-e69690e6b3a2e982a3e5a591e695b0e58897e5beaae78eafe88a82/<p><a href="http://acdream.info/problem?pid=1124" target="_blank" rel="noreferrer">题目链接</a></p> <p>题意：</p> <figure><img class="my-0 rounded-md" loading="lazy" decoding="async" fetchpriority="low" alt="" src="http://acdream.info/img/prob/1124/1.jpeg" ></figure> <p>F0 = 1 , F1 = 1 , F2 = 2 , Fn = Fn-1+Fn-2</p> <p>求：</p> <p>FFFn Mod P</p> <p>( 也就是 F[ F[ F[n] ] ]  % P )</p>https://111qqz.com/en/post/acm-icpc/2016/2016-11-02-hdu-3978/Wed, 02 Nov 2016 10:08:16 +0000https://111qqz.com/en/post/acm-icpc/2016/2016-11-02-hdu-3978/<p>题意：now he let you calculate G(n,k) .Here G(n,0) = f(n) , G(n,i) = f( G(n,i-1) ) (k &gt;= i &gt;= 1).其中f是斐波那契数列。</p>https://111qqz.com/en/post/acm-icpc/2016/2016-11-01-hdu-2522/Tue, 01 Nov 2016 12:21:12 +0000https://111qqz.com/en/post/acm-icpc/2016/2016-11-01-hdu-2522/<p><a href="http://acm.hdu.edu.cn/showproblem.php?pid=2522" target="_blank" rel="noreferrer">题目链接</a></p> <p>题意：求一个小数的循环节&hellip;</p> <p>思路：其实直接模拟就好&hellip;</p>https://111qqz.com/en/post/acm-icpc/2016/2016-10-31-hdu-4291/Mon, 31 Oct 2016 08:59:53 +0000https://111qqz.com/en/post/acm-icpc/2016/2016-10-31-hdu-4291/<p><a href="http://acm.hdu.edu.cn/showproblem.php?pid=4291" target="_blank" rel="noreferrer">题目链接</a></p> <p>题意：</p> <p>Given n (1 &lt;= n &lt;= 1018), You should solve for</p> <p>g(g(g(n))) mod 109 + 7 where</p> <p>g(n) = 3g(n - 1) + g(n - 2)</p> <p>g(1) = 1</p>https://111qqz.com/en/post/acm-icpc/2016/2016-10-30-hdu-3977/Sun, 30 Oct 2016 21:01:37 +0000https://111qqz.com/en/post/acm-icpc/2016/2016-10-30-hdu-3977/<p><a href="http://acm.hdu.edu.cn/showproblem.php?pid=3977" target="_blank" rel="noreferrer">题目链接</a></p> <p>题意：f[0] = 1,f[1] = 1,f[i] = f[i-1] + f[i-2] (i&gt;=2)，问最小的m满足f[n]%p==f[n+m]%p</p>