<?xml version="1.0" encoding="utf-8" standalone="yes"?><rss version="2.0" xmlns:atom="http://www.w3.org/2005/Atom"><channel><title>Lucas定理 on 111qqz's blog</title><link>https://111qqz.com/en/tags/lucas%E5%AE%9A%E7%90%86/</link><description>Recent content in Lucas定理 on 111qqz's blog</description><generator>Hugo -- gohugo.io</generator><language>en</language><copyright>© 2015-2026 111qqz</copyright><lastBuildDate>Thu, 03 Mar 2016 11:50:32 +0000</lastBuildDate><atom:link href="https://111qqz.com/en/tags/lucas%E5%AE%9A%E7%90%86/index.xml" rel="self" type="application/rss+xml"/><item><title>hdu 3929 Big Coefficients (递归形式的容斥原理+lucas定理的结论)</title><link>https://111qqz.com/en/post/acm-icpc/2016/2016-03-03-hdu3929/</link><pubDate>Thu, 03 Mar 2016 11:50:32 +0000</pubDate><guid>https://111qqz.com/en/post/acm-icpc/2016/2016-03-03-hdu3929/</guid><description>&lt;p&gt;题意：F(x) = (1+x)^a1 + (1+x)^a2 + … + (1+x)^am，求系数是奇数的项的个数。
思路：&lt;a href="http://blog.csdn.net/qian99/article/details/22692443" target="_blank" rel="noreferrer"&gt;解题报告&lt;/a&gt; 涉及到的由lucas定理得到的推论的证明&lt;a href="http://blog.csdn.net/qian99/article/details/22691571" target="_blank" rel="noreferrer"&gt;lucas定理证明&lt;/a&gt; 以及这篇理解里有递归形式的容斥定理的一般写法。。&lt;a href="http://www.acmerblog.com/hdu-3929-big-coefficients-6931.html" target="_blank" rel="noreferrer"&gt;递归形式的容斥定理&lt;/a&gt;&lt;/p&gt;</description></item><item><title>codeforces 451E Devu and Flowers (指数型母函数)</title><link>https://111qqz.com/en/post/acm-icpc/2016/2016-03-02-cf451e/</link><pubDate>Wed, 02 Mar 2016 13:41:51 +0000</pubDate><guid>https://111qqz.com/en/post/acm-icpc/2016/2016-03-02-cf451e/</guid><description>&lt;p&gt;&lt;a href="http://codeforces.com/problemset/problem/451/E" target="_blank" rel="noreferrer"&gt;http://codeforces.com/problemset/problem/451/E&lt;/a&gt;
题意;有n个花坛，要选s支花，每个花坛有f[i]支花，同一个花坛的花颜色相同，不同花坛的花颜色不同，问说可以有多少种组合。
思路：典型的母函数&amp;hellip;然而s有点大，根据泰勒展开什么的&amp;hellip;先转一下官方题解。&lt;/p&gt;</description></item><item><title>codeforces 560 E. Gerald and Giant Chess (dp+lucas定理,求大组合数 mod p,p为质数)</title><link>https://111qqz.com/en/post/acm-icpc/2015/2015-08-18-codeforces560e-geraldandgiantchessdplucase5ae9ae79086e6b182e5a4a7e7bb84e59088e695b0modppe4b8bae8b4a8e695b0/</link><pubDate>Tue, 18 Aug 2015 11:36:00 +0000</pubDate><guid>https://111qqz.com/en/post/acm-icpc/2015/2015-08-18-codeforces560e-geraldandgiantchessdplucase5ae9ae79086e6b182e5a4a7e7bb84e59088e695b0modppe4b8bae8b4a8e695b0/</guid><description>&lt;p&gt;dp方程想错了.果然还是欠练啊.&lt;/p&gt;
&lt;p&gt;如果我们不考虑坏点,那么从 (0,0)到(x,y)的方案数是c(x+y,x)或者c(x+y,y)&lt;/p&gt;</description></item></channel></rss>