<?xml version="1.0" encoding="utf-8" standalone="yes"?><rss version="2.0" xmlns:atom="http://www.w3.org/2005/Atom"><channel><title>逆元 on 111qqz's blog</title><link>https://111qqz.com/en/tags/%E9%80%86%E5%85%83/</link><description>Recent content in 逆元 on 111qqz's blog</description><generator>Hugo -- gohugo.io</generator><language>en</language><copyright>© 2015-2026 111qqz</copyright><lastBuildDate>Wed, 19 Oct 2016 10:03:52 +0000</lastBuildDate><atom:link href="https://111qqz.com/en/tags/%E9%80%86%E5%85%83/index.xml" rel="self" type="application/rss+xml"/><item><title>hdu 1211 RSA (扩展欧几里得算法求逆元 +快速幂)</title><link>https://111qqz.com/en/post/acm-icpc/2016/2016-10-19-hdu-1211-rsa-e689a9e5b195e6aca7e587a0e9878ce5be97e7ae97e6b395e6b182e98086e58583-e5bfabe9809fe5b982/</link><pubDate>Wed, 19 Oct 2016 10:03:52 +0000</pubDate><guid>https://111qqz.com/en/post/acm-icpc/2016/2016-10-19-hdu-1211-rsa-e689a9e5b195e6aca7e587a0e9878ce5be97e7ae97e6b395e6b182e98086e58583-e5bfabe9809fe5b982/</guid><description>&lt;p&gt;&lt;a href="http://acm.hdu.edu.cn/showproblem.php?pid=1211" target="_blank" rel="noreferrer"&gt;题目链接&lt;/a&gt;&lt;/p&gt;
&lt;p&gt;题意：给出p, q, e, l,令n = p * q, fn = (p-1) * (q-1)&lt;/p&gt;
&lt;p&gt;给出l个c,计算m = D(c) = c**&lt;em&gt;d&lt;/em&gt;** mod n,其中m为要输入的明文对应的ascii编码，d的计算方法：&amp;gt; calculate d, making d × e mod F(n) = 1 mod F(n), and d will be the private key。&lt;/p&gt;</description></item><item><title>逆元学习笔记</title><link>https://111qqz.com/en/post/acm-icpc/2016/2016-10-19-e98086e58583e5ada6e4b9a0e7ac94e8aeb0/</link><pubDate>Wed, 19 Oct 2016 09:17:15 +0000</pubDate><guid>https://111qqz.com/en/post/acm-icpc/2016/2016-10-19-e98086e58583e5ada6e4b9a0e7ac94e8aeb0/</guid><description>&lt;p&gt;&lt;a href="http://blog.csdn.net/acdreamers/article/details/8220787" target="_blank" rel="noreferrer"&gt;acdreamer_逆元学习笔记&lt;/a&gt;&lt;/p&gt;
&lt;p&gt;摘重点：&lt;/p&gt;
&lt;p&gt;ksm(a,mod-2)的方法求逆元只适用于mod为质数且 gcd(a,mod)==1&lt;/p&gt;</description></item><item><title>codeforces 594 D. REQ (树状数组+欧拉函数+逆元)</title><link>https://111qqz.com/en/post/acm-icpc/2016/2016-09-22-cf594d/</link><pubDate>Thu, 22 Sep 2016 07:48:00 +0000</pubDate><guid>https://111qqz.com/en/post/acm-icpc/2016/2016-09-22-cf594d/</guid><description>&lt;p&gt;&lt;a href="http://codeforces.com/problemset/problem/594/D" target="_blank" rel="noreferrer"&gt;题目链接&lt;/a&gt;&lt;/p&gt;
&lt;p&gt;题意：给出n个数,q个查询，每组一个区间，询问区间中所有数的乘积的欧拉函数9+7的答案是多少。&lt;/p&gt;</description></item><item><title>bc #77 div 2 B ||hdu 5651 xiaoxin juju needs help (排列组合，逆元)</title><link>https://111qqz.com/en/post/acm-icpc/2016/2016-03-27-hdu-5651/</link><pubDate>Sun, 27 Mar 2016 02:33:27 +0000</pubDate><guid>https://111qqz.com/en/post/acm-icpc/2016/2016-03-27-hdu-5651/</guid><description>&lt;p&gt;&lt;a href="http://acm.hdu.edu.cn/showproblem.php?pid=5651" target="_blank" rel="noreferrer"&gt;题目链接&lt;/a&gt;
题意；给出一个字符串，只由小写字母组成，可以任意排列，但是不能减少字符，问最多能得到多少个回文串，答案9+7&lt;/p&gt;</description></item><item><title>hdu 5145 NPY and girls</title><link>https://111qqz.com/en/post/acm-icpc/2016/2016-02-17-hdu5145/</link><pubDate>Wed, 17 Feb 2016 07:34:05 +0000</pubDate><guid>https://111qqz.com/en/post/acm-icpc/2016/2016-02-17-hdu5145/</guid><description>&lt;p&gt;&lt;a href="http://acm.hdu.edu.cn/showproblem.php?pid=5145" target="_blank" rel="noreferrer"&gt;http://acm.hdu.edu.cn/showproblem.php?pid=5145&lt;/a&gt;
题意：有n个女孩，编号1..n,第i个女孩在第a[i]个教室,m次访问，每次访问编号[L,R]的女孩，处于同一个教室的女孩一次只能访问一个，问有多少种访问方案。两个不同的方案当且仅当访问的顺序有所不同。&lt;/p&gt;</description></item></channel></rss>