<?xml version="1.0" encoding="utf-8" ?><rss version="2.0"><channel><title><![CDATA[BUAA_Alchemist的博客]]></title><description><![CDATA[Brick by brick, we set it all up!]]></description><link>https://blog.youkuaiyun.com/BUAA_Alchemist</link><language>zh-cn</language><generator>https://blog.youkuaiyun.com/</generator><copyright><![CDATA[Copyright &copy; BUAA_Alchemist]]></copyright><item><title><![CDATA[计蒜客2020蓝桥杯大学A组模拟赛题解]]></title><link>https://blog.youkuaiyun.com/BUAA_Alchemist/article/details/104070161</link><guid>https://blog.youkuaiyun.com/BUAA_Alchemist/article/details/104070161</guid><author>BUAA_Alchemist</author><pubDate>Wed, 22 Jan 2020 15:10:43 +0800</pubDate><description><![CDATA[计蒜客2020蓝桥杯大学A组模拟赛题解
蓝桥杯的话，去年拿了C++组的国二。今年报名了新成立的Python组，不知道能不能摸到国一的鱼
模拟赛链接如下：
https://www.jisuanke.com/contest/6510?view=challenges
打蓝桥的策略依旧很简单：前面60~70min 搞填空，后面刷大题，注意先看完所有题，不要只按顺序

A. 计算周长
一个很显然的想法是a和...]]></description><category></category></item><item><title><![CDATA[如何用分布式Pollard-Rho法对椭圆曲线离散对数问题（ECDLP）进行攻击（下）]]></title><link>https://blog.youkuaiyun.com/BUAA_Alchemist/article/details/95949098</link><guid>https://blog.youkuaiyun.com/BUAA_Alchemist/article/details/95949098</guid><author>BUAA_Alchemist</author><pubDate>Mon, 15 Jul 2019 21:39:15 +0800</pubDate><description><![CDATA[ 在上一篇中，我主要介绍了对赛题的初步攻击——单机Pollard-Rho。接下来介绍分布式算法。
 让多个机子，从点集合G中不同的起点进行出发，用QD法记录的桩点会被记录到一个公共表里，然后每个机子共享数据，也就是说每个机子可以知道一个桩点，是否曾经被自己或者其它机子产生出来过。
 这里面有一些隐含的信息。首先，并不是有n台机子，产生答案的时间就会缩短到原来的1/n，事实上，随机产生的Rho图数量...]]></description><category></category></item><item><title><![CDATA[如何用分布式Pollard-Rho法对椭圆曲线离散对数问题（ECDLP）进行攻击（上）]]></title><link>https://blog.youkuaiyun.com/BUAA_Alchemist/article/details/94437596</link><guid>https://blog.youkuaiyun.com/BUAA_Alchemist/article/details/94437596</guid><author>BUAA_Alchemist</author><pubDate>Mon, 15 Jul 2019 11:44:54 +0800</pubDate><description><![CDATA[ 上个学期，竞赛因为种种原因没有继续进行而搁浅。这里仅仅分享一下临时成果。
 椭圆曲线离散对数问题赛题介绍如下

 任何一个熟悉密码学的人基本上都会了解椭圆曲线离散对数问题。它比较普通素数域离散对数问题更具有难解性。因此相当多的公钥密码体制都构建在这个问题的基础上，如EC-Elgamal系统，以及数字签名、密钥交换协议等

...]]></description><category></category></item><item><title><![CDATA[训练复盘记录]]></title><link>https://blog.youkuaiyun.com/BUAA_Alchemist/article/details/91353200</link><guid>https://blog.youkuaiyun.com/BUAA_Alchemist/article/details/91353200</guid><author>BUAA_Alchemist</author><pubDate>Sun, 09 Jun 2019 15:08:48 +0800</pubDate><description><![CDATA[ 前言：我，zwr，qzz三人组成Hard_to_Name临时队从南昌邀请赛回来以后，由于我们Alchemist队学姐无法参加集训，正好前阵子qzz学长也是和自己配合过一段时间，又有意愿再打一年，于是正好加入我们队。于是现在这个队是我，qzz，lpy三人。计划是每周训练一次，假期的话，除了集训时间外每周集训两次。团队配合是十分重要的，每场训练都要依照惯例写复盘，反思，补题。希望我们最后能克服万难，...]]></description><category></category></item><item><title><![CDATA[谈一谈Windows子系统Linux(WSL)相关]]></title><link>https://blog.youkuaiyun.com/BUAA_Alchemist/article/details/90480061</link><guid>https://blog.youkuaiyun.com/BUAA_Alchemist/article/details/90480061</guid><author>BUAA_Alchemist</author><pubDate>Thu, 23 May 2019 14:34:57 +0800</pubDate><description><![CDATA[ 身边很多人喜欢在linux下做代码开发工作，但是想做其他事情比如玩游戏或者处理视频之类的，可能还是得回到Win下。我没有在Linux下写代码的必要性，所以一直没想装双系统，不过偶尔必要时需要用到Linux环境。我之前一直用的是Vmware+CentOS。这里介绍另一种比虚拟机更棒的解决方式——WSL。
 微软在两年前便已经悄悄上线了Linux子系统功能。它允许在一个命令行窗口中使用Linux命令...]]></description><category></category></item><item><title><![CDATA[【动态规划】lydsy4498 魔法的碰撞 lydsy4664 Count 特殊的序列dp方法]]></title><link>https://blog.youkuaiyun.com/BUAA_Alchemist/article/details/90107837</link><guid>https://blog.youkuaiyun.com/BUAA_Alchemist/article/details/90107837</guid><author>BUAA_Alchemist</author><pubDate>Sat, 11 May 2019 12:00:43 +0800</pubDate><description><![CDATA[ 这是我没有接触过的dp类型。因此记录一下。
 先看lydsy4498（这题现在OJ上已经消失了）。
 首先容易想到，对于一个布阵方式，我们可以把多余的格子抽掉，还原出一个魔法师紧凑的局面。那么每一个魔法师紧凑的局面，如果长度为lll，往其中填充L−lL-lL−l个格子，通过插板原理，我们知道它对应着Cn+L−lnC_{n+L-l}^nCn+L−ln​个布阵，于是问题转化为求长度为lll的紧凑局面...]]></description><category></category></item><item><title><![CDATA[北航ICPC集训队第二次春训（2019.4.28）]]></title><link>https://blog.youkuaiyun.com/BUAA_Alchemist/article/details/89813527</link><guid>https://blog.youkuaiyun.com/BUAA_Alchemist/article/details/89813527</guid><author>BUAA_Alchemist</author><pubDate>Sat, 04 May 2019 14:50:20 +0800</pubDate><description><![CDATA[https://codeforces.com/problemset/problem/626/C
n+m个人用积木搭堆（每个人搭一堆），n个人只能用高度为2的积木， m个人只能用高度为3的积木。两种积木均有无限多个。
这些人都不希望自己搭出的堆的高度与其他任何人的相同。请找出在所有人搭出的积木高度都不同的情况下，这些人中搭出的最高的堆的高度最小值。
Input
第一行输入包含两个用空格隔开的整数n ...]]></description><category></category></item><item><title><![CDATA[北航ICPC集训队第一次春训（2019.4.27）]]></title><link>https://blog.youkuaiyun.com/BUAA_Alchemist/article/details/89631789</link><guid>https://blog.youkuaiyun.com/BUAA_Alchemist/article/details/89631789</guid><author>BUAA_Alchemist</author><pubDate>Sat, 04 May 2019 13:12:32 +0800</pubDate><description><![CDATA[#include&lt;cstdio&gt;
#define mo 12345678910
using namespace std;
using LL=long long;

int top,n;
LL val[200005],ans;
char S[200005];

int main()
{
	scanf("%d",&amp;n);
	for(int i=1,d;i&lt;=n;i++)
	{...]]></description><category></category></item><item><title><![CDATA[【树链剖分】【模板】洛谷3384]]></title><link>https://blog.youkuaiyun.com/BUAA_Alchemist/article/details/89710186</link><guid>https://blog.youkuaiyun.com/BUAA_Alchemist/article/details/89710186</guid><author>BUAA_Alchemist</author><pubDate>Tue, 30 Apr 2019 15:16:13 +0800</pubDate><description><![CDATA[题目描述
如题，已知一棵包含N个结点的树（连通且无环），每个节点上包含一个数值，需要支持以下操作：
操作1： 格式： 1 x y z 表示将树从x到y结点最短路径上所有节点的值都加上z
操作2： 格式： 2 x y 表示求树从x到y结点最短路径上所有节点的值之和
操作3： 格式： 3 x z 表示将以x为根节点的子树内所有节点值都加上z
操作4： 格式： 4 x 表示求以x为根节点的子树内所有节点...]]></description><category></category></item><item><title><![CDATA[【动态规划】codeforces1149B Three Religions]]></title><link>https://blog.youkuaiyun.com/BUAA_Alchemist/article/details/89706388</link><guid>https://blog.youkuaiyun.com/BUAA_Alchemist/article/details/89706388</guid><author>BUAA_Alchemist</author><pubDate>Tue, 30 Apr 2019 11:45:52 +0800</pubDate><description><![CDATA[B. Three Religions
time limit per test3 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
During the archaeological research in the Middle East you found the traces ...]]></description><category></category></item><item><title><![CDATA[【贪心】【动态规划】【数学思维】codeforces1152D Neko and Aki's Prank]]></title><link>https://blog.youkuaiyun.com/BUAA_Alchemist/article/details/89516975</link><guid>https://blog.youkuaiyun.com/BUAA_Alchemist/article/details/89516975</guid><author>BUAA_Alchemist</author><pubDate>Thu, 25 Apr 2019 15:25:14 +0800</pubDate><description><![CDATA[D. Neko and Aki’s Prank
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
Neko is playing with his toys on the backyard of Aki’s house. Aki decid...]]></description><category></category></item><item><title><![CDATA[【动态规划】codeforces1155D Beautiful Array]]></title><link>https://blog.youkuaiyun.com/BUAA_Alchemist/article/details/89497060</link><guid>https://blog.youkuaiyun.com/BUAA_Alchemist/article/details/89497060</guid><author>BUAA_Alchemist</author><pubDate>Wed, 24 Apr 2019 16:40:31 +0800</pubDate><description><![CDATA[D. Beautiful Array
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
You are given an array a consisting of n integers. Beauty of array is the m...]]></description><category></category></item><item><title><![CDATA[作业图片链接引用区域]]></title><link>https://blog.youkuaiyun.com/BUAA_Alchemist/article/details/89467192</link><guid>https://blog.youkuaiyun.com/BUAA_Alchemist/article/details/89467192</guid><author>BUAA_Alchemist</author><pubDate>Tue, 23 Apr 2019 10:20:10 +0800</pubDate><description><![CDATA[]]></description><category></category></item><item><title><![CDATA[【数学思维】【动态规划】codeforces1153F Serval and Bonus Problem]]></title><link>https://blog.youkuaiyun.com/BUAA_Alchemist/article/details/89314191</link><guid>https://blog.youkuaiyun.com/BUAA_Alchemist/article/details/89314191</guid><author>BUAA_Alchemist</author><pubDate>Mon, 15 Apr 2019 16:12:03 +0800</pubDate><description><![CDATA[Getting closer and closer to a mathematician, Serval becomes a university student on math major in Japari University. On the Calculus class, his teacher taught him how to calculate the expected length...]]></description><category></category></item><item><title><![CDATA[【贪心】【二分答案】codeforces1119E Pavel and Triangles]]></title><link>https://blog.youkuaiyun.com/BUAA_Alchemist/article/details/89066482</link><guid>https://blog.youkuaiyun.com/BUAA_Alchemist/article/details/89066482</guid><author>BUAA_Alchemist</author><pubDate>Sun, 07 Apr 2019 10:52:29 +0800</pubDate><description><![CDATA[Pavel has several sticks with lengths equal to powers of two.
He has a0 sticks of length 20=1, a1 sticks of length 212^121 2, …, an−1 sticks of length 2n−12^{n−1}2n−1.
Pavel wants to make the maximum ...]]></description><category></category></item><item><title><![CDATA[【递推】codeforces1140E Palindrome-less Arrays]]></title><link>https://blog.youkuaiyun.com/BUAA_Alchemist/article/details/89048751</link><guid>https://blog.youkuaiyun.com/BUAA_Alchemist/article/details/89048751</guid><author>BUAA_Alchemist</author><pubDate>Fri, 05 Apr 2019 22:52:57 +0800</pubDate><description><![CDATA[1140E Palindrome-less Arrays
Let’s denote that some array b is bad if it contains a subarray bl,bl+1,…,br of odd length more than 1 (l&lt;r and r−l+1 is odd) such that ∀i∈{0,1,…,r−l} bl+i=br−i.
If an ...]]></description><category></category></item><item><title><![CDATA[【并查集】用水填坑]]></title><link>https://blog.youkuaiyun.com/BUAA_Alchemist/article/details/88898000</link><guid>https://blog.youkuaiyun.com/BUAA_Alchemist/article/details/88898000</guid><author>BUAA_Alchemist</author><pubDate>Fri, 29 Mar 2019 17:24:24 +0800</pubDate><description><![CDATA[题目链接：https://ac.nowcoder.com/acm/contest/403/A

 先想想一维的降雨积水问题。很容易发现可以考虑每个地两侧的高度最大值的最小值，就是这个地方的积水量，只需要把这些积水量累和即可。但是二维的就没法这么做了。继续想想一维的问题，发现可以有一种并查集的解法，进而推广到二维。
 我们把所有的高度排序，从小到大添加入原图，用并查集维护连通块，如果新加入的值a[i...]]></description><category></category></item><item><title><![CDATA[【动态规划】北航程序设计竞赛决赛B题]]></title><link>https://blog.youkuaiyun.com/BUAA_Alchemist/article/details/88881640</link><guid>https://blog.youkuaiyun.com/BUAA_Alchemist/article/details/88881640</guid><author>BUAA_Alchemist</author><pubDate>Fri, 29 Mar 2019 00:11:38 +0800</pubDate><description><![CDATA[ 决赛那天感觉确实没发挥好。本来有机会逃离三题大队的。结果中途很多人搞了几何构造题，忍不住跟风。事实证明以我的想象力果然搞不出来。还不如专心搞搞B这题。
 B的题意很简单，给一个长度不超过5000的字符串，字符集大小不超过5000，求字典序第k小的子串（子串可以不连续），k规模不超过子串总数以及1E18。
 先考虑每个字符都不同的情况，很容易想到一个一个把答案的字符定下来。我们每一次求出下一步接字...]]></description><category></category></item><item><title><![CDATA[第十四届北航程序设计竞赛预赛题解（非官方）]]></title><link>https://blog.youkuaiyun.com/BUAA_Alchemist/article/details/88875451</link><guid>https://blog.youkuaiyun.com/BUAA_Alchemist/article/details/88875451</guid><author>BUAA_Alchemist</author><pubDate>Thu, 28 Mar 2019 18:43:29 +0800</pubDate><description><![CDATA[ 最近沙河好多人生病了，我也不能幸免，今天终于舒服了。补一补上周的老坑吧。
 接下来尽量每天刷题。另外要加油学习汇编语言，优化我们的密码学竞赛程序。
官方题解参见：http://clatisus.com/BCPC2018_qualification_round
A 两等分的Alice
https://buaacoding.cn/problem/1914/index
 非常容易设计出粗暴的dp。考虑...]]></description><category></category></item><item><title><![CDATA[【费用流】洛谷1251 餐巾计划问题]]></title><link>https://blog.youkuaiyun.com/BUAA_Alchemist/article/details/88746705</link><guid>https://blog.youkuaiyun.com/BUAA_Alchemist/article/details/88746705</guid><author>BUAA_Alchemist</author><pubDate>Fri, 22 Mar 2019 17:36:55 +0800</pubDate><description><![CDATA[ 今年参加BCPC暨ACM选拔的同学非常多。看了下预赛榜单，my和ljy率领一大批18级dalao，tky等buaacoding刷题狂魔也都参加了，更可怕的是还有一堆CF黄名橙名故意只写了一点点题隐藏实力。这对于你航ACM事业当然是好事，但对于我们这些菜鸡就很黑暗了。希望明天能加油，Alchemist加油~

 不废话了，来看这题。如果不告诉我是网络流，我估计还真的很难想到它的算法。但告诉我是...]]></description><category></category></item></channel></rss>