<?xml version="1.0" encoding="utf-8" ?><rss version="2.0"><channel><title><![CDATA[Floraqiu的博客]]></title><description><![CDATA[有志者事竟成，破釜沉舟，百二秦关终属楚；苦心人天不负，卧薪尝胆，三千越甲可吞吴。]]></description><link>https://blog.youkuaiyun.com/Floraqiu</link><language>zh-cn</language><generator>https://blog.youkuaiyun.com/</generator><copyright><![CDATA[Copyright &copy; Floraqiu]]></copyright><item><title><![CDATA[1.Regression]]></title><link>https://blog.youkuaiyun.com/Floraqiu/article/details/107893620</link><guid>https://blog.youkuaiyun.com/Floraqiu/article/details/107893620</guid><author>Floraqiu</author><pubDate>Sun, 09 Aug 2020 14:21:04 +0800</pubDate><description><![CDATA[Regression
文章目录Regression零、Regression回归一、Linear Model线性回归1.Model（function set）2.Loss Function损失函数（goodness of function）损失函数形式为什么选用均方误差作为损失函数？直观解释概率解释3.Pick the best function1.采用Normal Equation正规方程求解前提已知求解总结2.采用Gradient Descent梯度下降求解优点步骤批量梯度下降算法batch gradie]]></description><category></category></item><item><title><![CDATA[打印优快云、博客园文章]]></title><link>https://blog.youkuaiyun.com/Floraqiu/article/details/107198298</link><guid>https://blog.youkuaiyun.com/Floraqiu/article/details/107198298</guid><author>Floraqiu</author><pubDate>Wed, 08 Jul 2020 09:43:49 +0800</pubDate><description><![CDATA[console中执行
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	$("#comment_title, #comment_list, #comment_bar, #comment_form, .announce, #ad_]]></description><category></category></item><item><title><![CDATA[【KMP】个人理解总结]]></title><link>https://blog.youkuaiyun.com/Floraqiu/article/details/105222603</link><guid>https://blog.youkuaiyun.com/Floraqiu/article/details/105222603</guid><author>Floraqiu</author><pubDate>Tue, 31 Mar 2020 15:44:47 +0800</pubDate><description><![CDATA[KMP总结
这几天复习kmp算法，发现自己看别人的讲解又看糊涂了，所以还是需要将别人的思想理解之后，自己再输出一边。
KMP的来源
KMP的出现是为了优化朴素模式匹配算法，那么朴素模式匹配算法是什么？又存在什么缺点呢？
朴素模式匹配算法 ，就是将主串中与模式串长度相同的子串提取出来，挨个和模式串对比，当子串与模式串某个对应字符不匹配时，就立即放弃当前子串，检索下一个子串。
/* 字符串下标始于0 ...]]></description><category></category></item><item><title><![CDATA[网络流入门]]></title><link>https://blog.youkuaiyun.com/Floraqiu/article/details/87980555</link><guid>https://blog.youkuaiyun.com/Floraqiu/article/details/87980555</guid><author>Floraqiu</author><pubDate>Wed, 27 Feb 2019 17:07:35 +0800</pubDate><description><![CDATA[网络流入门
1. 最大流问题

残余网络：由残余容量（每条边上的容量与流量之差）构成的叫残余网络。
增广路定理：如果残余网络中不存在增广路，则当前流就是最大流。


最小割最大流定理：
最大流的流量 = 最小割的容量

1.1 EK算法（用BFS寻找增广路）O(nm2)O(nm^2)O(nm2)

思路：
寻找增广路，记录从s到每个节点的路径上的最小残量a[i]，则a[t]就是s到t的最小残量。由...]]></description><category></category></item><item><title><![CDATA[数论总结]]></title><link>https://blog.youkuaiyun.com/Floraqiu/article/details/86660541</link><guid>https://blog.youkuaiyun.com/Floraqiu/article/details/86660541</guid><author>Floraqiu</author><pubDate>Sat, 26 Jan 2019 20:12:55 +0800</pubDate><description><![CDATA[数论
1.1 Lucas定理 （扩展Lucas定理）【组合数取模】

Lucas定理：
要计算Cnm%pC_n^m \%pCnm​%p的值
n=nkpk+nk−1pk−1+...+n1p+n0
n = n_kp^k+n_{k-1}p^{k-1}+...+n_1p+n_0
n=nk​pk+nk−1​pk−1+...+n1​p+n0​
m=mkpk+mk−1pk−1+...+m1p+m0
m = m_...]]></description><category></category></item><item><title><![CDATA[【线段树】1108E2	 Array and Segments (Hard version) & 1108E1	 Array and Segments (Easy version)]]></title><link>https://blog.youkuaiyun.com/Floraqiu/article/details/86632558</link><guid>https://blog.youkuaiyun.com/Floraqiu/article/details/86632558</guid><author>Floraqiu</author><pubDate>Thu, 24 Jan 2019 20:11:23 +0800</pubDate><description><![CDATA[The only difference between easy and hard versions is a number of elements in the array.
You are given an array a consisting of n integers. The value of the i-th element of the array is ai.
You are al...]]></description><category></category></item><item><title><![CDATA[【MST kruskal】 MST Unification CodeForces - 1108F]]></title><link>https://blog.youkuaiyun.com/Floraqiu/article/details/86630053</link><guid>https://blog.youkuaiyun.com/Floraqiu/article/details/86630053</guid><author>Floraqiu</author><pubDate>Thu, 24 Jan 2019 17:08:03 +0800</pubDate><description><![CDATA[MST Unification CodeForces - 1108F
You are given an undirected weighted connected graph with n vertices and m edges without loops and multiple edges.
The i-th edge is ei=(ui,vi,wi); the distance betwe...]]></description><category></category></item><item><title><![CDATA[【FWT快速沃尔什变换】讲解 + 例题  BZO-J4589   Hard Nim（FWT、博弈论、快速幂）]]></title><link>https://blog.youkuaiyun.com/Floraqiu/article/details/86622837</link><guid>https://blog.youkuaiyun.com/Floraqiu/article/details/86622837</guid><author>Floraqiu</author><pubDate>Thu, 24 Jan 2019 12:55:39 +0800</pubDate><description><![CDATA[参考：FWT快速沃尔什变换学习笔记(证明等都非常详细，虽然我没看证明 )
一、FWT的作用


FFT 、NTT：计算多项式卷积
Ck=∑i+j=kAi∗BjC_k=\sum_{i+j=k}A_i*B_jCk​=i+j=k∑​Ai​∗Bj​


FWT ：计算多项式的位运算卷积

Ck=∑i∣j=kAi∗BjC_k=\sum_{i|j=k}A_i*B_jCk​=∑i∣j=k​Ai​∗Bj​
Ck=...]]></description><category></category></item><item><title><![CDATA[【博弈论】常见总结]]></title><link>https://blog.youkuaiyun.com/Floraqiu/article/details/86616850</link><guid>https://blog.youkuaiyun.com/Floraqiu/article/details/86616850</guid><author>Floraqiu</author><pubDate>Wed, 23 Jan 2019 21:39:31 +0800</pubDate><description><![CDATA[每次看到博弈论的题目就很脑阔疼

平衡状态，又称作奇异局势。当面对这个局势时则会失败。
任意非平衡态经过一次操作可以变为平衡态。
先手能够在初始为非平衡的游戏中取胜，后手能够在初始为平衡的游戏中取胜。

SG函数待补
两个博客
1
2
来源
1
2

一、 巴仕博弈（Bash Game）1堆


题目 HDU1846
有一堆物品，两个人轮流从这堆物品中取物，规定每次至少取一个，最多取m个。最后取光...]]></description><category></category></item><item><title><![CDATA[【FFT】Kattis - aplusb A+B Problem  2016年ACM香港网络赛]]></title><link>https://blog.youkuaiyun.com/Floraqiu/article/details/86611210</link><guid>https://blog.youkuaiyun.com/Floraqiu/article/details/86611210</guid><author>Floraqiu</author><pubDate>Wed, 23 Jan 2019 16:13:49 +0800</pubDate><description><![CDATA[题目链接
题目  ：Kattis - aplusb A+B Problem
Given N integers in the range [−50000,50000], how many ways are there to pick three integers ai, aj, ak, such that i, j, k are pairwise distinct and ai+aj=ak? Two...]]></description><category></category></item><item><title><![CDATA[【矩阵快速幂】]]></title><link>https://blog.youkuaiyun.com/Floraqiu/article/details/86599801</link><guid>https://blog.youkuaiyun.com/Floraqiu/article/details/86599801</guid><author>Floraqiu</author><pubDate>Tue, 22 Jan 2019 19:09:55 +0800</pubDate><description><![CDATA[struct Matrix
{
    LL m[ms][ms];
} ans, base;

Matrix multi(Matrix a, Matrix b)
{
    Matrix tmp;
    for(int i = 0; i &amp;lt; ms; i++)
        for(int j = 0; j &amp;lt; ms; j++)
        {
            tmp.m...]]></description><category></category></item><item><title><![CDATA[【中国剩余定理】]]></title><link>https://blog.youkuaiyun.com/Floraqiu/article/details/86599786</link><guid>https://blog.youkuaiyun.com/Floraqiu/article/details/86599786</guid><author>Floraqiu</author><pubDate>Tue, 22 Jan 2019 19:08:44 +0800</pubDate><description><![CDATA[const int N = 15;
int n;
LL yu[N], mo[N];

LL exgcd(LL a, LL b, LL &amp;amp;x, LL &amp;amp;y)
{
    if(b == 0)
    {
        x = 1;
        y = 0;
        return a;
    }
    LL r = exgcd(b, a % b, x, y);
   ...]]></description><category></category></item><item><title><![CDATA[【高斯消元】]]></title><link>https://blog.youkuaiyun.com/Floraqiu/article/details/86599772</link><guid>https://blog.youkuaiyun.com/Floraqiu/article/details/86599772</guid><author>Floraqiu</author><pubDate>Tue, 22 Jan 2019 19:07:06 +0800</pubDate><description><![CDATA[double a[maxn][maxn];
double x[maxn];
double EPS = 1e-6;
int n, m;


int Gauss()
{
    int i, j, k;
    int max_r;
    double tmp;

    for(i = 1; i &amp;lt;= n; i++)
    {
        max_r = i;
        for(...]]></description><category></category></item><item><title><![CDATA[【Eular质数筛】]]></title><link>https://blog.youkuaiyun.com/Floraqiu/article/details/86599756</link><guid>https://blog.youkuaiyun.com/Floraqiu/article/details/86599756</guid><author>Floraqiu</author><pubDate>Tue, 22 Jan 2019 19:05:46 +0800</pubDate><description><![CDATA[const int N = 1e6 + 10;
bool prime[N];
int p[N], tot, num[N];
void init()
{
    tot = 0;
    clr(prime, true);
    for(int i = 2; i &amp;lt; N; i++)
    {
        if(prime[i])
            p[tot++] = i;
  ...]]></description><category></category></item><item><title><![CDATA[【欧拉降幂公式】【欧拉函数】]]></title><link>https://blog.youkuaiyun.com/Floraqiu/article/details/86598283</link><guid>https://blog.youkuaiyun.com/Floraqiu/article/details/86598283</guid><author>Floraqiu</author><pubDate>Tue, 22 Jan 2019 17:27:35 +0800</pubDate><description><![CDATA[一、降幂公式
AK≡AK%ϕ(m)+ϕ(m)(mod&amp;amp;ThinSpace;&amp;amp;ThinSpace;m) A^K \equiv A^{K\%\phi(m) + \phi(m)} (\mod m)AK≡AK%ϕ(m)+ϕ(m)(modm)
其中φ(即phi)是欧拉函数其中φ(即phi)是欧拉函数其中φ(即phi)是欧拉函数
二、欧拉函数

定义：对于一个正整数n，小于n且和n互质的正整数...]]></description><category></category></item><item><title><![CDATA[【FFT快速傅里叶变换】讲解]]></title><link>https://blog.youkuaiyun.com/Floraqiu/article/details/86590350</link><guid>https://blog.youkuaiyun.com/Floraqiu/article/details/86590350</guid><author>Floraqiu</author><pubDate>Tue, 22 Jan 2019 13:42:55 +0800</pubDate><description><![CDATA[参考来源：
十分简明易懂的FFT（快速傅里叶变换）
小学生都能看懂的FFT！！！
一、FFT的介绍


FFT是什么
快速傅里叶变换（FFT）是一种能在 O(nlog⁡n)O(n\log{n})O(nlogn) 的时间内将一个多项式转换成它的点值表示的算法。


FFT的作用
快速计算多项式乘法（即卷积）
（还可以用到字符串的模糊匹配）
前置知识


多项式的表示

点值表示法：



实际上就是...]]></description><category></category></item><item><title><![CDATA[【左偏树】例题]]></title><link>https://blog.youkuaiyun.com/Floraqiu/article/details/86563647</link><guid>https://blog.youkuaiyun.com/Floraqiu/article/details/86563647</guid><author>Floraqiu</author><pubDate>Sun, 20 Jan 2019 17:03:31 +0800</pubDate><description><![CDATA[例题一： dispatching HYSBZ - 2809
在一个忍者的帮派里，一些忍者们被选中派遣给顾客，然后依据自己的工作获取报偿。在这个帮派里，有一名忍者被称之为 Master。除了 Master以外，每名忍者都有且仅有一个上级。为保密，同时增强忍者们的领导力，所有与他们工作相关的指令总是由上级发送给他的直接下属，而不允许通过其他的方式发送。现在你要招募一批忍者，并把它们派遣给顾客。你需要为...]]></description><category></category></item><item><title><![CDATA[【左偏树】讲解]]></title><link>https://blog.youkuaiyun.com/Floraqiu/article/details/86563419</link><guid>https://blog.youkuaiyun.com/Floraqiu/article/details/86563419</guid><author>Floraqiu</author><pubDate>Sun, 20 Jan 2019 16:43:39 +0800</pubDate><description><![CDATA[一、左偏树介绍


左偏树(Leftist Tree) 是一种可并堆(Mergeable Heap) ，它除了支持优先队列的三个基本操作（插入，删除，取最小节点），还支持一个很特殊的操作——合并操作。


左偏树是一棵堆有序(Heap Ordered)二叉树。


左偏树满足左偏性质(Leftist Property)。


二、左偏树性质


定义一棵左偏树中的外节点(External Node...]]></description><category></category></item><item><title><![CDATA[【并查集】（并查集的合并中 使用了启发式合并）数据分割 HDU - 6109 （相等关系：并查集   不等关系：set）]]></title><link>https://blog.youkuaiyun.com/Floraqiu/article/details/86563082</link><guid>https://blog.youkuaiyun.com/Floraqiu/article/details/86563082</guid><author>Floraqiu</author><pubDate>Sun, 20 Jan 2019 16:16:54 +0800</pubDate><description><![CDATA[Key Point:
相等关系：可传递的，使用并查集进行检验
不等关系：不可传递，使用set进行检验
（如 X1 !=  X2   X1 != X3  但是X2 = X3 是可以成立的）

数据分割 HDU - 6109
小w来到百度之星的赛场上，准备开始实现一个程序自动分析系统。
这个程序接受一些形如xi=xj 或 xi≠xj
的相等/不等约束条件作为输入，判定是否可以通过给每个 w 赋适当的值...]]></description><category></category></item><item><title><![CDATA[【启发式合并】例题 CodeForces-600E（子树颜色 树上众数）+ CodeForces - 1009F（每层节点数的众数）+  CSU - 1811(去边后，求颜色并集大小)]]></title><link>https://blog.youkuaiyun.com/Floraqiu/article/details/86562503</link><guid>https://blog.youkuaiyun.com/Floraqiu/article/details/86562503</guid><author>Floraqiu</author><pubDate>Sun, 20 Jan 2019 15:58:59 +0800</pubDate><description><![CDATA[例题一  Lomsat gelral CodeForces - 600E
You are given a rooted tree with root in vertex 1. Each vertex is coloured in some colour.
Let’s call colour c dominating in the subtree of vertex v if there are n...]]></description><category></category></item></channel></rss>