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<a href="/ST%E8%A1%A8%EF%BC%88%E5%80%8D%E5%A2%9E%E8%A1%A8%EF%BC%89/" target="_blank">
ST表（倍增表）
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</h1>
<a class="index-excerpt index-excerpt__noimg" href="/ST%E8%A1%A8%EF%BC%88%E5%80%8D%E5%A2%9E%E8%A1%A8%EF%BC%89/" target="_blank">
<div>
看上次做ST表示整整一年前啊，当时刚初三下期。 高中确实没有初中快乐，更少的朋友，更多的压力。 害，往事不堪回首。 ST表就是倍增思想，和倍增求 LCA 非常像。具体看一看代码就行了 P3865 【模板】ST 表 1234567891011121314151617181920212223242526272829303132333435#include<bits/stdc++.h>usi
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<time datetime="2023-02-28 17:35" pubdate>
2023-02-28
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<span class="category-chains">
<span class="category-chain">
<a href="/categories/%E4%BF%A1%E6%81%AF%E5%AD%A6/" class="category-chain-item">信息学</a>
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<a href="/categories/%E4%BF%A1%E6%81%AF%E5%AD%A6/C-%E5%9F%BA%E7%A1%80/" class="category-chain-item">C++基础</a>
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<a href="/tags/%E4%BF%A1%E6%81%AF%E5%AD%A6/">#信息学</a>
<a href="/tags/ST%E8%A1%A8/">#ST表</a>
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<a href="/%E5%AD%97%E7%AC%A6%E4%B8%B2%E5%93%88%E5%B8%8C/" target="_blank">
字符串哈希
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</h1>
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<div>
字符串哈希非常简洁 （暴力），它根本不管冲不冲突，直接硬莽 具体的，对于一个字符串 s，它的哈希值为： 通过使用两个数 p,mp,mp,m hash=(s[n]+s[n−1]⋅p+s[n−2]⋅p2+⋯+s[1]⋅pn−1)mod m=∑i=1ns[i]⋅pn−imod m\begin{aligned} \mathrm{hash} &= (s[n]+s[n-1] \cdot p+s[
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2023-02-28
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<a href="/categories/%E4%BF%A1%E6%81%AF%E5%AD%A6/" class="category-chain-item">信息学</a>
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<a href="/categories/%E4%BF%A1%E6%81%AF%E5%AD%A6/C-%E5%9F%BA%E7%A1%80/" class="category-chain-item">C++基础</a>
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<a href="/tags/%E5%AD%97%E7%AC%A6%E4%B8%B2%E5%93%88%E5%B8%8C/">#字符串哈希</a>
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<h1 class="index-header">
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哈希
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</h1>
<a class="index-excerpt index-excerpt__noimg" href="/%E5%93%88%E5%B8%8C/" target="_blank">
<div>
通常情况下 STL 的 map 可香了，但是一个缺点就是常数有些大。 因此手写哈希依然必要。 最首先的思路就是将 要哈希的值 对 一个质数 取模。对一个质数取模的好处是可以降低冲突的次数。（感觉的，没有证） 但是冲突终于还是会有的，所以对于冲突，有下面两种方法。 开放寻址法 12345678910111213141516#include <cstring>#include <io
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<time datetime="2023-02-28 16:37" pubdate>
2023-02-28
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<a href="/categories/%E4%BF%A1%E6%81%AF%E5%AD%A6/C-%E5%9F%BA%E7%A1%80/" class="category-chain-item">C++基础</a>
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<a href="/KMP/" target="_blank">
KMP
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太多要复习了，脑阔痛。 复习半天双向搜索，还是来复习点简单的缓缓 整个 KMP 唯一难搞的地方就是 ne[]。其意义如下： ne[i] 表示当模式串和主串在模式串位置 iii 处匹配发生失配时，模式串的下标应该会退到的位置。 ne[i] 也可以表示在模式串截止到位置 iii 的前缀子串中，从头开始和从末尾开始的最长相同子串的长度。 如图: 长的是主串，短的是模式串。当主串与模式串在 iii
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<time datetime="2023-02-27 17:52" pubdate>
2023-02-27
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<a href="/categories/%E4%BF%A1%E6%81%AF%E5%AD%A6/" class="category-chain-item">信息学</a>
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<a href="/categories/%E4%BF%A1%E6%81%AF%E5%AD%A6/%E5%AD%97%E7%AC%A6%E4%B8%B2/" class="category-chain-item">字符串</a>
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<a href="/categories/%E4%BF%A1%E6%81%AF%E5%AD%A6/%E5%AD%97%E7%AC%A6%E4%B8%B2/KMP/" class="category-chain-item">KMP</a>
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<a href="/tags/KMP/">#KMP</a>
<a href="/tags/%E5%AD%97%E7%AC%A6%E4%B8%B2/">#字符串</a>
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双向搜索
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当竞赛生眼高手低，瞧不起基础算法的时候，基础算法就会来反杀你。
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<time datetime="2023-02-25 16:59" pubdate>
2023-02-25
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<a href="/categories/%E4%BF%A1%E6%81%AF%E5%AD%A6/" class="category-chain-item">信息学</a>
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<a href="/categories/%E4%BF%A1%E6%81%AF%E5%AD%A6/%E5%9B%BE%E8%AE%BA%E4%B8%8E%E6%90%9C%E7%B4%A2/" class="category-chain-item">图论与搜索</a>
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<a href="/%E6%95%B0%E8%AE%BA%E9%9B%86%E9%94%A6/" target="_blank">
数论集锦
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赶快杀青数论，没时间了 欧拉定理 欧拉函数 φ(n)\varphi(n)φ(n)，表示小于等于 nnn 和 nnn 互质的数的个数。 朴素计算 φ(n)=n⋅∏质数p∣n(p−1p)\varphi(n)=n \cdot \prod_{质数 p|n} (\frac{p-1}{p}) φ(n)=n⋅质数p∣n∏(pp−1) （证明：容斥原理） 性质 ∀n>1\forall n>1
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<time datetime="2023-02-23 16:24" pubdate>
2023-02-23
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<a href="/categories/%E4%BF%A1%E6%81%AF%E5%AD%A6/%E6%95%B0%E5%AD%A6/" class="category-chain-item">数学</a>
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<a href="/categories/%E4%BF%A1%E6%81%AF%E5%AD%A6/%E6%95%B0%E5%AD%A6/%E6%95%B0%E8%AE%BA/" class="category-chain-item">数论</a>
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<a href="/tags/%E6%95%B0%E8%AE%BA/">#数论</a>
<a href="/tags/%E6%AC%A7%E6%8B%89%E5%AE%9A%E7%90%86/">#欧拉定理</a>
<a href="/tags/%E6%89%A9%E5%B1%95%E6%AC%A7%E6%8B%89%E5%AE%9A%E7%90%86/">#扩展欧拉定理</a>
<a href="/tags/%E8%B4%B9%E9%A9%AC%E5%B0%8F%E5%AE%9A%E7%90%86/">#费马小定理</a>
<a href="/tags/%E5%A8%81%E5%B0%94%E9%80%8A%E5%AE%9A%E7%90%86/">#威尔逊定理</a>
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<a href="/%E6%9C%80%E7%9F%AD%E8%B7%AF%E9%9B%86%E9%94%A6/" target="_blank">
最短路集锦
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好久不用，最简单的也会忘记 Floyd 就是三层循环 123456789void floyd(){ for(int k=1;k<=n;k++){ for(int i=1;i<=n;i++){ for(int j=1;j<=n;j++){ d[i][j]=min(d[i]
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<a href="/tags/%E6%9C%80%E7%9F%AD%E8%B7%AF/">#最短路</a>
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Class 类
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<time datetime="2023-02-21 17:33" pubdate>
2023-02-21
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<a href="/categories/%E4%BF%A1%E6%81%AF%E5%AD%A6/C-%E5%9F%BA%E7%A1%80/" class="category-chain-item">C++基础</a>
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<a href="/tags/%E7%B1%BB/">#类</a>
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CRT和exCRT
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时间不多了，把数论复习一遍。 P1495 【模板】中国剩余定理（CRT）/ 曹冲养猪 给定 nnn 个整数 m1,m2,…,mnm_1,m_2,\dots,m_nm1,m2,…,mn 保证它们两两互质，又给定任意 nnn 个整数 a1,a2,…,ana_1,a_2,\dots,a_na1,a2,…,an，现在要求一个整数 xxx，使其满足下面的线性同余方程组： {x≡a1(modm1)
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<time datetime="2023-02-20 16:32" pubdate>
2023-02-20
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<a href="/tags/exCRT/">#exCRT</a>
<a href="/tags/CRT/">#CRT</a>
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裴蜀定理和exgcd和线性同余方程
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数论本身是优美的，但是很多人选择用繁琐又不严谨的证明遮盖它的优美。 描述 《算法竞赛进阶指南》上对裴蜀定理的描述是错误的——真正的裴蜀定理应该这么描述： 对于任意两个整数 a,ba,ba,b，不定方程 ax+by=cax+by=cax+by=c 有解 当且仅当 ccc 可以由 gcd(a,b)\gcd(a,b)gcd(a,b) 整除。 证明 我们设 A={ax+by∣x,y∈Z}\mathb
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<time datetime="2023-02-18 16:49" pubdate>
2023-02-18
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