GOOD BYE OI

**

大米饼正式退役了，OI给我带来很多东西

我会的数学知识基本都在下面了

博客园的评论区问题如果我看到了应该是会尽力回答的．．．

这也是我作为一个OIer最后一次讲课的讲稿

20190731

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多项式乘法

FFT

基本概念

1.多项式的两种表达(拉格朗日插值法)

多项式：\(A(x) = \sum_{i=0}^{n-1}a_ix^i\)，最高项次数为\(n-1\)，次数界为\(n\)

\((a_0,\cdots,a_{n-1})\)为多项式的系数表达, \((x_0,y_0),\cdots,(x_{n-1},y_{n-1})\) 为多项式的点值表达

2.\(n\)次单位根(泰勒展开(对多项式也适用)，欧拉公式，单位根，引理)

泰勒展开：

函数f(x)在\(x=x_0\)除的展开为：

$f(x) = f^0(x_0) + f^1(x_0)(x-x_0) + f^{2(x_0)\frac{(x-x_0)}2}{2!} + \cdots + f^i (x_0)\frac{(x-x_0)^i}{i!} +\cdots $

其中\(f^i(x)\)表示\(f(x)\)的\(i\)次导数(所以要运用首先你得记住初等函数的求导公式)：

\[练习：\\
e^x \ = \ \sum_{i}\frac{x^i}{i!}\\
ln(1-x) \ = \ - \sum_{i\ge 1} \frac{x}{i} \\
cos(x) \ = \ \sum_{i} \frac{(-1)^ix^{2i}}{(2i)!} \\
sin(x) \ = \ \sum_{i} \frac{(-1)^ix^{2i+1}}{(2i+1)!} \\
\frac{1}{(1-x)^k} \ = \ \sum_{i}(^{k+i-1}_i)x^i \\
\]

可以证明一个表达式：

\[e^{\pm ix} = cos(x) \pm isin(x)
\]

这指出\(e^{ix}\)对应的复平面上的点是 \((cos(x),sin(x))\) ，另外得到欧拉公式：

\[e^{\pi i } + 1 = 0
\]

有：

\[e^{2\pi i} = 1
\]

定义n次单位根\(x^n = 1\) 为满足的\(n\)个复数

一个可行的方式是$\omega_n \ = \ e^{2 \pi i /n} $ ，在复数平面上，形成了一个乘法的循环，即\(\omega_n^k = \omega_n^{k \ mod \ n}\)

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" alt="">

性质：

1.消去引理：$ \omega_{dn}^{dk} = \omega_n^k ，d \neq 0$

​ 2.折半引理： n次单位跟平方的集合 = n/2次单位根平方的集合

​ 3.求和引理(等比数列求和即可证明) ：

\[\frac{1}{n}\sum_{i}^{n-1}\omega^{ij}_n \ =
\begin{cases}
1 & j \ mod \ n = 0\\
0 & j \ mod \ n !=0\\
\end{cases}
\]

DFT

1.概念：

DFT,IDFT,矩阵(定义，转置，乘法，初等变换(换法，倍法，消法)，逆矩阵(存在性))，逆矩阵(求和引理)

​ 定义 $A(x) = \sum_{i}^{n-1} a_ix^i $ 的DFT为点值表达：\(((w_n^0,y_0),\cdots,(w_n^{n-1},y_{n-1} ))\)

​ 反过来可以对一个点值表达定义求表达，称为IDFT

​ 用矩阵表达:$Y = VA \Rightarrow A = V^{-1}Y $

​ $A_i = a_i, Y_i = y_i , V_{i,j} \ = \ w_n^{ij} $

根据求和引理：$V^{-1}{ij} =\frac{w{n}^{-ij} }{n} $

完整写一下DFT和IDFT就是

\[y_j \ = \ \sum_{i=0}^{n-1} a_i w_n^{ij} \\
y_j \ = \ \frac{1}{n} \sum_{i=0}^{n-1} a_i w_n^{-ij} \\
\]

2.求法：(\(A_0,A_1\)，折半引理，蝴蝶操作)

两个式子本质上是一样的，考虑求第一个：

对于A(x)，构造

\[A_0 (x) = a_0 + a_2 x^2 + \cdots + a_{n-2}x^{n/2-1} \\
A_1 (x) = a_1 + a_3 x^2 + \cdots + a_{n-1}x^{n/2-1} \\
\Rightarrow
A(x) = A_0(x^2) + xA_1(x^2) \\
根据折半引理，问题规模大小就被我们减到了一半 \\
具体实现，只需要翻转系数的二进制位就可以了 \\
\]

//len为多项式次数界，2^L = len  , C 为自定义复数
for(int i = 0;i < len; i++) R[i] = R[i>>1]>>1 | (i&1) << (L-1) ;
void FFT(C *a,int f) {//f = 1/-1 ,正/逆变换
     for(int i = 0; i < len; i++) if(i < R[i]) swap(a[i],a[R[i]]);
     for(int i = 1; i < len; i<<=1) {
         C wn(cos(pi/i),f * sin(pi/i));
         for(int j = 0; j < len; j+=(i<<1)) {
             C w(1,0);
             for(int k = 0; k < i; k++,w*=wn) {
                 C x = a[j + k],y = w * a[j + i + k];
                 a[j + k] = x + y,a[j + i + k] = x - y;
             }
         }
     }
     if(f==-1) for(int i = 0; i < len; i++) a[i] /= len;
 }

NTT

使用于运算在mod P 的意义下进行

如果n | P -1 , 令 G 为 P 的原根，

MTT

处理不是ntt模数的情况

拆系数ntt:

按照二进制拆成前15位和后15位，分别相乘

构造 \(P(x) = A(x) + iB(x) , Q(x) = A(x) - iB(x)\)

简单推导结论：\(dft \ Q_k \ = \ dft \ P_{n-k}\)

另外还有三模数ntt(不会。。。)

例题

1.通串的匹配问题

2.loj6434 神仙的游戏

3.bzoj3992 序列统计 (原根)

分治FFT　

1.普通形式模板

2.f = f 卷 f 的形式，直接做会有问题

只会在l = 左边界的时候有点问题，不妨直接不确定的那部分系数去掉

当 l 是不是边界的时候将贡献/2，在分治的最底层乘回来

3.两道可以感受分治ntt的好题：

cd 848E Days of Floral Colours　(1)

uoj401 青蕈领主 (2)

**FWT FMT FST **

1.问题，基本思想，构造的东西

可以参考

AGC034F RNG and XOR

cf1119H Triple

uoj348 周区划分

循环卷积与n进制异或fwt

(考虑FFT系数超出设置的len会怎么样？)　

JZOJ6239　

同样我们根据最高位分类：\(A_0,A_1,\cdots,A_{n-1}\)

考虑只有一维的情况，需要找到一个变换

使得变换->点乘->逆变换之后得到的是mod n的加法

可以联想到循环卷积，直接令变换的矩阵为V即可

和fwt同理扩展到很多维就可以了

生成函数和多项式

1.定义

​ 策爷的课件

​ OGF EGF ，加法乘法的意义

2.例子：

二进制数,背包的计数理解OGF的拼接

置换的计数,无向图计数理解序列EGF的拼接

3.多项式相关：

0.预备知识：高中数学，部分高等数学

​ 多项式求导，多项式积分

​ 1.多项式求逆，多项式开根

​ bzoj3625 小朋友和二叉树

​ 分治fft很多时候都多项式求逆做

​ 2.多项式ln

​ luogu U79666 树上深度博弈

​ 3.牛顿迭代，多项式exp ，多项式求幂

​ luogu 4389 付公主的背包

​ uoj50 链式反应

​ 4.多项式除法，取模

​ 常系数线性齐次递推 (*)

​ (行列式定义，的五点性质：转置，换法(推)，倍法(推)，分拆(推出展开)，消法)

​ luogu 4723 线性递推模板

​ 5.多点求值和快速插值( *)

​ uoj182 a^-1 + b problem

​ 6.拉格朗日反演 (*)

\[f(g(x)) = g(f(x)) = x ，即：f是g的复合逆\\
[x^n]f(x) = \frac{1}{n}[x^{n-1}](\frac{x}{f(x)})^n \\
推广形式：
[x^n]h(g(x)) = \frac{1}{n}[x^{n-1}]h'(x)(\frac{x}{f(x)})^n \\
\]

​ bzoj3684 大朋友和二叉树

这里有一篇十分通俗的微分doc :

一阶微分方程解法

数论

基本定理

完全剩余系性质

威尔逊定理：

\[p是质数 \Leftrightarrow (p - 1)! \equiv -1 (mod \ p)\\
\]

费马小定理：

\[p 是质数,a不整除p \Rightarrow a^{p-1} \equiv 1 (mod p)
\]

欧拉定理：(积性函数，线筛)

\[phi(p)定义为1 \to p-1和p 互质的数的个数　\\
如果p = \prod_{i=1}^{m} a_i^{b_i} , \phi(p) = p \prod_{i=1}^{m} (1-\frac{1}{a_i}) \\
欧拉定理描述为：\\
(a,p) = 1 \Rightarrow a^{\phi(p) } \equiv 1 (mod \ p ) \\
\]

勒让得定理：

\[n!的唯一分解中质因p的个数为：\\
\sum_{k\ge1}\lfloor \frac{n}{p^k} \rfloor \\
\]

中国剩余定理(CRT)：

\[\begin{cases}
x \equiv a_1 (mod \ m_1) \\
x \equiv a_2 (mod \ m_2) \\
\vdots \\
x \equiv a_n (mod \ m_n) \\
\end{cases} \\
当m_1,m_2,\cdots,m_n两两互质一定有解为：\\
x \equiv \sum_{i=1}^{m}a_iM_iM_i^{-1} (mod \ M) \\
其中M=\prod_{i=1}^{n}m_i ，M_i = \prod_{j=1,j \neq i}^{n}m_j\\
M_i^{-1}M_i \equiv 1 (mod \ m_i) \\
\]

扩展CRT：\(m_1,m_2,\cdots,m_n\)不满足两两互质；

(百度的第一篇博客推法思路大致是的，但是里面有很多多余的步骤?)

\[只需要考虑如何解：
\begin{cases}
x \equiv a_1 (mod \ m_1) \\
x \equiv a_2 (mod \ m_2) \\
\end{cases}　\\
x = k m_1 + a_1 \\
\Rightarrow km_1 = a_2 - a_1 (mod \ m_2) \\
有解的充要条件是:(m_1,m_2) | a_2 - a_1 \\
\Rightarrow k\frac{m_1}{(m_1,m_2)} = \frac{a_2 - a_1}{(m_1,m_2)} (mod \ \frac{m_2}{(m_1,m_2)})　\\
\Rightarrow k = \frac{a_2 - a_1}{(m_1,m_2)}(\frac{m_1}{(m_1,m_2)})^{-1} (mod \ \frac{m_2}{(m_1,m_2)})
\\
我们右边 = t ,则有k = k'\frac{m_2}{(m_1,m_2)} + t \\
x = k m_1 + a_1 ＝　k'\frac{m_1m_2}{(m_1,m_2)} + tm_1 + a_1 \\
即: x \equiv (inv({m_1\over (m_1,m_2)},{m_2\over (m_1,m_2)})*{(a_2-a_1)\over (m_1,m_2)})\%{m_2\over (m_1,m_2)}*m_1+a_1 (mod \ \frac{m_1m_2}{(m_1,m_2)}) \\
\]

素数判定和大整数分解

Miller-Rabin

1.理论基础：

​ 1)费马小定理的逆定理

​ 有一定概率不成立，我们称这样的p为伪素数

​ 2)二次探测定理：

​ $p是素数，x^2 \equiv 1 (mod \ p ) \Rightarrow x \equiv -1 \ or \ 1 (mod \ p) $

2.算法流程 miller_rabin(long long n)​：

​ 1)随机一个数字$a < n $, 设 \(u\) 是 \(n - 1\) 除去\(2\)的次幂剩下的数

​ 2)对 \(a^{u}\) 和 \(a^{2u}\) 做二次探测并不断令\(u = 2u\) 直到$ u = n -1$

​ 3)对\(a^u\) 即 \(a^{n-1}\)检验费马小定理

​ 可以多做几次．

Pollard-Rho

1.理论基础：

1)生日悖论

​ 2)floyd判环法

​ 相遇后，令一个指针不动，另一个回到起点二者同速运动可巧妙找到环的起点

2.算法流程 pollard-rho(long long n) :

1)miller-rabin(n)检测成功则退出，否则进入2)尝试找到一个非平凡因子

​ 2)用$ x_{i} = (x_{i-1}^2 + c) \ mod \ (n-1) + 1 $ 生成伪随机数

​ 判断\(gcd(x_i - y ,n )\)是否是非平凡因子，是则退出

​ 在\(i\)是２的幂处记录\(y = x_i\)，在\(2i\)处更新\(y\)，其间如果有\(x_{(i+1) \to 2i} = y\)则说明出现了环

​ 这样做据说是\(O(n^{\frac{1}{4}})\)的 ，我不会证

另：O(1)快速乘法

inline long long multi(long long x,long long y,long long mod)
{
    long long tmp=(x*y-(long long)((long double)x/mod*y+eps)*mod);
    return tmp<0 ? tmp+mod : tmp;
}
eps 是用来设置精度的
虽然快但是比较容易炸精度，慎用．．．

二次剩余

1.判定：

​ 二次剩余的个数为\(\frac{p-1}{2}\)

​ 勒让徳符号 \((\frac{a}{p}) = a^{\frac{p-1}{2}}\) (a<p)

​ 欧拉判别法：

\[a 是 0 的充分必要条件是(\frac{a}{p}) = 0\\
a 是 p 的二次剩余的充分必要条件是(\frac{a}{p}) \equiv 1 (mod \ p) \\
a 是 p 的二次非剩余的充分必要条件是(\frac{a}{p}) \equiv -1 (mod \ p) \\
\]

(证明：先用费马小证=１的必要以及二分性，再配对反证充分性)

2.(*)求法：cipolla算法

​ 解：$x^2 \equiv n (mod \ p) $

​ 考虑随机一个\(p\)的二次非剩余$a^2 - n $

​ 定义一个新的数域(类似复数域)令\(w = \sqrt{a^2 - n}\)

​ 则\(x = \pm (a+w)^{\frac{p+1}{2}}\)

\[\begin{align}
&(a+w)^{p+1}\\
&= (a+w)^p (a+w) \\
&= (a^p+w^p)(a+w) \\
&= (a-w)(a+w) \\
&= a^2 - w^2 \\
&= n \\
\end{align}
\]

最后结果里面的x虚部一定是0，据说是因为由拉格朗日定理\(f(x) \equiv 0\)在mod p 的数域下最多只有f(x)的最高次数那么多个根，我也不知道为什么．．．．．

组合数

1.部分组合恒等式：

\[\begin{align}
&定义可得:\\
&1) \ (^n_m) = (^{n-1}_{m-1}) + (^{n-1}_{m})\\
&2) \ (^{n+1}_m) = \frac{n+1}{n-m+1}(^{n}_{m}) \\
&3) \ (^n_{m+1}) = \frac{n-m}{m+1}(^{n}_{m}) \\
&4) \ (^{n+1}_{m+1}) = \frac{n+1}{m+1}(^n_m) \\
&下面要讲的二项式定理可以推：\sum_{i=0}^n(^n_i) = 2^n \\
&\sum_{i=0}^{min(a,b)}(^a_i)(^b_i) = (^{a+b}_b) \Rightarrow \sum_{i=0}^n(^n_i)^2 = C_{2n}^n\\
&组合数的一个竖列和斜列都是可以化简的：\\
&1)\sum_{i=0}^{k}(^{n+i}_{m+i}) = (^{n+k+1}_{m+k}) - (^n_{m-1}) \\
&2)\sum_{i=0}^{k}(^{n+i}_{m}) = (^{n+k+1}_{m+1})-(^n_{m+1}) \\
\end{align}
\]

２.二项式定理

\[(a＋b)^n = \sum_{i=0}^{n} (^n_i)a^ib^{n-i} \\
\]

这个可以归纳证.

３.lucas定理

\[p是素数，设n = sp+q,m = tp+r (q,r<p)\\
则：\\
(^n_m) \equiv (^s_t)(^q_r) \\
\]

这个可以在p进制下扩展

可以构造\((1+x)^n\) 利用二项式定理证明

４.扩展lucas :

bzoj4830 抛硬币

bsgs

\[\begin{align}
&求：A^x \equiv B (mod \ C)的最小非负整数解，满足(A,C)=1\\
&(A,C)=1 \Rightarrow A^x\equiv A^{x \ mod \ \phi(c)} (mod \ C)\\
&那么令m = \lceil \sqrt C \rceil ，A^{im-j} \equiv B(mod \ C)\\
&(A^{m})^i \equiv A^jB(mod \ C)\\
&用hash表存下右边,枚举左边即可求解 \\
\end{align}
\]

时间复杂度：\(O(\sqrt C 或 \sqrt C log \sqrt C)\)

扩展bsgs

\[问题一样，只是没有(A,C) = 1 ，这是就没有欧拉定理了 \\
首先判掉B = 1,0的情况 ,考虑不断除去a 和模数的gcd：\\
设D = \prod_{i=1}^{k}d_i,其中d_i表示第i次求得a和模数的gcd\\
A^{x-k} \frac{A^k}{D} \equiv \frac{B}{D} (mod \ \frac{C}{D})\\
枚举x = 1 \to k ，x > k 的移项之后可以直接bsgs
\]

**线性代数 **

1.矩阵

​ 高斯-亚当消元求逆矩阵(满秩)

​ 线性相关/无关，张成，基

​ 线性基

​ 插入，合并，询问最大/最小值，查询排名，查询第k小

​ bzoj 2844 albus就是要第一个出场

​ bzoj3569 DZY Loves Chinese II

​ 你们哪天看懂了证明记得告诉我

2.行列式

\[det(A) = \sum_{P} (-1)^{\tau(P)} \prod_{i=1}^{n} A_{i,p_i}\\
\]

​ 性质：

​ 1.初等变换：1 2 3

​ 2.\(det(A^\tau) = det(A)\)

​ 3.\(det(A) = det(A)det(B)\)

​ 证明：拉普拉斯定理

​ Matrix - tree 定理 (K=D-A)

​ 2016级的*ZJ学长

​ BEST定理

群论

1.群的判定：

​ (a)封闭性：任意\(a,b\in G\) ，存在\(c \in G,a \cdot b=c\)

​ (b)结合律：任意$a,b,c\in G \(，满足\)(a \cdot b)\cdot c = a \cdot (b \cdot c) $

​ (c)单位元：存在\(e \in G\)，对任意\(a \in G\)，满足\(a \cdot e = e \cdot a = a\)

​ (d)逆元：对任意\(a \in G\)，存在\(b \in G\)，\(a \cdot b = b \cdot a = e\)，记做\(b = a^{-1}\)

​ (置换，置换的乘法，置换群)

2.burnside 引理

\[\begin{align}
&L = \frac{1}{|G|}\sum_{j=1}^{s}D(a_j)\\
&其中L表示本质不同的元素个数，G表示置换群为\{a_1,\cdots,a_s\},\\
&D(a_j)表示置换a_j下的不动点个数\\
&证明：\\
&引理1：定义稳定核G(c)为集合\{f|f \cdot c = c ,c \in G\} ，它是一个群\\
&引理2: 和c本质相同的方案数：\frac{|G|}{G(c)}\\
&证明：\\
& \ \ 若f,g,c \in G \ 则 \ f \cdot c = g \cdot c \Leftrightarrow f^{-1}g \in G(c) \\
& \ \ 那么对于一个f ，和它作用相同的置换g的集合是\{f \cdot h | h \in G(c)\} \\
& \ \ 那么和c 本质相同的方案数为\frac{|G|}{|G(c)|}\\
&立即推出：|G(c)| = \frac{|G|}{|等价类(c)|}\\
&对f\cdot c=c的(f,c)计数\\
&1) \sum_{j=1}^{s}D(a_j)\\
&2) \sum_{i=1}^{n}|G(c)| = \sum_{i=1}^{n}\frac{|G|}{|等价类|} = L|G|\\
&最后一步是一起考虑一个等价类的贡献，1)和2)相等即推出burnside引理！
\end{align}
\]

3.polya 定理

​ 每个置换都可以写成若干个互不相交的循环的乘积

​ 设G是一个n阶置换的群，用m种颜色去涂：

\[L\ = \ \frac{1}{G}(m^{c(g_1)} + m^{c(g2)}+\cdots+m^{c(g_n)}) \\
\]

其中\(g_i \in G\)，\(c(g_i)\)为\(g_i\)中循环的个数

luogu 4916 魔力环

prufer 序列

有标号无根树

生成，性质，还原

bzoj1005 明明的烦恼

概率论

我理论部分基本空白，只知道一些结论：

\(f(x)\)在\([L,R]\)的期望

\[\begin{align}
如果x是一个[L,R]均匀分布的随机变量\\
E(f(x)) \ = \ \frac{\int_{L}^{R} f(x) \ dx}{R-L} \\
\end{align}
\]

贝叶斯公式

\[P(A|B) = \frac{P(B|A)P(A)}{P(B)} \\
\]

uoj299 游戏

博弈论

．．．．

重点理解sg函数

容斥和反演

基本形式：

\[\begin{align}
ans = \sum_if(i)a_i \\
\end{align}
\]

组合数容斥

\[\begin{align}
&f(i) = (-1)^i\\
&\sum_{i=0}^{n}(-1)^i(^n_i) = [n=0] \\
&因为：(或者直接考虑(1-1)^n)\\
&\begin{cases}
当n = 0,显然= 1\\
当n \gt 0:原式= \sum_{i=0}^{n-1}(-1)^{i+1}(^{n-1}_i) + \sum_{i=0}^{n-1}(-1)^i(^{n-1}_i) = 0\\
\end{cases}
&\end{align}
\]

loj2351 毒蛇越狱

**二项式反演 **

\[f(n) = \sum_{i=0}^{n}(^n_i)g(i) \Leftrightarrow g(n) = \sum_{i=0}^{n}(-1)^{n-i}(^n_i)f(i)\\
另一种更好看的形式：\\
f(n) = \sum_{i=0}^{n}(-1)^{i}(^n_i)g(i) \Leftrightarrow g(n) = \sum_{i=0}^{n}(-1)^{i}(^n_i)f(i)\\
另一种不太好看的形式：\\
f(i) = \sum_{j\ge i}(^j_i)g(j) \Leftrightarrow g(i) = \sum_{j \ge i}(-1)^{j-i}(^j_i) f(j) \\
\]

经典的错排问题

loj3120 珍珠

斯特林容斥

\[\begin{align}
&f(i) = (-1)^{i-1}(i-1)! \\
&\sum_{i=1}^{n}\{^n_i\}(-1)^{i-1}(i-1)! = [n=1] \\
&因为：\\
&\begin{cases}
当n = 1，显然 = 1\\
当n \gt 1,原式 = \sum_{i=1}^{n-1}(-1)^{i-1}i!\{^{n-1}_{i}\}+\sum_{i=1}^{n-1}(-1)^ii!\{^{n-1}_{i}\} = 0\\
\end{cases}
\end{align}
\]

斯特林数(定义，递推式，求法 : I 倍增FFT,||卷积，用于转换幂，反转公式)

bzoj 异或图

斯特林数

1.定义：

\[1.两类数的递推式：\\
[^n_i] = [^{n-1}_{i-1}]+(n-1)[^{n-1}_i]\\
\{^n_i\} = \{^{n-1}_{i-1}\} + i\{^{n-1}_i\}\\
根据定义 \sum_{i=0}^{n}[^n_i] = n!\\
\]

2.斯特林数一行的求法：

第一类：(倍增fft)

\[[_m^n] = [x^m] x^{\overline n}\\
倍增FFT即可
\]

\(O(m \ log \ m )\)

第二类：(存在卷积形式)

\[根据组合意义可以知道：\\
n^m = \sum_{i=0}^{m}(^m_i)\{^n_i\}i!\\
对上面的式子二项式反演或者直接考虑容斥可以得到：\\
\{^n_m\} = \frac{1}{m!}\sum_{i=0}^{m}(-1)^{m-i}(^m_i)i^{n}\\
立即推出卷积形式：\\
\{^n_m\} = \sum_{i=0}^{m}\frac{(-1)^{m-i}}{(m-i)!}\frac{i^{n}}{i!}\\
\]

\(O(m \ log \ m )\)

bzoj2159 crash的文明世界

**斯特林反演 **

\[f(n) = \sum_{i=1}^{n} \{^n_i\} g(i) \Longleftrightarrow g(n) = \sum_{i=1}^{n} (-1)^{n-i} [^n_i]f(i)\\
f(n) = \sum_{i=1}^{n} [^n_i] g(i) \Longleftrightarrow g(n) = \sum_{i=1}^{n} (-1)^{n-i} \{^n_i\}f(i)\\
证明可以应用反转公式:\\
\displaystyle \sum_{k=m}^n (-1)^{n-k}\begin{bmatrix}n\\k\end{bmatrix} \begin{Bmatrix}k\\m\end{Bmatrix}=[m=n]\\
\sum_{k=m}^n (-1)^{n-k}\begin{Bmatrix}n\\k\end{Bmatrix} \begin{bmatrix}k\\m\end{bmatrix}=[m=n]\\
反转公式的证明可以应用斯特林数和幂的关系:\\
x^n = \sum_{i=0}^{n}\{^n_i\}x^{\underline i}\\
x^{\overline n} = \sum_{i=0}^{n}[^n_i]x^i\\
这个关系归纳可证，再带入幂的转化关系可证反转\\
\]

我只做过裸题

**容斥系数 **

\[本质是我们只需要满足\\
\sum_{i}^{n}f(i)g(i) = \delta_n \\
g，\delta是确定的，所以我们可以直接O(n^2)解出f\\
或者可以直接反演求出f,某些情况或者可以用生成函数推出f\\
或者：直接打表鸭！\\
\]

玲珑杯线上赛河南专场　Round#17 B

min-max容斥

\[\max(S)=\sum_{T\subseteq S} (-1)^{|T|-1}\min(T)\\
E(\max S)=\sum_{T\subseteq S} (-1)^{|T|-1}E(\min T)\\
\text{lcm}(S)=\prod_{T\subseteq S} (-1)^{|T|-1}\gcd(T)\\
min和max当然是对称的\\
\]

loj2542 随机游走

luogu4704 重返现世

kth min-max容斥

\[推容斥系数的技巧可以用了：\\
[i=k] = \sum_{j=1}^{i}(^{i-1}_{j-1})f(j) \\
二项式反演得：\\
f(i) = \sum_{j=1}^{i}(-1)^{i-j}(^{i-1}_{j-1})[j==k] \\
即f(i) = (-1)^{i-k}(^{i-1}_{k-1}) \\
kthmax(S) = \sum_{T \subset S} (-1)^{|T|-k}(^{|T|-1}_{k-1})min(T)
\]

子集反演

\[f(S) = \sum_{T \subset S}g(T) \Leftrightarrow
g(S) = \sum_{T \subset S}(-1)^{|S|-|T|}f(T)\\
f(S) = \sum_{S \subset T}g(T) \Leftrightarrow
g(S) = \sum_{S \subset T}(-1)^{|T|-|S|}f(T)\\
\]

loj 2983 数树

**单位根反演 **

\[\frac{1}{n}\sum_{j=0}^{n-1}\omega_n^{j(i-t)} = [i \% n=t]\\
\]

loj6485 LJJ学二项式定理

loj3058 白兔之舞

**莫比乌斯反演 **

Po姐的ppt

莫比乌斯函数：

1.定义

\[\mu(n) =
\begin{cases}
1 &n=1\\
(-1)^r & n=p_1\cdots p_r,p_1\cdots p_r 是两两不同的质数\\
0 &其它，即n有大于1的平方因子\\
\end{cases}\\
\]

2.性质：

\[\begin{align}
&1)积性\\
&2)\sum_{d|n} \mu(d) = [n=1] \\
&3)\sum_{d|n}\frac{\mu(d)}{d}=\frac{\phi(n)}{n}\\
&(考虑\sum_{i=1}^{n}[gcd(i,n)=d] = \phi(\frac n d) 即可证3)\\
\end{align}
\]

反演的两种形式：

\[f(n) = \sum_{d|n} g(d) \Leftrightarrow g(n) = \sum_{d|n}\mu(\frac{n}{d})f(d)\\
f(n) = \sum_{n|d} g(d) \Leftrightarrow g(n) = \sum_{n|d}\mu(\frac{d}{n})f(d)\\
\]

bzoj2820 YY的GCD

bzoj3434 时空穿梭

loj 2085 循环之美

OIwiki !

(大家可能只有自行学习杜教筛，洲阁筛，和min25筛了)

杜教筛的复杂度，我当时学的时候一直没有搞清楚有一点，后来看rqy的博客看到了：

\[O(\sum_{i=1}^{\sqrt n} \sqrt \frac{n}{i}) \approx O(2\sqrt{n}\int_0^{\sqrt n}i^{\frac{1}{2}}) = O(n^{\frac{3}{4}})\\
\]

(由于没时间讲计算几何了，大家记得学一下闵可夫斯基和)