Solution Set - 数学

A - Perpetual Subtraction

题意：一个数有pi的概率为i，一次操作将数随机变为小于等于它的数，为m次操作后变为每个数的概率。给出的最大数$N(1\hspace{0.17em}\le \hspace{0.17em}N\hspace{0.17em}\le \hspace{0.17em}{10}^5)$，操作数$M(0\hspace{0.17em}\le \hspace{0.17em}M\hspace{0.17em}\le \hspace{0.17em}{10}^{18})$。

首先有$O(nm)$的dp，$f_{i,j}$为经过i此操作后为数j的概率，有转移$f_{i,j}=\sum _{k>=j}{f}_{i-1,k}\times \frac{1}{k+1}$。明显可以矩阵优化为$O(n^3\log m)$，有转移矩阵
答案为$A^{m}P$

注意这个上三角矩阵，特征值为对角线

对应特征值${\lambda }_i=\frac{1}{i+1}$的特征向量为$v_i=((-1{)}^{i+j}\left(\begin{array}{c}i\\ j\end{array}\right){)}_{j=0}^n$（纵向量）

证明1

设$W=A{v}_i$ $$\begin{array}{rl}{W}_k=& \sum _{j=0}^n{A}_{k,j}{v}_{i,j}\\ =& \sum _{j=k}^i(-1{)}^{i+j}\left(\begin{array}{c}i\\ j\end{array}\right)\frac{1}{j+1}\\ =& \sum _{j=k}^i(-1{)}^{i+j}\frac{i!}{(i-j)!j!(j+1)}\frac{i+1}{i+1}\\ =& \frac{1}{i+1}\sum _{j=k}^i(-1{)}^{i+j}\left(\begin{array}{c}i+1\\ j+1\end{array}\right)\\ =& \frac{1}{i+1}\sum _{j=k}^i(-1{)}^{i+j}(\left(\begin{array}{c}i\\ j\end{array}\right)+\left(\begin{array}{c}i\\ j+1\end{array}\right))\\ =& \frac{1}{i+1}(\sum _{j=k}^i(-1{)}^{i+j}\left(\begin{array}{c}i\\ j\end{array}\right)+\sum _{j=k}^i(-1{)}^{i+j}\left(\begin{array}{c}i\\ j+1\end{array}\right))\\ =& \frac{1}{i+1}(\sum _{j=k}^i(-1{)}^{i+j}\left(\begin{array}{c}i\\ j\end{array}\right)-\sum _{j=k+1}^{i+1}(-1{)}^{i+j}\left(\begin{array}{c}i\\ j\end{array}\right))\\ =& \frac{1}{i+1}(-1{)}^{i+k}\left(\begin{array}{c}i\\ k\end{array}\right)\\ =& {\lambda }_i{v}_{i,k}\end{array}$$

于是，我们有，每一纵列是一个特征向量

我们还需要$Q$的逆元，直接给出

证明2

设$W=Q{Q}^{-1}$ $$\begin{array}{rl}{W}_{i,j}=& \sum _{k=0}^n{Q}_{i,k}{Q}_{k,j}^{-1}\\ =& \sum _{k=0}^n(-1{)}^{i+k}\left(\begin{array}{c}k\\ i\end{array}\right)\left(\begin{array}{c}j\\ k\end{array}\right)\\ =& \sum _{k=i}^j(-1{)}^{i+k}\frac{j!}{(j-k)!(k-i)!i!}\\ =& \frac{j!}{i!(j-i)!}\sum _{k=i}^j(-1{)}^{i+k}\left(\begin{array}{c}j-i\\ k-i\end{array}\right)\\ =& \left(\begin{array}{c}j\\ i\end{array}\right)\sum _{k=0}^{j-i}(-1{)}^k\left(\begin{array}{c}j-i\\ k\end{array}\right)\\ =& \left(\begin{array}{c}j\\ i\end{array}\right)0^{j-i}\end{array}$$ $Q_{i,k},Q_{k,j}^{-1}$都非0，要求$i\le k\le j$，所以当$i>j$时，$W_{i,j}=0$； 当$i<j$时，$W_{i,j}=0$；当$i=j$时，$W_{i,j}=1$；

特征分解后我们有了$A^m=Q{\mathrm{\Lambda }}^m{Q}^{-1}$，其中${\mathrm{\Lambda }}_{i,i}^m=\frac{1}{(i+1{)}^m}$可以$O(n\log m)$算出。答案为$Q({\mathrm{\Lambda }}^m(Q^{-1}P))$，用结合律从后向前算，因为$P$为纵向量，矩阵乘法为$O(n^2)$。

继续优化，发现$Q$和$Q^{-1}$可卷积，使用FFT可以达到$O(n\log n)$的复杂度

证明3

设$W=Qx$ $$\begin{array}{rl}{W}_i=& \sum _{j=0}^n{Q}_{i,j}{x}_j\\ =& \sum _{j=i}^n(-1{)}^{i+j}\left(\begin{array}{c}j\\ i\end{array}\right)x_j\\ =& \frac{(-1{)}^i}{i!}\sum _{j=i}^n(-1{)}^{j}j!x_j\frac{1}{(j-i)!}\end{array}$$ 设$f(i)=(-1{)}^{i}i!x_i,g(i)=\frac{1}{i!}.W_i=\frac{(-1{)}^i}{i!}\sum _{j=i}^{n}f(j)g(j-i)$