q模拟入门 ARC139F Solution

ARC139F

等价于 $F_2^m$ 里选出 $n$ 个向量，求每种选择方案之和

枚举线性基大小 $k$​，设其主元是 $a_1\sim a_k$​，等价于让 $n$​ 个向量张成 $k$​ 维空间，

等价于数有多少个 $n$ 行 $k$ 列的满秩 $01$ 矩阵，转置矩阵后变成有多少个 $k$ 行 $n$ 列 $01$ 矩阵线性无关

也等价于 $n$​ 维空间选出 $k$​ 个向量彼此线性无关，方案数：

$$\prod _{i=0}^{k-1}(2^n-2^i)$$

线性基最大异或和期望：主元必选，非主元可选可不选

$$\sum 2^{a_i}+\sum _{i=0}^{a_k-1}[\mathrm{\forall }j,a_j\ne i]\frac{1}{2}{2}^i=\frac{1}{2}(2^{a_k+1}-1+\sum 2^{a_i})$$

再有主元之后有多少个线性基，那么考虑填入，从低到高依次填入，那么有填法：

$$\prod 2^{a_i-(i-1)}$$

那么答案就是：

$$\begin{array}{rl}& \sum _{k=0}^{min(n,m)}\sum _{a_1\sim a_k}\prod _{i=0}^{k-1}(2^n-2^i)\frac{1}{2}(\sum 2^{a_i}+2^{a_k+1}-1)\prod _{i=1}^k{2}^{a_i-(i-1)}\\ =& \sum _{k=0}^{min(n,m)}\sum _{a_1\sim a_k}\prod _{i=0}^{k-1}(2^{n-i}-1)\frac{1}{2}(\sum 2^{a_i}+2^{a_k+1}-1)\prod _{i=1}^k{2}^{a_i}\\ =& \sum _{k=0}^{min(n,m)}\prod _{i=0}^{k-1}(2^{n-i}-1)\sum _{a_1\sim a_k}(\sum 2^{a_i}+2^{a_k+1}-1)2^{-1+\sum a_i}\end{array}$$

问题就在于求出后面部分

对于 $\sum _{a_1\sim a_k}{2}^{\sum a_i}=[x^k]\prod _{i=0}^{m-1}(1+2^{i}x)=2^{k(k-1)/2}·{(\genfrac{}{}{0ex}{}{m}{k})}_2$

设 $F(x)=\prod _{i=0}^{m-1}(1+2^{i}x)$，则有：

$$F(2x)=\prod _{i=1}^m(1+2^{i}x)=\frac{F(x)·(1+2^{m}x)}{1+x}$$

也就是 $(1+x)F(2x)=F(x)(1+2^{m}x)⟹2^k{f}_k+2^{k+1}{f}_{k+1}=2^m{f}_k+f_{k+1}$

我们得到 $f_{k+1}=\frac{2^m-2^k}{2^{k+1}-1}{f}_k$，又因为 $f_0=1$，所以有

$$f_n=\prod _{k=0}^{n-1}\frac{2^m-2^k}{2^{k+1}-1}=\prod _{k=0}^{n-1}\frac{2^k(2^{m-k}-1)}{2^{k+1}-1}=2^{n(n-1)/2}\prod _{k=0}^{n-1}\frac{2^{m-k}-1}{2^{n-k}-1}=2^{n(n-1)/2}{(\genfrac{}{}{0ex}{}{m}{n})}_2$$

所以 $-1$ 可以拿出来了。

再考虑 $\sum _{a_1\sim a_k}{2}^{\sum a_i}·\sum 2^{a_i}$ 怎么弄，不妨设其为 $F(m,k)$

可以将其表达为 $\sum 2^m-1-\sum _{\mathrm{\forall }j,a_j\ne i}{2}^i$

而后面这一部分，事实上可以考虑按位统计贡献，也就是考虑选出 $k+1$ 个元素时候，保留原本的 $a_1\sim a_k$，加入一个新的 $j$，式子也就变成了 $\sum 2^{j+\sum a_i}$

也就是：

$$\begin{array}{rl}F(m,k)& =(2^m-1)2^{k(k-1)/2}{(\genfrac{}{}{0ex}{}{m}{k})}_2-\sum _{j=0}^{m-1}\sum _{a_1\sim a_k,j\ne a_i,\mathrm{\forall }i}{2}^{j+\sum a_i}\\ & =(2^m-1)2^{k(k-1)/2}{(\genfrac{}{}{0ex}{}{m}{k})}_2-\sum _{a_1\sim a_{k+1}}(\genfrac{}{}{0ex}{}{k+1}{k})2^{\sum _{i=1}^{k+1}{a}_i}\\ & =(2^m-1)2^{k(k-1)/2}{(\genfrac{}{}{0ex}{}{m}{k})}_2-(k+1)2^{k(k+1)/2}{(\genfrac{}{}{0ex}{}{m}{k+1})}_2\end{array}$$

然后考虑 $2^{a_k+1}{2}^{\sum a}$

可以考虑枚举 $a_k$，有：

$$\begin{array}{rl}& =\sum _{t=1}^m{2}^t\sum _{a_1\sim a_k,a_k=t-1}{2}^{\sum a_i}\\ & =\sum _{t=1}^m{2}^{2t-1}\sum _{a_1\sim a_{k-1},a_{k-1}<t-1}{2}^{\sum a_i}\\ & =\sum _{t=1}^m{2}^{2t-1}{2}^{(k-1)(k-2)/2}{(\genfrac{}{}{0ex}{}{t-1}{k-1})}_2\\ & =2^{(k-1)(k-2)/2+1}\sum _{t=0}^{m-1}{2}^{2t}{(\genfrac{}{}{0ex}{}{t}{k-1})}_2\\ & =2^{(k-1)(k-2)/2+1}\sum _{t=0}^{m-1}(2^{t+1}·2^{t-k+1}·2^{k-2}){(\genfrac{}{}{0ex}{}{t}{k-1})}_2\\ & =2^{k(k-1)/2}\sum _{t=0}^{m-1}(2^{t+1}-1)2^{t-k+1}{(\genfrac{}{}{0ex}{}{t}{k-1})}_2+2^{t-k+1}{(\genfrac{}{}{0ex}{}{t}{k-1})}_2\\ & =2^{k(k-1)/2}({(\genfrac{}{}{0ex}{}{m}{k})}_2+\sum _{t=0}^{m-1}(2^{t+1}-1)2^{t-k+1}{(\genfrac{}{}{0ex}{}{t}{k-1})}_2)\end{array}$$

又因为：

$${(\genfrac{}{}{0ex}{}{t}{k})}_q=\prod _{i=0}^{k-1}\frac{q^{t-i}-1}{q^{k-i}-1}⟹{(\genfrac{}{}{0ex}{}{t+1}{k+1})}_q={(\genfrac{}{}{0ex}{}{t}{k})}_q\frac{q^{t+1}-1}{q^{k+1}-1}$$

所以有：

$$\begin{array}{rl}& =2^{k(k-1)/2}({(\genfrac{}{}{0ex}{}{m}{k})}_2+\sum _{t=0}^{m-1}(2^{t+1}-1)2^{t-k+1}{(\genfrac{}{}{0ex}{}{t}{k-1})}_2)\\ & =2^{k(k-1)/2}({(\genfrac{}{}{0ex}{}{m}{k})}_2+(2^k-1)\sum _{t=0}^{m-1}{2}^{t-k+1}{(\genfrac{}{}{0ex}{}{t+1}{k})}_2)\\ & =2^{k(k-1)/2}({(\genfrac{}{}{0ex}{}{m}{k})}_2+(2^k-1)\sum _{t=1}^m{2}^{t-k}{(\genfrac{}{}{0ex}{}{t}{k})}_2)\\ & =2^{k(k-1)/2}({(\genfrac{}{}{0ex}{}{m}{k})}_2+(2^k-1){(\genfrac{}{}{0ex}{}{m+1}{k+1})}_2)\end{array}$$

全部回代，答案是：

$$\begin{array}{rl}& =\sum _{k=0}^{min(n,m)}\prod _{i=0}^{k-1}(2^{n-i}-1)\sum _{a_1\sim a_k}(\sum 2^{a_i}+2^{a_k+1}-1)2^{-1+\sum a_i}\\ & =\sum _{k=0}^{min(n,m)}\frac{1}{2}\prod _{i=0}^{k-1}(2^{n-i}-1)\times \\ & ((2^m-1)2^{k(k-1)/2}{(\genfrac{}{}{0ex}{}{m}{k})}_2-(k+1)2^{k(k+1)/2}{(\genfrac{}{}{0ex}{}{m}{k+1})}_2-2^{k(k-1)/2}·{(\genfrac{}{}{0ex}{}{m}{k})}_2+2^{k(k-1)/2}({(\genfrac{}{}{0ex}{}{m}{k})}_2+(2^k-1){(\genfrac{}{}{0ex}{}{m+1}{k+1})}_2))\\ & =\frac{1}{2}\sum _{k=0}^{min(n,m)}\prod _{i=0}^{k-1}(2^{n-i}-1)2^{k(k-1)/2}({(\genfrac{}{}{0ex}{}{m}{k})}_2(2^{m+1}-2^{m-k}-1)-{(\genfrac{}{}{0ex}{}{m}{k+1})}_2(k{2}^k+1))\end{array}$$