20241114 NOIP训练赛 T3

简化题面:

有一个 $n+1$ 行 $k$ 列的 $01$ 矩阵，行标号 $0\sim n$，列标号 $1\sim k$，求满足一下条件的矩阵个数，对 ${10}^9+7$ 取模：

1. 对于 $0\sim n-1$ 行子矩阵中，没有一列全为 $1$。
2. 对于 $1\sim n$ 行子矩阵中，存在一列全为 $0$。

$3\le k,n\le {10}^{18}$。

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考虑中间 $n-1$ 行有多少列全是 $0$，多少列全是 $1$。

设有 $i$ 列全是 $0$，$j$ 列全是 $1$ 选择哪些列的方案是 $(\genfrac{}{}{0ex}{}{k}{i})(\genfrac{}{}{0ex}{}{k-i}{j})$

其他每列的填法是 $2^{n-1}-2$，所以方案数是 $(2^{n-1}-2{)}^{k-i-j}$

对于第一行，$j$ 列全是 $1$ 的只能填 $0$，其他格子随便填，方案数是 $2^{k-j}$

对于最后一行，$i$ 列全是 $0$ 的至少有一个填 $0$，其他格子随便填，方案数是 $2^{k-i}(2^i-1)$

综上答案为

$$\begin{array}{rl}& \sum _{i=0}^k\sum _{j=0}^{k-i}(\genfrac{}{}{0ex}{}{k}{i})(\genfrac{}{}{0ex}{}{k-i}{j})\cdot (2^{n-1}-2{)}^{k-i-j}\cdot 2^{k-j}\cdot 2^{k-i}(2^i-1)\\ & =\sum _{i=0}^k(\genfrac{}{}{0ex}{}{k}{i})2^{k-i}(2^i-1)\sum _{j=0}^{k-i}(\genfrac{}{}{0ex}{}{k-i}{j})(2^{n-1}-2{)}^{k-i-j}{2}^{k-j}\\ & =\sum _{i=0}^k(\genfrac{}{}{0ex}{}{k}{i})2^{k-i}{2}^i\sum _{j=0}^{k-i}(\genfrac{}{}{0ex}{}{k-i}{j})(2^{n-1}-2{)}^{k-i-j}{2}^{k-j}-\sum _{i=0}^k(\genfrac{}{}{0ex}{}{k}{i})2^{k-i}\sum _{j=0}^{k-i}(\genfrac{}{}{0ex}{}{k-i}{j})(2^{n-1}-2{)}^{k-i-j}{2}^{k-j}\\ & =\sum _{i=0}^k(\genfrac{}{}{0ex}{}{k}{i})2^k\sum _{j=0}^{k-i}(\genfrac{}{}{0ex}{}{k-i}{j})(2^{n-1}-2{)}^{k-i-j}{2}^{k-j}-\sum _{i=0}^k(\genfrac{}{}{0ex}{}{k}{i})2^k{2}^{-i}\sum _{j=0}^{k-i}(\genfrac{}{}{0ex}{}{k-i}{j})(2^{n-1}-2{)}^{k-i-j}{2}^{k-j}\\ & =2^k(\sum _{i=0}^k(\genfrac{}{}{0ex}{}{k}{i})\sum _{j=0}^{k-i}(\genfrac{}{}{0ex}{}{k-i}{j})(2^{n-1}-2{)}^{k-i-j}{2}^{k-j}-\sum _{i=0}^k(\genfrac{}{}{0ex}{}{k}{i})2^{-i}\sum _{j=0}^{k-i}(\genfrac{}{}{0ex}{}{k-i}{j})(2^{n-1}-2{)}^{k-i-j}{2}^{k-j})\\ & =2^k(\sum _{i=0}^k(\genfrac{}{}{0ex}{}{k}{i})2^i{2}^{-i}\sum _{j=0}^{k-i}(\genfrac{}{}{0ex}{}{k-i}{j})(2^{n-1}-2{)}^{k-i-j}{2}^{k-j}-\sum _{i=0}^k(\genfrac{}{}{0ex}{}{k}{i})2^{-i}\sum _{j=0}^{k-i}(\genfrac{}{}{0ex}{}{k-i}{j})(2^{n-1}-2{)}^{k-i-j}{2}^{k-j})\end{array}$$

单看 $2^{-i}\sum _{j=0}^{k-i}(\genfrac{}{}{0ex}{}{k-i}{j})(2^{n-1}-2{)}^{k-i-j}{2}^{k-j}$：

$$\begin{array}{rl}& 2^{-i}\sum _{j=0}^{k-i}(\genfrac{}{}{0ex}{}{k-i}{j})(2^{n-1}-2{)}^{k-i-j}{2}^{k-j}\\ & =\sum _{j=0}^{k-i}(\genfrac{}{}{0ex}{}{k-i}{j})(2^{n-1}-2{)}^{k-i-j}{2}^{k-i-j}\\ & =\sum _{j=0}^{k-i}(\genfrac{}{}{0ex}{}{k-i}{j})(2^n-4{)}^{k-i-j}{1}^{k-i-j}\\ & =((2^n-4)+1{)}^{k-i}\\ & =(2^n-3{)}^{k-i}\end{array}$$

倒数第二步是二项式定理。

原式为

$$\begin{array}{rl}& 2^k(\sum _{i=0}^k(\genfrac{}{}{0ex}{}{k}{i})2^i(2^n-3{)}^{k-i}-\sum _{i=0}^k(\genfrac{}{}{0ex}{}{k}{i})(2^n-3{)}^{k-i})\\ & =2^k(\sum _{i=0}^k(\genfrac{}{}{0ex}{}{k}{i})2^i(2^n-3{)}^{k-i}-\sum _{i=0}^k(\genfrac{}{}{0ex}{}{k}{i})(2^n-3{)}^{k-i}{1}^i)\\ & =2^k((2+(2^n-3){)}^k-(2^n-3+1{)}^k)\\ & =2^k((2^n-1{)}^k-(2^n-2{)}^k)\end{array}$$

答案即为

$$2^k((2^n-1{)}^k-(2^n-2{)}^k)$$

使用快速幂计算即可，时间复杂度 $O(\log )$。