Day10
考試
T3
形式化題意，給定 \(n,m\)，求\(\sum^n_{i=1} \sum^m_{j=1} \displaystyle \begin{pmatrix}n\\i\\\end{pmatrix}\displaystyle \begin{pmatrix}i\\j\\\end{pmatrix}\)
推式子:
\[\sum^n_{i=1} \sum^m_{j=1}
\displaystyle \begin{pmatrix}
n\\
i\\
\end{pmatrix}
\displaystyle \begin{pmatrix}
i\\
j\\
\end{pmatrix}
\]
\[\Longrightarrow
\sum^n_{i=1} \sum^m_{j=1}
\displaystyle \begin{pmatrix}
n\\
j\\
\end{pmatrix}
\displaystyle \begin{pmatrix}
n-j\\
i-j\\
\end{pmatrix}
\]
\[\Longrightarrow
\sum^n_{j=1} \sum^m_{i=1}
\displaystyle\begin{pmatrix}
n\\
i\\
\end{pmatrix}
\displaystyle \begin{pmatrix}
n-i\\
j-i\\
\end{pmatrix}
\]
\[\Longrightarrow
\sum^m_{i=1}
\displaystyle\begin{pmatrix}
n\\
i\\
\end{pmatrix}
\sum^n_{j=1}
\displaystyle \begin{pmatrix}
n-i\\
j-i\\
\end{pmatrix}
\]
\[\Longrightarrow
\sum^m_{i=1}
\displaystyle\begin{pmatrix}
n\\
i\\
\end{pmatrix}
\displaystyle \begin{pmatrix}
\displaystyle \begin{pmatrix}
n-i\\
1-i\\
\end{pmatrix}
+
\displaystyle \begin{pmatrix}
n-i\\
2-i\\
\end{pmatrix}
+\dots+
\displaystyle \begin{pmatrix}
n-i\\
n-i\\
\end{pmatrix}
\end{pmatrix}
\]
\[\Longrightarrow
\sum^m_{i=1}
\displaystyle\begin{pmatrix}
n\\
i\\
\end{pmatrix}
2^{n-i}
\]
再根據\(C^{i+1}_n=\frac{n-i}{i+1} \times C^i_n\)遞推組合數即可
T4
同餘最短路板子題