0. 二项式反演
0.1. 形式
\[F(n)=\sum_{i=0}^{n}(-1)^{i}{n \choose i}G(i) \iff G(n)=\sum_{i=0}^{n}(-1)^{i}{n \choose i}F(i)
\]
\[F(n)=\sum_{i=0}^{n}{n \choose i}G(i) \iff G(n)=\sum_{i=0}^{n}(-1)^{n-i}{n \choose i}F(i)
\]
\[F(x)=\sum_{i=x}^{n}(-1)^{i}{i \choose x}G(i) \iff G(x)=\sum_{i=x}^{n}(-1)^{i}{n \choose i}F(i)
\]
\[F(x)=\sum_{i=x}^{n}{i \choose x}G(i) \iff G(x)=\sum_{i=x}^{n}(-1)^{i-x}{i \choose x}F(i)
\]