【学习笔记】二项式反演
公式一
f ( n ) = ∑ i = m n ( n i ) g ( i ) ⇒ g ( n ) = ∑ i = m n ( − 1 ) n − i ( n i ) f ( i ) \begin{aligned} f (n) &= \sum_{i = m}^{n} \binom{n}{i} g(i) \\ \Rightarrow g(n) &= \sum_{i = m}^{n} (-1)^{n - i} \binom{n}{i}f(i) \end{aligned} f(n)⇒g(n)=i=m∑n(in)g(i)=i=m∑n(−1)n−i(in)f(i)
公式二
f ( n ) = ∑ i = n m ( i n ) g ( i ) ⇒ g ( n ) = ∑ i = n m ( − 1 ) i − n ( i n ) f ( i ) \begin{aligned} f (n) &= \sum_{i = n}^{m} \binom{i}{n} g(i) \\ \Rightarrow g(n) &= \sum_{i = n}^{m} (-1)^{i - n} \binom{i}{n}f(i) \end{aligned} f(n)⇒g(n)=i=n∑m(ni)g(i)=i=n∑m(−1)i−n(ni)f(i)
