二项式反演
[egin{cases}
F(n) =sumlimits_{i=0}^nmathbf C_n^iG(i)\
G(n) =sumlimits_{i=0}^n(-1)^{n-i}mathbf C_n^iF(i)
end{cases}
]
单位根反演
[[n|a]=frac{1}{n}sumlimits_{k=0}^{n-1}omega_n^{ak}
]
可以导出
[[aequiv bpmod n]=[a-bequiv0pmod n]=[n|(a-b)]=frac{1}{n}sumlimits_{i=1}^{n-1}omega_n^{k(a-b)}=frac{1}{n}sumlimits_{i=1}^{n-1}omega_n^{ak}omega_n^{-bk}
]
Min-Max反演
[max(S)=sumlimits_{Tsubseteq S}(-1)^{|T|-1}min(T)
]