Be the memo ......
Binomial inversion:
So \ (f_i \) represents select at least \ (i \) a, \ (G_i \) represents a good choice knocked \ (i \) a
\(f_i=\sum_{j=0}^i \dbinom{n}{j}*g_j\)
\(g_i=\sum_{j=0}^{i}(-1)^{i-j}\dbinom {n}{j}f_j\)
Vandermonde convolution:
\(\dbinom{n}{k}=\sum_{i=0}^k\dbinom{m}{i}*\dbinom{k-i}{n-m}\)
Combination of several unnamed formula:
\(\dbinom{m}{i}*\dbinom{i}{j}=\dbinom{m}{i}*\dbinom{m-j}{i-j}\)
Chinese remainder theorem:
\(M_i=lcm/m_i, x_i=M_i^{-1}(mod\ m_i)\)
\(Ans=\sum_{i=1}^n x_i*a_i*M_i(mod\ lcm)\)
Stirling Number of the second kind:
\(S(n, m)=\dfrac{1}{m!}*\sum_{k=0}^m(-1)^k*\dbinom{m}{l}*(m-k)^n\)
\(n^k=\sum_{i=0}^kS(k, i) * i! *\dbinom{n}{i}\)
Kang expand open:
\(\sum_{i=1}^n(s_i-\sum_{j=1}^{i-1}[s_j<s_i])*(n-i+1)!\)
Lagrange Interpolation:
\(f(k)=\sum_{i=0}^ny_i* \prod_{i!=j}\dfrac{k-x_j}{x_i-x_j}\)