细节备忘录
带模数的n次单位根求法
\(w_n^0=1\)
\(w_n^1=(模数的原根)^{(mod-1)/n}\)
\(w_n^m=w_n^{m-1} \cdot w_n^1\)
stirling数
https://www.cnblogs.com/owenyu/p/6724661.html
容斥原理和广义容斥原理
https://www.cnblogs.com/Parsnip/p/11530658.html
\(w_n^0=1\)
\(w_n^1=(模数的原根)^{(mod-1)/n}\)
\(w_n^m=w_n^{m-1} \cdot w_n^1\)
https://www.cnblogs.com/owenyu/p/6724661.html
https://www.cnblogs.com/Parsnip/p/11530658.html
