• 2024-06-19QOJ1285 Stirling Number
    QOJ传送门因为\(x^{\overlinep}\equivx^p-x\pmodp\),所以设\(n=pq+r\),其中\(r\in[0,p-1]\),则有:\[\begin{aligned}x^{\overlinen}&=(\prod\limits_{i=0}^{q-1}(x+ip)^{\overlinep})(x+pq)^{\overliner}\\&=(x^p