具体解析见：https://blog.csdn.net/qq_17807067/article/details/127231568   #include <bits/stdc++.h> using namespace std; using ll = long long; int n,p; ll inv[20000528]; int main(){ scanf("%d %d",&n,&p); inv[1]=1;//1的逆元是1 printf("%ld\n",inv[1]); ...