线性求逆元

Posted 2021-01-25 033000-

const int maxn=3000000;
ll inv[maxn+5];
void marktable_inv(int p){
    inv[0]=inv[1]=1;
    for(int i=2;i<maxn;i++)
    inv[i]=(p-p/i)*inv[p%i]%p;
        
}