数学：拓展Lucas定理

拓展Lucas定理解决大组合数取模并且模数为任意数的情况

大概的思路是把模数用唯一分解定理拆开之后然后去做

然后要解决的一个子问题是求模质数的k次方

将分母部分转化成逆元再去做就好了

 #include<bits/stdc++.h>
 using namespace std;
 const int maxn =  + ;
 typedef long long LL;
 
 LL Pow(LL n, LL m, LL mod) {
     LL ans = ;
     while(m > ) {
         if(m & ) ans = (LL)ans * n % mod;
         n = (LL)n * n % mod; m >>= ;
     }
     return ans;
 }
 LL Pow(LL n,LL m) {
     LL ans = ;
     while(m > ) {
         if(m & ) ans = ans * n;
         n = n * n; m >>= ;
     }
     return ans;
 }
 LL x, y;
 LL exgcd(LL a, LL b) {
     if(a == ) {
         x = , y = ;
         return b;
     }LL r = exgcd(b%a, a);
     LL t = x; x = y - (b/a)*x; y = t;
     return r;
 }
 LL rev(LL a, LL b) { exgcd(a, b); return ((x % b) + b) % b; }
 LL Calc(LL n, LL p, LL t) {
     if(n == ) return ;
 
     LL s = Pow(p, t), k = n / s, tmp = ;
     for(LL i=; i<=s; i++) if(i % p) tmp = (LL)tmp * i % s;
 
     LL ans = Pow(tmp, k, s);
     for(LL i=s*k + ; i<=n; i++) if(i % p) ans = (LL)ans * i % s;
 
     return (LL)ans * Calc(n / p, p, t) % s;
 }
 LL C(LL n, LL m, LL p, LL t) {
     LL s = Pow(p, t), q = ;
     for(LL i=n; i; i/=p) q += i / p;
     for(LL i=m; i; i/=p) q -= i / p;
     for(LL i=n-m; i; i/=p) q -= i / p;
 
     LL ans = Pow(p, q);
     LL a = Calc(n, p, t), b = Calc(m, p, t), c = Calc(n-m, p, t);
     return (LL)(ans * a * rev(b, s) * rev(c, s)) % s;
 }
 LL China(LL A[], LL M[], LL cnt) {
     LL ans = , m, n = ;
     for(LL i=; i<=cnt; i++) n *= M[i];
     for(LL i=; i<=cnt; i++) {
         m = n / M[i];
         exgcd(M[i], m);
         ans = (ans + (LL)y * m * A[i]) % n;
     }
     return (ans + n) % n;
 }
 LL A[maxn], M[maxn], cnt;
 LL Lucas(LL n, LL m, LL mod) {
     for(LL i=; i*i <= mod; i++) if(mod % i == ) {
         LL t = ;
         while(mod % i == ) t++, mod /= i;
         M[++cnt] = Pow(i, t);
         A[cnt] = C(n, m, i, t);
     }if(mod > ) {
         M[++cnt] = mod;
         A[cnt] = C(n, m, mod, );
     }
     return China(A, M, cnt);
 }
 LL n, k, p;
 int main() {
     cin >> n >> k >> p;
     cout << Lucas(n, k, p) << endl;
     return ;
 }

然后补充一个内容，线性时间复杂度内求出所有的逆元

A[i] = -(p / i) * A[p % i];