51nod1538： 一道难题（特征多项式+多项式取模/求逆）

传送门

题解：
观察下这个等式，其实就是 $f_n = \sum_{i}f_{n-a_i}$。

前面的求逆预处理一下，后面的特征多项式倍增和取模就行了，时间复杂度 $O(k \log k \log m)$。

注意 $a_1$可能大于23333，RE了半天。。

#include <bits/stdc++.h>
using namespace std;

const int RLEN=1<<18|1;
inline char nc() {
	static char ibuf[RLEN],*ib,*ob;
	(ib==ob) && (ob=(ib=ibuf)+fread(ibuf,1,RLEN,stdin));
	return (ib==ob) ? -1 : *ib++;
}
inline int rd() {
	char ch=nc(); int i=0,f=1;
	while(!isdigit(ch)) {if(ch=='-')f=-1; ch=nc();}
	while(isdigit(ch)) {i=(i<<1)+(i<<3)+ch-'0'; ch=nc();}
	return i*f;
}

const int N=2e6+50, mod=104857601, lim=23333;
inline int add(int x,int y) {return (x+y>=mod) ? (x+y-mod) : (x+y);}
inline int dec(int x,int y) {return (x-y<0) ? (x-y+mod) : (x-y);}
inline int mul(int x,int y) {return (long long)x*y%mod;}
inline int power(int a,int b,int rs=1) {for(;b;b>>=1,a=mul(a,a)) if(b&1) rs=mul(rs,a); return rs;}

namespace FFT {
	int w[N*8],pos[N*8],k;
	int A[N*8],B[N*8];
	inline void pre() {
		for(int bl=1,i=0;bl<=2e5;i+=bl,bl<<=1) {
			int wn=power(3,(mod-1)/bl/2);
			w[i]=1;
			for(int j=1;j<bl;j++) w[i+j]=mul(w[i+j-1],wn);
		}
	}
	inline void init(int nn) {
		for(k=1;k<=nn;k<<=1);
		for(int i=1;i<k;i++) pos[i]=(i&1) ? ((pos[i>>1]>>1)^(k>>1)) : (pos[i>>1]>>1);
		memset(A,0,sizeof(int)*k);
		memset(B,0,sizeof(int)*k);
	}
	inline void dft(int *a) {
		for(int i=1;i<k;i++) if(pos[i]>i) swap(a[pos[i]],a[i]);
		for(int bl=1,i=0;bl<k;i+=bl,bl<<=1) {
			int tl=bl<<1, wn=power(3,(mod-1)/tl);
			for(int bg=0;bg<k;bg+=tl)
				for(int j=0;j<bl;j++) {
					int &t1=a[bg+j], &t2=a[bg+j+bl], t=mul(t2,w[i+j]);
					t2=dec(t1,t); t1=add(t1,t);
				}
		}
	}
	inline void func() {
		dft(A); dft(B);
		for(int i=0;i<k;i++) B[i]=mul(A[i],B[i]);
		dft(B); reverse(B+1,B+k);
		const int inv=power(k,mod-2);
		for(int i=0;i<k;i++) B[i]=mul(B[i],inv);
	}
}
int n,a[N],A,B; long long m;
struct poly {
	vector <int> a;
	inline int deg() const {return a.size()-1;}
	poly(int d=0,int t=0) {a.resize(d+1); a[d]=t;}
	inline int& operator [](const int &b) {return a[b];}
	inline const int& operator [](const int &b) const {return a[b];}
	friend inline poly operator *(const poly &a,const poly &b) {
		FFT::init(a.deg()+b.deg());
		for(int i=0;i<=a.deg();i++) FFT::A[i]=a[i];
		for(int i=0;i<=b.deg();i++) FFT::B[i]=b[i];
		FFT::func(); 
		poly c(a.deg()+b.deg(),0);
		for(int i=0;i<=c.deg();i++) c[i]=FFT::B[i];
		return c;
	}
	friend inline poly operator *(const poly &a,const int &b) {
		poly c=a; 
		for(int i=0;i<=c.deg();i++) c.a[i]=mul(c.a[i],b);
		return c;	
	}
	friend inline poly operator -(const poly &a,const poly &b) {
		poly c(max(a.deg(),b.deg()));
		for(int i=0;i<=a.deg();i++) c[i]=a[i];
		for(int i=0;i<=b.deg();i++) c[i]=dec(c[i],b[i]);
		return c;
	}	
	friend inline poly operator +(const poly &a,const poly &b) {
		poly c(max(a.deg(),b.deg()));
		for(int i=0;i<=a.deg();i++) c[i]=a[i];
		for(int i=0;i<=b.deg();i++) c[i]=add(c[i],b[i]);
		return c;
	}
	inline poly calc_inv(poly f,int len) {
		if(len==1) return poly(0,power(f[0],mod-2));
		poly f0=calc_inv(f.extend(len>>1),len>>1);
		return (f0*2-(f0*f0).extend(len-1)*f).extend(len-1);
	}
	inline poly calc_inverse(int nn) {
		FFT::init(nn);
		return calc_inv(this->extend(nn),FFT::k).extend(nn);
	}
	inline poly rev() {poly c=*this; reverse(c.a.begin(),c.a.end()); return c;}
	friend inline poly operator %(const poly &a,const poly &b) {
		if(a.deg()<b.deg()) return a;
		poly A=a, B=b; 
		int res=a.deg()-b.deg();
		A=A.rev().extend(res); 
		B=B.rev().extend(res);
		poly C=B.calc_inverse(res);
		poly D=(A*C).extend(res).rev(); 
		return (a-b*D).extend(b.deg()-1);
	}
	inline void dg() {
		if(!deg()) a[0]=0;
		else {
			for(int i=0;i<deg();i++) a[i]=mul(i+1,a[i+1]);
			a.pop_back();
		}
	}
	inline poly extend(int nn) {
		poly c=*this;
		c.a.resize(nn+1);
		return c;
	}
	inline int getval(int x) {
		int res=0, w=1;
		for(int i=0;i<=deg();i++)
			res=add(res,mul(w,a[i])), w=mul(w,x);
		return res;
	}
};

int main() {
	FFT::pre();
	cin>>n>>m;
	cin>>a[1]>>A>>B;
	for(int i=2;i<=n;i++) a[i]=((long long)a[i-1]*A+B)%lim+1;
	poly g(lim,0);
	for(int i=1;i<=n;i++) if(a[i]<=lim) g[a[i]]++;
	poly f(0,1);
	f=(f-g).calc_inverse(lim);
	g=poly(lim,1);
	for(int i=1;i<=n;i++) if(a[i]<=lim) g[lim-a[i]]--;
	poly rs(0,1), h(1,1);
	for(;m;m>>=1,h=h*h%g)
		if(m&1) rs=rs*h%g;
	int ans=0;
	for(int i=0;i<=lim && i<=rs.deg();i++)
		ans=add(ans,mul(rs[i],f[i]));
	cout<<ans<<'\n';
}