【BZOJ3392】【洛谷P3321】【SDOI2015】—序列统计（NTT）

传送门

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考虑如果是加的话就是板子题了
构建生成函数$n$次方就可以了

考虑如果转成对数的话乘就变成加了
在模意义下取对数就可以变成加法

注意这里的多项式乘法是循环卷积

每次对$x^{m-1}$取模即可

由于不满足$Ln$的条件
只能直接$O(nlog^2n)$快速幂求

#include<bits/stdc++.h>
using namespace std;
const int RLEN=1<<20|1;
inline char gc(){
	static char ibuf[RLEN],*ib,*ob;
	(ob==ib)&&(ob=(ib=ibuf)+fread(ibuf,1,RLEN,stdin));
	return (ob==ib)?EOF:*ib++;
}
#define gc getchar
inline int read(){
	char ch=gc();
	int res=0,f=1;
	while(!isdigit(ch))f^=ch=='-',ch=gc();
	while(isdigit(ch))res=(res+(res<<2)<<1)+(ch^48),ch=gc();
	return f?res:-res;
}
#define ll long long	
#define re register
#define pii pair<int,int>
#define fi first
#define se second
#define pb push_back
#define cs const
#define poly vector<int>
#define bg begin
const int mod=1004535809,G=3;
inline int add(int a,int b){return a+b>=mod?a+b-mod:a+b;}
inline void Add(int &a,int b){a=add(a,b);}
inline int dec(int a,int b){return a>=b?a-b:a-b+mod;}
inline void Dec(int &a,int b){a=dec(a,b);}
inline int mul(int a,int b){return 1ll*a*b>=mod?1ll*a*b%mod:a*b;}
inline void Mul(int &a,int b){a=mul(a,b);}
inline int ksm(int a,int b){int res=1;for(;b;b>>=1,a=mul(a,a))(b&1)?(res=mul(res,a)):0;return res;}
inline int ksm(int a,int b,int Mod){int res=1;
	for(;b;b>>=1,a=1ll*a*a%Mod)if(b&1)(res=1ll*res*a%Mod);return res;
}
inline void chemx(int &a,int b){a<b?a=b:0;}
inline void chemn(int &a,int b){a>b?a=b:0;}
int pr[100],tot;
inline bool check(int x,int phi){
	for(int i=1;i<=tot;i++)if(ksm(x,phi/pr[i],phi+1)==1)return true;
	return false;
}
inline int find_G(int x){
	int phi=x-1;tot=0;
	for(int i=2;i*i<=x;i++){
		while(phi%i==0)pr[++tot]=i,phi/=i;
	}
	if(phi>1)pr[++tot]=phi;
	int g=2;
	while(check(g,x-1))g++;
	return g;
}
cs int N=8005,C=16;
int *w[C+1];
int rev[N<<2];
inline void init_w(){
	for(int i=1;i<=C;i++)
	w[i]=new int[1<<(i-1)];
	int wn=ksm(G,(mod-1)/(1<<C));
	w[C][0]=1;
	for(int i=1;i<(1<<(C-1));i++)w[C][i]=mul(w[C][i-1],wn);
	for(int i=C-1;i;i--)
	for(int j=0;j<(1<<(i-1));j++)
	w[i][j]=w[i+1][j<<1];
}

inline void init_rev(int lim){
	for(int i=0;i<lim;i++)rev[i]=(rev[i>>1]>>1)|((i&1)*(lim>>1));
}
inline void ntt(poly &f,int lim,int kd){
	for(int i=0;i<lim;i++)if(i>rev[i])swap(f[i],f[rev[i]]);
	for(int mid=1,l=1,a0,a1;mid<lim;mid<<=1,l++)
	for(int i=0;i<lim;i+=(mid<<1))
	for(int j=0;j<mid;j++)
	a0=f[i+j],a1=mul(w[l][j],f[i+j+mid]),f[i+j]=add(a0,a1),f[i+j+mid]=dec(a0,a1);
	if(kd==-1){
		reverse(f.bg()+1,f.bg()+lim);
		for(int i=0,inv=ksm(lim,mod-2);i<lim;i++)Mul(f[i],inv);
	}
}
int n,m,mp[N],x,s[N],num;
inline poly operator *(poly a,poly b){
	int deg=a.size()+b.size()-1,lim=1;
	while(lim<deg)lim<<=1;
	init_rev(lim);
	a.resize(lim),ntt(a,lim,1);
	b.resize(lim),ntt(b,lim,1);
	for(int i=0;i<lim;i++)Mul(a[i],b[i]);
	ntt(a,lim,-1);
	for(int i=0;i<m-1;i++)Add(a[i],a[i+m-1]);
	a.resize(m-1);
	return a;
}
poly res;
int main(){
	init_w();
	n=read(),m=read(),x=read(),num=read();
	res.resize(m);
	int g=find_G(m);
	for(int p=1,i=0;i<m-1;i++,p=p*g%m)mp[p]=i;
	for(int i=1;i<=num;i++){
		int p=read();
		if(p)res[mp[p]]++;
	}
	poly a;
	a.resize(m);
	a[mp[1]]=1;
	for(;n;n>>=1)if(n&1)a=a*res;res=res*res;
	cout<<a[mp[x]];
}