[luogu4389][生成函數][NTT]付公主的揹包

luogu4389

考慮OGF，$ans=[x^n]\prod_{i=1}^n(\frac{1}{1-x^i})^{a_i}$，其中$a_i$表示大小爲$i$的物品的個數
看到這種一般想到用exp：
$\prod_{i=1}^n(\frac{1}{1-x^i})^{a_i}$
$=exp(-\sum_{i=1}^na_iln(1-x^i))$
求導：
$=exp(-\sum_{i=1}^na_i\sum_{j\ge1}\frac{x^{ij}}{j})$
交換求和順序：
$=exp(\sum_{j\ge1}\frac{1}{j}A(x^j))$
所以可以枚舉$j$然後暴力算後面，最後exp一次，複雜度是調和級數

Code：

#include<bits/stdc++.h>
#define ll long long
#define pb push_back
#define mod 998244353
#define g 3
#define poly vector<int>
using namespace std;
inline int read(){
	int res=0,f=1;char ch=getchar();
	while(!isdigit(ch)) {if(ch=='-') f=-f;ch=getchar();}
	while(isdigit(ch)) {res=(res<<1)+(res<<3)+(ch^48);ch=getchar();}
	return res*f;
}
const int N=1e5+5;
inline int add(int x,int y){x+=y;if(x>=mod) x-=mod;return x;}
inline int dec(int x,int y){x-=y;if(x<0) x+=mod;return x;}
inline int mul(int x,int y){return 1ll*x*y%mod;}
inline void inc(int &x,int y){x+=y;if(x>=mod) x-=mod;}
inline void Dec(int &x,int y){x-=y;if(x<0) x+=mod;}
inline void Mul(int &x,int y){x=1ll*x*y%mod;}
inline int ksm(int a,int b){int res=1;for(;b;b>>=1,a=mul(a,a)) if(b&1) res=mul(res,a);return res;}
namespace Ntt{
	const int C=18;
	int *w[19],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 n){for(int i=0;i<n;i++) rev[i]=(rev[i>>1]>>1)|((i&1)*(n>>1));}
	inline void ntt(poly &f,int n,int kd){
		for(int i=0;i<n;i++) if(i>rev[i]) swap(f[i],f[rev[i]]); 
		for(int mid=1,l=1;mid<n;mid<<=1,l++)
			for(int i=0;i<n;i+=(mid<<1))
				for(int j=0,a0,a1;j<mid;j++){
					a0=f[i+j],a1=mul(f[i+j+mid],w[l][j]);
					f[i+j]=add(a0,a1);f[i+j+mid]=dec(a0,a1);
				}
		if(kd==-1){
			reverse(f.begin()+1,f.begin()+n);
			for(int inv=ksm(n,mod-2),i=0;i<n;i++) Mul(f[i],inv);
		}
	}
}
using namespace Ntt;
namespace poly_operator{
	inline poly operator - (poly a,poly b){
		poly res;int lim=max(a.size(),b.size());res.resize(lim);
		for(int i=0;i<lim;i++) res[i]=dec(a[i],b[i]);return res;
	}
	inline poly operator * (poly a,poly b){
		int m=a.size()+b.size()-1,n=1;
		while(n<m) n<<=1;
		init_rev(n);
		a.resize(n);ntt(a,n,1);
		b.resize(n);ntt(b,n,1);
		for(int i=0;i<n;i++) Mul(a[i],b[i]);
		ntt(a,n,-1);a.resize(m);
		return a;
	}
	inline poly p_inv(poly a,int n){
		poly c,b(1,ksm(a[0],mod-2));
		for(int lim=4;lim<(n<<2);lim<<=1){
			init_rev(lim);
			c=a;c.resize(lim>>1);
			c.resize(lim);ntt(c,lim,1);
			b.resize(lim);ntt(b,lim,1);
			for(int i=0;i<lim;i++) Mul(b[i],dec(2,mul(b[i],c[i])));
			ntt(b,lim,-1);b.resize(lim>>1);
		}
		b.resize(n);return b;
	}
	ll inv[N<<2];
	inline poly deriv(poly a){
		for(int i=0;i<a.size()-1;i++) a[i]=mul(a[i+1],i+1);
		a.pop_back();return a;
	}
	inline poly integ(poly a){
		a.pb(0);
		for(int i=a.size()-1;i;i--) a[i]=mul(a[i-1],inv[i]);
		a[0]=0;return a;
	}
	inline poly ln(poly a,int n){
		a=integ(deriv(a)*p_inv(a,n));a.resize(n);
		return a;
	}
	inline poly exp(poly a,int n,int m){
		poly b(1,1),c;
		for(int lim=2;lim<(n<<1);lim<<=1){
			c=ln(b,lim);
			for(int i=0;i<lim;i++) c[i]=dec(i<=m?a[i]:0,c[i]);
			inc(c[0],1);b=b*c;
			b.resize(lim);
		}
		b.resize(n);return b;
	}
}
using namespace poly_operator;
int tong[N];
poly f;
int main(){
	inv[1]=1;init_w();
	for(int i=2;i<(N<<2);i++) inv[i]=mul((mod-mod/i),inv[mod%i]);
	int n=read(),m=read();f.resize(m+1);
	for(int i=1;i<=n;i++) ++tong[read()];
	for(int i=1;i<=m;i++) if(tong[i])
		for(int j=1;j*i<=m;j++) inc(f[i*j],mul(tong[i],inv[j]));
	f=exp(f,m+1,m);
	for(int i=1;i<=m;i++) cout<<f[i]<<"\n";
	return 0;
}