【洛谷P4389】—付公主的背包(多项式Exp+生成函数)

传送门


考虑直接构造生成函数f(x)=i=0ivmxiv=11xvf(x)=\sum_{i=0}^{iv\le m}x^{iv}=\frac 1 {1-x^v}
但是所有直接乘起来是O(nmlog)O(nmlog)

考虑对所有函数取对数,加起来之后再ExpExp回来
但是直接取LnLn也不可取

考虑令g(x)=Ln(f(x))g(x)=Ln(f(x))
取导后
g(x)=f(x)f(x)=(1xv)i=0ivmivxiv1g'(x)=\frac{f'(x)}{f(x)}=(1-x^v)\sum_{i=0}^{iv\le m}iv*x^{iv-1}

         =i=0ivmivxiv1i=0ivmivx(i+1)v1\ \ \ \ \ \ \ \ \ =\sum_{i=0}^{iv\le m}iv*x^{iv-1}-\sum_{i=0}^{iv\le m}iv*x^{(i+1)*v-1}

         =i=0ivmvxiv1\ \ \ \ \ \ \ \ \ =\sum_{i=0}^{iv\le m}vx^{iv-1}

还原则得到
g(x)=i=1ivmvivxivg(x)=\sum_{i=1}^{iv\le m}\frac{v}{iv}x^{iv}

        =i=0ivm1ixiv\ \ \ \ \ \ \ \ =\sum_{i=0}^{iv\le m}\frac{1}{i}x^{iv}

考虑对于一个价值的vv
只会出现mv\frac m v
所以把v=[1,m]v=[1,m]全部计算得到gg的复杂度是O(mlogm)O(mlogm)

然后做一次ExpExp还原回来就是了

复杂度O(mlogm)O(mlogm)

#include<bits/stdc++.h>
using namespace std;
#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 re register
#define pb push_back
#define cs const
#define pii pair<int,int>
#define fi first
#define se second
#define ll long long
cs int mod=998244353,G=3;
inline int add(int a,int b){return (a+=b)>=mod?a-mod:a;}
inline void Add(int &a,int b){(a+=b)>=mod?(a-=mod):0;}
inline int dec(int a,int b){return (a-=b)<0?a+mod:a;}
inline void Dec(int &a,int b){(a-=b)<0?(a+=mod):0;}
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));return res;
}
cs int N=100005;
#define poly vector<int>
int rev[N<<2],inv[N<<2];
int n,m,v[N];
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 a0,a1,mid=1;mid<lim;mid<<=1){
		int wn=ksm(G,(mod-1)/(mid<<1));
		for(int i=0;i<lim;i+=(mid<<1)){
			int w=1;
			for(int j=0;j<mid;j++,Mul(w,wn)){
				a0=f[i+j],a1=mul(w,f[i+j+mid]);
				f[i+j]=add(a0,a1),f[i+j+mid]=dec(a0,a1);
			}
		}
	}
	if(kd==-1){
		reverse(f.begin()+1,f.begin()+lim);
		for(int inv=ksm(lim,mod-2),i=0;i<lim;i++)Mul(f[i],inv);
	}
}
inline void init_rev(int lim){
	for(int i=0;i<lim;i++)rev[i]=(rev[i>>1]>>1)|((i&1)*(lim>>1));
}
inline poly mul(poly a,poly b){
	int deg=a.size()+b.size()-1,lim=1;
	while(lim<deg)lim<<=1;
	init_rev(lim);
	a.resize(lim),b.resize(lim);
	ntt(a,lim,1),ntt(b,lim,1);
	for(int i=0;i<lim;i++)Mul(a[i],b[i]);
	ntt(a,lim,-1),a.resize(deg);
	return a;
}
inline poly Inv(poly a,int deg){
	poly b(1,ksm(a[0],mod-2)),c;
	for(int lim=4;lim<(deg<<2);lim<<=1){
		c=a,c.resize(lim>>1);
		init_rev(lim);
		c.resize(lim),b.resize(lim);
		ntt(b,lim,1),ntt(c,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(deg);return b;
}
inline poly deriv(poly a){
	for(int i=0;i<a.size();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 lim){
	a=integ(mul(deriv(a),Inv(a,lim))),a.resize(lim);
	return a;
}
inline poly exp(poly a,int deg){
	poly b(1,1),c;
	for(int lim=2;lim<(deg<<1);lim<<=1){
		c=Ln(b,lim);
		for(int i=0;i<lim;i++)c[i]=dec(i<=m?a[i]:0,c[i]);
		Add(c[0],1),b=mul(b,c),b.resize(lim);
	}b.resize(deg);return b;
}
inline void init(){
	inv[1]=1;
	for(int i=2;i<N*4;i++)inv[i]=dec(0,mul(mod/i,inv[mod%i]));
}
poly f;
int main(){
	init();
	n=read(),m=read(),f.resize(m+1);
	for(int i=1;i<=n;i++)v[read()]++;
	for(int i=1;i<=m;i++){
		if(v[i]){
			for(int j=1;j*i<=m;j++)
			Add(f[j*i],mul(v[i],inv[j]));
		}
	}
	f=exp(f,m+1);
	for(int i=1;i<=m;i++)cout<<f[i]<<'\n';
}
posted @ 2019-07-31 16:17  Stargazer_cykoi  阅读(121)  评论(0编辑  收藏  举报