• 【洛谷 P5850】 calc加强版(生成函数+NTT)


    传送门

    先看做无序的最后乘上n!n!
    显然可以构造生成函数(1+ix)prod(1+ix)
    分治nttntt好像也可以过?不过不知道为啥RERE
    应该是炸空间了

    先取对在expexp

    =exp((i1+ix)dx)=exp(int(sum{frac i{1+ix}})mathrm{dx})
    expexp内的
    =(i=1kj=0(1)jijxj)=int(sum_{i=1}^ksum_{j=0}^{infty}(-1)^ji^jx^j)
    =i=1kj=1(1)j1ijjxj=sum_{i=1}^ksum_{j=1}^{infty}(-1)^{j-1}frac{i^j}{j}x^j
    =j=1(1)j1j!xji=1kij=sum_{j=1}^{infty}frac{(-1)^{j-1}}{j!}x^jsum_{i=1}^ki^j

    后面就是要求等幂和
    考虑等幂和的EGFEGF

    i=1j=1kjixii!sum_{i=1}^{infty}sum_{j=1}^kfrac{j^ix^i}{i!}
    =j=1ki=1jixii!=sum_{j=1}^ksum_{i=1}^{infty}frac{j^ix^i}{i!}
    =j=1kejx=sum_{j=1}^ke^{jx}
    =j=0kejx1=sum_{j=0}^ke^{jx}-1
    =e(k+1)x1ex11=frac{e^{(k+1)x}-1}{e^x-1}-1

    复杂度O(nlogn)O(nlogn)

    #include<bits/stdc++.h>
    using namespace std;
    #define cs const
    #define re register
    #define pb push_back
    #define pii pair<int,int>
    #define ll long long
    #define fi first
    #define se second
    #define bg begin
    cs int RLEN=1<<20|1;
    inline char gc(){
        static char ibuf[RLEN],*ib,*ob;
        (ib==ob)&&(ob=(ib=ibuf)+fread(ibuf,1,RLEN,stdin));
        return (ib==ob)?EOF:*ib++;
    }
    inline int read(){
        char ch=gc();
        int res=0;bool 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;
    }
    template<class tp>inline void chemx(tp &a,tp b){a<b?a=b:0;}
    template<class tp>inline void chemn(tp &a,tp b){a>b?a=b:0;}
    cs int mod=998244353;
    inline int add(int a,int b){return (a+=b)>=mod?(a-mod):a;}
    inline int dec(int a,int b){a-=b;return a+(a>>31&mod);}
    inline int mul(int a,int b){static ll r;r=1ll*a*b;return (r>=mod)?(r%mod):r;}
    inline void Add(int &a,int b){(a+=b)>=mod?(a-=mod):0;}
    inline void Dec(int &a,int b){a-=b,a+=a>>31&mod;}
    inline void Mul(int &a,int b){static ll r;r=1ll*a*b;a=(r>=mod)?(r%mod):r;}
    inline int ksm(int a,int b,int res=1){for(;b;b>>=1,Mul(a,a))(b&1)&&(Mul(res,a),1);return res;}
    inline int Inv(int x){return ksm(x,mod-2);}
    inline int fix(int x){return (x<0)?x+mod:x;}
    cs int N=500005;
    int fac[N],ifac[N],iv[N];
    inline void init_inv(cs int len=N-5){
    	iv[0]=iv[1]=fac[0]=ifac[0]=1;
    	for(int i=1;i<=len;i++)fac[i]=mul(fac[i-1],i);
    	ifac[len]=Inv(fac[len]);
    	for(int i=len-1;i;i--)ifac[i]=mul(ifac[i+1],i+1);
    	for(int i=2;i<=len;i++)iv[i]=mul(mod-mod/i,iv[mod%i]);
    }
    typedef vector<int> poly;
    namespace Poly{
    	cs int G=3,C=21,M=(1<<C)+1;
    	int *w[C+1];
    	int rev[M];
    	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 init_w(){
    		for(int i=1;i<=C;i++)w[i]=new int[(1<<(i-1))+1];
    		int wn=ksm(G,(mod-1)/(1<<C));w[C][0]=1;
    		for(int i=1,l=1<<(C-1);i<l;i++)w[C][i]=mul(w[C][i-1],wn);
    		for(int j=C-1;j;j--)
    		for(int i=0;i<(1<<(j-1));i++)w[j][i]=w[j+1][i<<1];
    	}
    	inline void ntt(int *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(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+1,f+lim);
    			for(int i=0,iv=Inv(lim);i<lim;i++)Mul(f[i],iv);
    		}
    	}
    	inline poly operator *(poly a,poly b){
    		int deg=a.size()+b.size()-1;
    		if(deg<=32){
    			poly c(deg,0);
    			for(int i=0;i<a.size();i++)
    			for(int j=0;j<b.size();j++)
    			Add(c[i+j],mul(a[i],b[j]));
    			return c;
    		}
    		int lim=1;
    		while(lim<deg)lim<<=1;
    		init_rev(lim);
    		a.resize(lim),ntt(&a[0],lim,1);
    		b.resize(lim),ntt(&b[0],lim,1);
    		for(int i=0;i<lim;i++)Mul(a[i],b[i]);
    		ntt(&a[0],lim,-1),a.resize(deg);
    		return a;
    	}
    	inline poly Inv(poly a,int deg){
    		poly b(1,::Inv(a[0])),c;
    		for(int lim=4;lim<(deg<<2);lim<<=1){
    			init_rev(lim);
    			c.resize(lim>>1);
    			for(int i=0;i<(lim>>1);i++)c[i]=(i<a.size()?a[i]:0);
    			c.resize(lim),ntt(&c[0],lim,1);
    			b.resize(lim),ntt(&b[0],lim,1);
    			for(int i=0;i<lim;i++)Mul(b[i],dec(2,mul(b[i],c[i])));
    			ntt(&b[0],lim,-1),b.resize(lim>>1);
    		}
    		b.resize(deg);return b;
    	}
    	inline poly deriv(poly a){
    		for(int i=0;i<(int)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],iv[i]);
    		a[0]=0;return a;
    	}
    	inline poly Ln(poly a,int deg){
    		a=integ(deriv(a)*Inv(a,deg)),a.resize(deg);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<a.size()?a[i]:0,c[i]);
    			Add(c[0],1),b=b*c,b.resize(lim);
    		}b.resize(deg);return b;
    	}
    } 
    using namespace Poly;
    int k,m;
    poly e,ek,f,g;
    int main(){
    	#ifdef Stargazer
    	freopen("lx.in","r",stdin);
    	#endif
    	k=read(),m=read();
    	init_w(),init_inv();
    	e.resize(m+1),ek.resize(m+1),f.resize(m+1);
    	for(int i=0,mt=k+1;i<=m;i++,Mul(mt,k+1))e[i]=ifac[i+1],ek[i]=mul(ifac[i+1],mt);
    	g=ek*Inv(e,m+1);
    	for(int i=1;i<=m;i++){
    		f[i]=mul(g[i],fac[i-1]);
    		if(!(i&1))f[i]=dec(0,f[i]);
    	}
    	f=Exp(f,m+1);
    	for(int i=1;i<=m;i++)cout<<mul(fac[i],f[i])<<'
    ';
    }
    
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  • 原文地址:https://www.cnblogs.com/stargazer-cyk/p/12328304.html
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