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  • 【洛谷P4705】—玩游戏(生成函数+NTT)

    传送门
    VanVan游戏?


    显然我们要求的是

    Ans=t!k=1ti=1naikk!j=1mbjtk(tk)!Ans=t!sum_{k=1}^{t}sum_{i=1}^{n}frac{a_i^k}{k!}sum_{j=1}^{m}frac{b_j^{t-k}}{(t-k)!}

    考虑求

    F(x)=k=1ti=1naikk!F(x)=sum_{k=1}^{t}sum_{i=1}^{n}frac{a_i^k}{k!}

    构建F(x)F(x)的生成函数

    f(x)=k=1i=1naikk!xk=i=1nk=11k!(aix)k f(x)=sum_{k=1}^{infty}sum_{i=1}^{n}frac{a_i ^ k}{k!}x^k \ =sum_{i=1}^{n}sum_{k=1}^{infty}frac{1}{k!}(a_ix)^k

    先不看1k!frac{1}{k!}

    后面就是一个等比数列
    就是
    i=1n11aixsum_{i=1}^{n}frac{1}{1-a_ix}

    =i=1n1xai1aix=sum_{i=1}^{n}1-xfrac{-a_i}{1-a_ix}

    =nxi=1naj1aix=n-xsum_{i=1}^nfrac{-a_j}{1-a_ix}

    =nx[Ln(i=1n(1aix))]=n-x[Ln(prod_{i=1}^n(1-a_ix))]'

    分治NTTNTT即可
    然后每一项乘一个1i!frac{1}{i!}就把k!k!弄回来了

    #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
    #define poly vector<int>
    #define bg begin
    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;
    }
    inline void chemx(ll &a,ll b){a<b?a=b:0;}
    inline void chemn(ll &a,ll b){a>b?a=b:0;}
    cs int N=(1<<20|1),C=20;
    poly w[C+1];
    inline void init_w(){
    	for(int i=1;i<=C;i++)w[i].resize(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];
    }
    int rev[N<<2],inv[N<<2],fac[N<<2],ifac[N<<2];
    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_inv(int len=N*4-5){
    	inv[0]=inv[1]=1;
    	for(int i=2;i<=len;i++)inv[i]=mul(mod-mod/i,inv[mod%i]);
    	ifac[0]=fac[0]=1;
    	for(int i=1;i<=len;i++)fac[i]=mul(fac[i-1],i);
    	ifac[len]=ksm(fac[len],mod-2);
    	for(int i=len-1;i;i--)ifac[i]=mul(ifac[i+1],i+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(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.bg()+1,f.bg()+lim);
    		for(int i=0;i<lim;i++)Mul(f[i],inv[lim]);
    	}
    }
    inline poly operator *(poly a,poly b){
    	int deg=a.size()+b.size()-1,lim=1;
    	if(deg<=128){
    		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;
    	}
    	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),a.resize(deg);
    	return a;
    }
    
    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 Inv(poly a,int deg){
    	poly c,b(1,ksm(a[0],mod-2));
    	for(int lim=4;lim<(deg<<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(deg);return b;
    }
    
    inline poly Ln(poly a,int lim){
    	a=integ(deriv(a)*Inv(a,lim)),a.resize(lim);
    	return a;
    }
    #define lc (u<<1)
    #define rc ((u<<1)|1)
    #define mid ((l+r)>>1)
    poly f[N<<2];
    inline void build(int u,int l,int r,int *v){
    	if(l==r){f[u].clear();f[u].pb(1),f[u].pb(mod-v[l]);return;}
    	build(lc,l,mid,v),build(rc,mid+1,r,v);
    	f[u]=f[lc]*f[rc];
    }
    #undef lc
    #undef rc
    #undef mid
    int t;
    inline poly Build(int n,int *v){
    	build(1,1,n,v);
    	poly res=deriv(Ln(f[1],t+1));
    	res.pb(0);
    	for(int i=res.size()-1;i;i--)res[i]=dec(0,res[i-1]);
    	res[0]=n;
    	for(int i=1;i<res.size();i++)Mul(res[i],ifac[i]);
    	return res;
    }
    int n,m;
    int a[N],b[N];
    int main(){
    	n=read(),m=read(),init_w(),init_inv();
    	for(int i=1;i<=n;i++)a[i]=read();
    	for(int i=1;i<=m;i++)b[i]=read();
    	t=read();
    	poly F=Build(n,a),G=Build(m,b);
    	F=F*G;int pt=mul(ksm(n,mod-2),ksm(m,mod-2));
    	for(int i=1;i<=t;i++)cout<<mul(F[i],mul(pt,fac[i]))<<'
    ';
    }
    
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  • 原文地址:https://www.cnblogs.com/stargazer-cyk/p/12328674.html
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