• 【BZOJ4555】【TJOI2016】【HEOI2016】—求和(第二类斯特林数+NTT)


    传送门

    题意:求i=0nj=0iS2(i,j)j!2jsum_{i=0}^nsum_{j=0}^{i}S_2(i,j)j!2^j

    由于S2(i,j)i<jS_2(i,j),i<jS2(i,j)=0S_2(i,j)=0

    ans=i=0nj=0nS2(i,j)j!2jans=sum_{i=0}^{n}sum_{j=0}^{n}S_2(i,j)j!2^j
    由于S2(i,j)=k=0j(1)jk(jk)kij!S_2(i,j)=frac{sum_{k=0}^{j}(-1)^{j-k}{jchoose k}k^i}{j!}
    ans=i=0nj=0nk=0j(1)jk(jk)ki2jans=sum_{i=0}^nsum_{j=0}^n{sum_{k=0}^{j}(-1)^{j-k}{jchoose k}k^i}2^j

    =i=0nj=0nk=0j(1)jk(jk)!×kik!×2jj!=sum_{i=0}^{n}sum_{j=0}^{n}sum_{k=0}^{j}frac{(-1)^{j-k}}{(j-k)!} imesfrac{k^i}{k!} imes 2^jj!

    =j=0n2jj!k=0j(1)jk(jk)!×i=0nkik!=sum_{j=0}^{n}2^jj!sum_{k=0}^{j}frac{(-1)^{j-k}}{(j-k)!} imesfrac{sum_{i=0}^{n}k^i}{k!}

    =j=0n2jj!k=0j(1)jk(jk)!×kn+11k!(k1)=sum_{j=0}^{n}2^jj!sum_{k=0}^{j}frac{(-1)^{j-k}}{(j-k)!} imesfrac{k^{n+1}-1}{k!(k-1)}

    f(x)=(1)xx!,g(x)=xn+11x!(x1)f(x)=frac{(-1)^x}{x!},g(x)=frac{x^{n+1}-1}{x!*(x-1)}
    注意特判ggx=0x=0x=1x=1的情况

    卷积一下就完了

    #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
    const 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 int Inv(int x){
    	return ksm(x,mod-2);
    }
    const int N=100005;
    int ifac[N],fac[N],n;
    int rev[N<<2];
    #define poly vector<int>
    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;mid<lim;mid<<=1){
            int now=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++,w=mul(w,now)){
                    int 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),1))for(int i=0,inv=ksm(lim,mod-2);i<lim;i++)f[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;
    }
    poly f,g;
    inline void init(){
    	fac[0]=ifac[0]=1;
    	for(int i=1;i<=n;i++)fac[i]=mul(fac[i-1],i);
    	ifac[n]=Inv(fac[n]);
    	for(int i=n-1;i;i--)ifac[i]=mul(ifac[i+1],i+1);
    }
    int main(){
    	n=read();
    	init();
    	for(int i=0;i<=n;i++)
    	f.pb((i&1)?mul(mod-1,ifac[i]):ifac[i]);
    	g.pb(1),g.pb(n+1);
    	for(int i=2;i<=n;i++)
    	g.pb(mul(dec(ksm(i,n+1),1),Inv(mul(i-1,fac[i]))));
    	poly t=mul(f,g);
    	int res=0;
    	for(int i=0,bin=1;i<=n;i++,Mul(bin,2))
    		Add(res,mul(mul(bin,fac[i]),t[i]));
    	cout<<res;
    }
    
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  • 原文地址:https://www.cnblogs.com/stargazer-cyk/p/12328730.html
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