题面

题解

如果您不知道伯努利数是什么可以去看看

首先我们把自然数幂和化成伯努利数的形式

\[\sum_{i=1}^{n-1}i^k={1\over k+1}\sum_{i=0}^k{k+1\choose i}B_in^{k+1-i}\]

然后接下来就是推倒了

\[ \begin{aligned} Ans &=\sum_{k=0}^na_kS_k(x)\\ &=\sum_{k=0}^na_k\left(x^k+{1\over k+1}\sum_{i=0}^k{k+1\choose i}B_ix^{k+1-i}\right)\\ &=\sum_{k=0}^na_kx^k+\sum_{k=0}^na_kk!\sum_{i=0}^k{B_ix^{k+1-i}\over i!(k+1-i)!}\\ \end{aligned} \]

然后我们枚举\(d=k+1-i\)

\[ \begin{aligned} Ans &=\sum_{k=0}^na_kx^k+\sum_{d=1}^{n+1}{x^d\over d!}\sum_{k=d-1}^na_kk!{B_{k+1-d}\over (k+1-d)!} \end{aligned} \]

我们令\(G_i=B_{n-i}\)

\[ \begin{aligned} Ans &=\sum_{k=0}^na_kx^k+\sum_{d=1}^{n+1}{x^d\over d!}\sum_{k=d-1}^na_kk!{G_{n-k-1+d}\over (n-k-1+d)!} \end{aligned} \]

那么我们把\(G(x)\)和\(A(x)\)做个卷积，那么它们的第\(n+i-1\)项系数加上\(A(x)\)的第\(i\)项系数就是\([x^i]Ans\)了

顺便注意\([x^0]Ans\)恒为\(a_0\)

//minamoto#include
    
     #define R register#define fp(i,a,b) for(R int i=(a),I=(b)+1;i
     
      I;--i)#define go(u) for(int i=head[u],v=e[i].v;i;i=e[i].nx,v=e[i].v)using namespace std;char buf[1<<21],*p1=buf,*p2=buf;inline char getc(){return p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++;}int read(){    R int res,f=1;R char ch;    while((ch=getc())>'9'||ch<'0')(ch=='-')&&(f=-1);    for(res=ch-'0';(ch=getc())>='0'&&ch<='9';res=res*10+ch-'0');    return res*f;}char sr[1<<21],z[20];int K=-1,Z=0;inline void Ot(){fwrite(sr,1,K+1,stdout),K=-1;}void print(R int x){    if(K>1<<20)Ot();if(x<0)sr[++K]='-',x=-x;    while(z[++Z]=x%10+48,x/=10);    while(sr[++K]=z[Z],--Z);sr[++K]=' ';}const int N=(1<<19)+5,P=998244353,Gi=332748118;inline int add(R int x,R int y){return x+y>=P?x+y-P:x+y;}inline int dec(R int x,R int y){return x-y<0?x-y+P:x-y;}inline int mul(R int x,R int y){return 1ll*x*y-1ll*x*y/P*P;}int ksm(R int x,R int y){    R int res=1;    for(;y;y>>=1,x=mul(x,x))if(y&1)res=mul(res,x);    return res;}int fac[N],ifac[N],inv[N],B[N],C[N],A[N],ans[N],O[N],r[N],a[N];int n,lim,l,len;void init(int len){    lim=1,l=0;while(lim
      
       <<=1,++l;    fp(i,0,lim-1)r[i]=(r[i>>1]>>1)|((i&1)<<(l-1));}void NTT(int *A,int ty){    fp(i,0,lim-1)if(i
       
        >1);    static int A[N],B[N];init(len<<1);    fp(i,0,len-1)A[i]=a[i],B[i]=b[i];    fp(i,len,lim-1)A[i]=B[i]=0;    NTT(A,1),NTT(B,1);    fp(i,0,lim-1)A[i]=mul(A[i],mul(B[i],B[i]));    NTT(A,-1);    fp(i,0,len-1)b[i]=dec(add(b[i],b[i]),A[i]);}void qwq(int len){    B[0]=inv[0]=inv[1]=fac[0]=fac[1]=ifac[0]=ifac[1]=1;    fp(i,2,len+5){        fac[i]=mul(fac[i-1],i),        inv[i]=mul(P-P/i,inv[P%i]),        ifac[i]=mul(ifac[i-1],inv[i]);    }    fp(i,0,len-1)A[i]=ifac[i+1];    Inv(A,B,len);    fp(i,0,len-1)B[i]=mul(B[i],fac[i]);}int main(){//  freopen("testdata.in","r",stdin);    n=read(),len=1;while(len<=n)len<<=1;    qwq(len);    fp(i,0,n)a[i]=read(),B[i]=mul(B[i],ifac[i]),C[i]=mul(a[i],fac[i]);    reverse(B,B+n+1);init((n<<1)+1);    fp(i,n+1,lim-1)B[i]=C[i]=0;    NTT(B,1),NTT(C,1);    fp(i,0,lim-1)B[i]=mul(B[i],C[i]);    NTT(B,-1);    fp(i,0,n+1)B[n+i-1]=mul(B[n+i-1],ifac[i]);    fp(i,0,n)B[n+i-1]=add(B[n+i-1],a[i]);    print(a[0]);    fp(i,1,n+1)print(B[n+i-1]);    return Ot(),0;}