洛谷P3711 仓鼠的数学题（伯努利数+多项式求逆）

题面

传送门

题解

如果您不知道伯努利数是什么可以去看看这篇文章

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<bits/stdc++.h> #define R register #define fp(i,a,b) for(R int i=(a),I=(b)+1;i<I;++i) #define fd(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<len)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<r[i])swap(A[i],A[r[i]]); for(R int mid=1;mid<lim;mid<<=1){ R int I=(mid<<1),Wn=ksm(ty==1?3:Gi,(P-1)/I),t;O[0]=1; fp(i,1,mid-1)O[i]=mul(O[i-1],Wn); for(R int j=0;j<lim;j+=I)fp(k,0,mid-1) A[j+k+mid]=dec(A[j+k],t=mul(O[k],A[j+k+mid])), A[j+k]=add(A[j+k],t); } if(ty==-1)for(R int i=0,inv=ksm(lim,P-2);i<lim;++i)A[i]=mul(A[i],inv); } void Inv(int *a,int *b,int len){ if(len==1)return b[0]=ksm(a[0],P-2),void(); Inv(a,b,len>>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; }