题面

题解

首先你得知道什么是拉格朗日反演->

我们列出树的个数的生成函数

\[T(x)=x+\prod_{i\in D}T^i(x)\]

\[T(x)-\prod_{i\in D}T^i(x)=x\]

我们记\(F(x)=T(x)\)，\(G(x)=x-\prod_{i\in D}x^i\)，那么有\(G(F(x))=x\)

根据拉格朗日反演，可得

\[[x^n]F(x)=\frac{1}{n}[x^{-1}]\frac{1}{G(x)^n}\]

//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;}const int N=5e5+5,P=950009857,g=7,Gi=135715694;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 r[N],O[N],F[N],G[N],inv[N],lim,l,n,m;void init(R 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]);    fp(i,len,lim-1)b[i]=0;}void Ln(int *a,int *b,int len){    static int A[N],B[N];    fp(i,1,len-1)A[i-1]=mul(a[i],i);A[len-1]=0;    Inv(a,B,len),init(len<<1);    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],B[i]);    NTT(A,-1);    fp(i,1,len-1)b[i]=mul(A[i-1],inv[i]);b[0]=0;    fp(i,len,lim-1)b[i]=0;}void Exp(int *a,int *b,int len){    if(len==1)return b[0]=1,void();    Exp(a,b,len>>1);static int A[N];    Ln(b,A,len),init(len<<1);    A[0]=dec(a[0]+1,A[0]);    fp(i,1,len-1)A[i]=dec(a[i],A[i]);    fp(i,len,lim-1)A[i]=b[i]=0;    NTT(A,1),NTT(b,1);    fp(i,0,lim-1)b[i]=mul(b[i],A[i]);    NTT(b,-1);    fp(i,len,lim-1)b[i]=0;}void ksm(int *a,int *b,int len,int k){    static int A[N];    Ln(a,A,len);    fp(i,0,len-1)A[i]=mul(A[i],k);    Exp(A,b,len);}int Lagrange(int *a,int len,int k){    static int A[N],B[N];    Inv(a,A,len),ksm(A,B,len,k);    return mul(B[k-1],inv[k]);}int main(){//  freopen("testdata.in","r",stdin);    n=read(),m=read();    int len=1;while(len<=n)len<<=1;    inv[0]=inv[1]=1;fp(i,2,len)inv[i]=1ll*(P-P/i)*inv[P%i]%P;    ++F[0];while(m--)--F[read()-1];    printf("%d\n",Lagrange(F,len,n));    return 0;}