LOJ6570 caterpillar count
tags: generating function, polynomial
The meaning of problems
hsezoi guy olinr like giant caterpillars van, he defined caterpillar is a tree, the tree exists to satisfy a chain tree, making the tree all the points from this tree to chain up to \ (1 \) . Given \ (n-\) \ ((n-\ ^ LE10. 5) \) . Now you find \ (n \) points, a labeled number of caterpillars. The answer to \ (998 244 353 \) modulo.
answer
Structure generating function
For a diameter of the intermediate node caterpillar, the total size of the i i discharge method, exponential generating function is
\[ A(x)=\sum_{i=1}^\infty\frac{ix^i}{i!} \]
For a node connected to both ends of the dot diameter, forced to hang at least one node, generating function is exponential
\[ B(x)=\sum_{i=2}^\infty\frac{ix^i}{i!} \]
Then the result is \ ((A (x) ^ 0 + A (x) ^ 1 + \ cdots) B (x) ^ 2 = \ frac {B (x) ^ 2} {1-A (x)} \)
Output:
\[ \frac{n!}2[x^n]\frac {B(x)^2}{1-A(x)} \]
Note that n = 2 to Japanese sentence
By the way I put down polynomial board need when you can pull
#include<cstdio>
#include<vector>
//#define debug(...) fprintf(stderr,__VA_ARGS__)
#define debug(...) ((void)0)
typedef std::vector<int> poly;
const int P=998244353;
int fpow(int a,int b){int res=1;for(;b;b>>=1,a=1ll*a*a%P)if(b&1)res=1ll*res*a%P;return res;}
void pt(const poly&a){for(int i=0;i<(int)a.size();++i)debug("%d ",a[i]);debug("\n");}
int getlim(int n){int x=1;while(x<=n)x<<=1;return x;}
void ntt(poly&a,int g,int lim){
a.resize(lim);
for(int i=0,j=0;i<lim;++i){
if(i<j)std::swap(a[i],a[j]);
for(int k=lim>>1;(j^=k)<k;k>>=1);
}
poly w(lim>>1);w[0]=1;
for(int i=1;i<lim;i<<=1){
for(int j=1,wn=fpow(g,(P-1)/(i<<1));j<i;++j)w[j]=1ll*w[j-1]*wn%P;
for(int j=0;j<lim;j+=i<<1)
for(int k=0;k<i;++k){
int x=a[j+k],y=1ll*a[i+j+k]*w[k]%P;
a[j+k]=(x+y)%P,a[i+j+k]=(x-y+P)%P;
}
}
if(g==332748118)for(int i=0,I=fpow(lim,P-2);i<(int)a.size();++i)a[i]=1ll*a[i]*I%P;
}
poly pmul(poly a,poly b){
int need=(int)a.size()+b.size()-1,lim=getlim(need);
ntt(a,3,lim),ntt(b,3,lim);
for(int i=0;i<lim;++i)a[i]=1ll*a[i]*b[i]%P;
ntt(a,332748118,lim);
return a.resize(need),a;
}
poly padd(poly a,poly b){
if(a.size()<b.size()){
for(int i=0;i<(int)a.size();++i)(b[i]+=a[i])%=P;
return b;
}else{
for(int i=0;i<(int)b.size();++i)(a[i]+=b[i])%=P;
return a;
}
}
poly pinv(const poly&a,int n=-1){
if(n==-1)n=a.size();
if(n==1)return poly(1,fpow(a[0],P-2));
poly b=pinv(a,(n+1)>>1),tmp=poly(a.begin(),a.begin()+n);
int lim=getlim(n*2-2);
ntt(b,3,lim),ntt(tmp,3,lim);
for(int i=0;i<lim;++i)b[i]=(2-1ll*b[i]*tmp[i]%P+P)%P*b[i]%P;
ntt(b,332748118,lim);
return b.resize(n),b;
}
poly pdao(const poly&a){
poly b((int)a.size()-1);
for(int i=1;i<(int)a.size();++i)b[i-1]=1ll*a[i]*i%P;
return b;
}
poly pji(const poly&a){
poly b((int)a.size()+1);
for(int i=0;i<(int)a.size();++i)b[i+1]=1ll*a[i]*fpow(i+1,P-2)%P;
return b;
}
poly pln(const poly&a){
poly b(pmul(pdao(a),pinv(a)));
b.resize((int)a.size()-1);
return pji(b);
}
poly pexp(const poly&a,int n=-1){
if(n==-1)n=a.size();
if(n==1)return poly(1,1);
poly b=pexp(a,(n+1)>>1),c(b);
c.resize(n),c=pln(c),--c[0];
for(int i=0;i<n;++i)c[i]=(a[i]-c[i]+P)%P;
poly d(pmul(b,c));
return d.resize(n),d;
}
const int N=100005;
int n,fac[N],inv[N];
int main(){
fac[0]=fac[1]=inv[0]=inv[1]=1;
for(int i=2;i<N;++i)fac[i]=1ll*fac[i-1]*i%P,inv[i]=1ll*(P-P/i)*inv[P%i]%P;
for(int i=2;i<N;++i)inv[i]=1ll*inv[i]*inv[i-1]%P;
scanf("%d",&n);if(n<=2)return puts("1"),0;
poly A(n+1),B(n+1);A[0]=1;
for(int i=1;i<=n;++i)A[i]=(P-1ll*i*inv[i]%P)%P;
for(int i=2;i<=n;++i)B[i]=1ll*i*inv[i]%P;
A=pinv(A),B=pmul(B,B),B.resize(n+1),A=pmul(A,B),A.resize(n+1);
printf("%lld\n",(n+1ll*A[n]*fac[n]%P*((P+1)>>1)%P)%P);
return 0;
}