AtCoder

Problem Statement

 

One day, Takahashi was given the following problem from Aoki:

• You are given a tree with N vertices and an integer K. The vertices are numbered 1 through N. The edges are represented by pairs of integers (a_i,b_i).
• For a set S of vertices in the tree, let f(S) be the minimum number of the vertices in a subtree of the given tree that contains all vertices in S.
• There are  ways to choose K vertices from the trees. For each of them, let S be the set of the chosen vertices, and find the sum of f(S) over all  ways.
• Since the answer may be extremely large, print it modulo 924844033(prime).

Since it was too easy for him, he decided to solve this problem for all K=1,2,…,N.

Constraints

 

• 2≦N≦200,000
• 1≦a_i,b_i≦N
• The given graph is a tree.

Input

 

The input is given from Standard Input in the following format:

N
a1 b1
a2 b2
:
aN−1 bN−1

Output

 

Print N lines. The i-th line should contain the answer to the problem where K=i, modulo 924844033.

Sample Input 1

 

3
1 2
2 3

Sample Output 1

 

3
7
3

The diagram above illustrates the case where K=2. The chosen vertices are colored pink, and the subtrees with the minimum number of vertices are enclosed by red lines.

Sample Input 2

 

4
1 2
1 3
1 4

Sample Output 2

 

4
15
13
4

Sample Input 3

 

7
1 2
2 3
2 4
4 5
4 6
6 7

Sample Output 3

 

7
67
150
179
122
45
7


    考虑每条边的贡献，对于k==i的答案显然会贡献 C(n,i) - C(s,i) - C(n-s,i) ，其中s是边一端的子树大小。
    这玩意暴力算显然是 O(N^2) 的，想一想还可以怎么优化。
    
    显然每条边的第一项 C(n,i) 是常量，我们最后对于每个i加上即可；
    后面的组合数可以拆成 阶乘和阶乘的逆的乘积的形式，所以我们就可以构造两个多项式：
A = ∑ cnt[s] * x^s   (cnt[s] 是 子树大小为s的子树个数)
B = Σ 1/(i!) * x^(-i)
    这两个多项式卷出来就可以得到每个k==i的答案(别忘了每个位置再乘一个阶乘的逆元)

     (只有我一个人被模数坑了吗QWQ)

/*
    对于一条把树分成s和n-s的边
	C(s,k) -> s! / k! / (s-k)!
	C(n-s,k) -> (n-s)! / k! / (n-s-k)! 
*/
#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int maxn=600005,ha=924844033,root=5;
inline int add(int x,int y){ x+=y; return x>=ha?x-ha:x;}
inline void ADD(int &x,int y){ x+=y; if(x>=ha) x-=ha;}

inline int ksm(int x,int y){
	int an=1;
	for(;y;y>>=1,x=x*(ll)x%ha) if(y&1) an=an*(ll)x%ha;
	return an;
}

int jc[maxn+5],ni[maxn+5],n,m,ans,to[maxn*2],ne[maxn*2],num,inv;
int a[maxn+5],b[maxn+5],r[maxn+5],siz[maxn+5],hd[maxn],N,l,INV;

inline void addline(int x,int y){ to[++num]=y,ne[num]=hd[x],hd[x]=num;}

inline int C(int x,int y){ return x<y?0:jc[x]*(ll)ni[y]%ha*(ll)ni[x-y]%ha;}

inline void init(){
	jc[0]=1;
	for(int i=1;i<=maxn;i++) jc[i]=jc[i-1]*(ll)i%ha;
	ni[maxn]=ksm(jc[maxn],ha-2);
	for(int i=maxn;i;i--) ni[i-1]=ni[i]*(ll)i%ha;
}

void dfs(int x,int fa){
	siz[x]=1;
	for(int i=hd[x];i;i=ne[i]) if(to[i]!=fa) dfs(to[i],x),siz[x]+=siz[to[i]];
	if(x!=1) ADD(a[siz[x]],jc[siz[x]]),ADD(a[n-siz[x]],jc[n-siz[x]]);
}

inline void NTT(int *c,int f){
	for(int i=0;i<N;i++) if(i<r[i]) swap(c[i],c[r[i]]);
	
	for(int i=1;i<N;i<<=1){
		int omega=ksm(f==1?root:inv,(ha-1)/(i<<1));
		for(int j=0,P=i<<1;j<N;j+=P){
			int now=1;
			for(int k=0;k<i;k++,now=now*(ll)omega%ha){
				int x=c[j+k],y=c[j+k+i]*(ll)now%ha;
				c[j+k]=add(x,y);
				c[j+k+i]=add(x,ha-y);
			}
		}
	}
	
	if(f==-1) for(int i=0;i<N;i++) c[i]=c[i]*(ll)INV%ha;
}

inline void solve(){
	dfs(1,1);
	for(int i=0;i<n;i++) b[n-i]=ni[i];
	
	for(N=1;N<=(2*n);N<<=1) l++;
	for(int i=0;i<N;i++) r[i]=(r[i>>1]>>1)|((i&1)<<(l-1));
	
	NTT(a,1),NTT(b,1);
	for(int i=0;i<N;i++) a[i]=a[i]*(ll)b[i]%ha;
	INV=ksm(N,ha-2),NTT(a,-1);
	
	for(int i=1;i<=n;i++) a[i+n]=add(ha-a[i+n]*(ll)ni[i]%ha,C(n,i)*(ll)n%ha);
}

int main(){
//	freopen("data.in","r",stdin);
//	freopen("data.out","w",stdout);
	
	init(),inv=ksm(5,ha-2);
	
	scanf("%d",&n);
	
	int uu,vv;
	for(int i=1;i<n;i++) scanf("%d%d",&uu,&vv),addline(uu,vv),addline(vv,uu);
	
	solve();
	
	for(int i=1;i<=n;i++) printf("%d
",a[i+n]);
	return 0;
}