Description
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 （ai,bi）.
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 C(n,k) 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 C(n,k) 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.
2≤N≤200000，1≤ai,bi≤N.The given graph is a tree.
Input
Output
Print N lines. The i-th line should contain the answer to the problem where K=i, modulo 924844033.题意：给定一棵 $n$ 个节点的树，选出 $k$ 个特殊点，假设点集为 $S$，令 $f(S)$ 为最小的包含这 $k$ 个节点的连通块，分别求出 $k=1...n$ 在所有情况下的 $f(S)$ 的和。
分析：
考虑暴力，一个点被统计在连通块内，即在以它为根时，选出来的 $k$ 个点都在它的同一个儿子的子树内。即节点 $x$ 被统计进答案的次数 $g(x)$ 为：
$$g(x)=inom{n}{k}-sum _{(x,i)subseteq E}inom{sz_{i}}{k}$$
令 $cnt_{x}$ 表示上述公式里有多少个 $sz_{i}=x$，那么可以得到：
$$ans_{k}=sum _{i=1}^{n}cnt_{i}cdotinom{i}{k}$$
整理可得：
$$k!cdot ans_{k}=sum _{i=1}^{n}frac{cnt_{i}cdot i!}{(i-k)!}$$
令 $a_{i}=cnt_{i}cdot i!$，$b_{i}=(n-i)!$，则可得：
$$k!cdot ans_{k}=sum _{i=1}^{n}a_{i}cdot b_{n-i+k}$$
最终答案为 $ncdot inom{n}{k}-ans_{k}$ 。
1 #include<cstdio> 2 #include<algorithm> 3 #include<cstring> 4 #define LL long long 5 using namespace std; 6 const int N=2e5+5; 7 const int M=524288+5; 8 const int mod=924844033; 9 int n,nn,cnt,u,v,ans,first[N],fac[N],inv[N]; 10 int num[N],sz[N],a[M],b[M]; 11 struct edge{int to,next;}e[N*2]; 12 int read() 13 { 14 int x=0,f=1;char c=getchar(); 15 while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();} 16 while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();} 17 return x*f; 18 } 19 void ins(int u,int v){e[++cnt]=(edge){v,first[u]};first[u]=cnt;} 20 void dfs(int x,int fa) 21 { 22 sz[x]=1; 23 for(int i=first[x];i;i=e[i].next) 24 { 25 int to=e[i].to; 26 if(to==fa)continue; 27 dfs(to,x); 28 sz[x]+=sz[to]; 29 } 30 if(fa!=-1)a[sz[x]]++,a[n-sz[x]]++; 31 } 32 int power(int a,int b) 33 { 34 int ans=1; 35 while(b) 36 { 37 if(b&1)ans=1ll*ans*a%mod; 38 a=1ll*a*a%mod;b>>=1; 39 } 40 return ans; 41 } 42 void ntt(int *a,int n,int f) 43 { 44 int k=0;while((1<<k)<n)k++; 45 for(int i=0;i<n;i++) 46 { 47 int t=0; 48 for(int j=0;j<k;j++) 49 if(i&(1<<j))t|=(1<<(k-j-1)); 50 if(i<t)swap(a[i],a[t]); 51 } 52 for(int l=2;l<=n;l<<=1) 53 { 54 int m=l>>1,nw=power(5,(mod-1)/l); 55 if(f==-1)nw=power(nw,mod-2); 56 for(int *p=a;p!=a+n;p+=l) 57 { 58 int w=1; 59 for(int i=0;i<m;i++) 60 { 61 int t=1ll*p[m+i]*w%mod; 62 p[m+i]=(p[i]-t+mod)%mod; 63 p[i]=(p[i]+t)%mod; 64 w=1ll*w*nw%mod; 65 } 66 } 67 } 68 if(f==-1) 69 { 70 int inv=power(n,mod-2); 71 for(int i=0;i<n;i++)a[i]=1ll*a[i]*inv%mod; 72 } 73 } 74 int main() 75 { 76 n=read(); 77 for(int i=1;i<n;i++) 78 { 79 u=read();v=read(); 80 ins(u,v);ins(v,u); 81 } 82 dfs(1,-1); 83 fac[0]=1; 84 for(int i=1;i<=n;i++)fac[i]=1ll*fac[i-1]*i%mod; 85 inv[n]=power(fac[n],mod-2); 86 for(int i=n;i>=1;i--)inv[i-1]=1ll*inv[i]*i%mod; 87 for(int i=1;i<=n;i++)a[i]=1ll*a[i]*fac[i]%mod; 88 for(int i=0;i<=n;i++)b[n-i]=inv[i]; 89 nn=1;while(nn<n+n+1)nn<<=1; 90 ntt(a,nn,1);ntt(b,nn,1); 91 for(int i=0;i<nn;i++)a[i]=1ll*a[i]*b[i]%mod; 92 ntt(a,nn,-1); 93 for(int i=1;i<=n;i++) 94 { 95 ans=1ll*fac[n]*inv[i]%mod*inv[n-i]%mod*n%mod; 96 printf("%lld ",(ans-1ll*a[n+i]*inv[i]%mod+mod)%mod); 97 } 98 return 0; 99 }