題面
題解
這當前處理的點集大小爲\(k\)，那麼考慮將每個點的貢獻拆開來算，那麼如果這\(K\)個點都在以\(x\)爲根的一棵子樹內，這個點就沒有貢獻
令\(size_x\)表示\(x\)子樹的大小，有
\[ f(k)={N \choose k} - \sum_{x=1}^N\sum_{(x,v)}{ size_v \choose K} \]
令\(cnt_i\)表示無根樹大小爲\(i\)的子樹個數，那麼有，
\[ f(k)=\sum_{i=k}^N cnt_i\times \frac{i!}{k!(i-k)!} \]
其中\(\frac {1}{k!}\)可以最後再算。
令\(g(i)=cnt_i\times i!\)，\(h(i)=\frac {1}{i!}\)，注意這個式子兩字母之差相等，考慮將\(g\)翻轉，
有
\[ \begin{aligned} f(k)&=\sum_{i=k}^{N}g(i)h(i-k)\\ &=\sum_{i=0}^{N-k}g(i+k)h(i)\\ &=\sum_{i=0}^{N-k}g_r(N-i-k)h_i \end{aligned} \]
然後\(\mathcal{NTT}\)就可以了。
代碼
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
inline int gi() {
register int data = 0, w = 1;
register char ch = 0;
while (!isdigit(ch) && ch != '-') ch = getchar();
if (ch == '-') w = -1, ch = getchar();
while (isdigit(ch)) data = 10 * data + ch - '0', ch = getchar();
return w * data;
}
const int Mod = 924844033, G = 5;
int fpow(int x, int y) {
int res = 1;
while (y) {
if (y & 1) res = 1ll * res * x % Mod;
x = 1ll * x * x % Mod;
y >>= 1;
}
return res;
}
const int iG = fpow(G, Mod - 2);
const int MAX_N = 2e5 + 5;
int Limit, rev[MAX_N << 2];
void FFT_prepare(int len) {
int p = 0;
for (Limit = 1; Limit <= len; Limit <<= 1) ++p;
for (int i = 1; i < Limit; i++) rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (p - 1));
}
void NTT(int *p, int op) {
for (int i = 1; i < Limit; i++) if (i < rev[i]) swap(p[i], p[rev[i]]);
for (int i = 1; i < Limit; i <<= 1) {
int rot = fpow(op == 1 ? G : iG, (Mod - 1) / (i << 1));
for (int j = 0; j < Limit; j += i << 1)
for (int k = 0, w = 1; k < i; k++, w = 1ll * w * rot % Mod) {
int x = p[j + k], y = 1ll * w * p[i + j + k] % Mod;
p[j + k] = (x + y) % Mod, p[i + j + k] = (x - y + Mod) % Mod;
}
}
if (!op) {
int inv = fpow(Limit, Mod - 2);
for (int i = 0; i < Limit; i++) p[i] = 1ll * p[i] * inv % Mod;
}
}
struct Graph { int to, next; } e[MAX_N << 1];
int fir[MAX_N], e_cnt;
void clearGraph() { memset(fir, -1, sizeof(fir)); e_cnt = 0; }
void Add_Edge(int u, int v) { e[e_cnt] = (Graph){v, fir[u]}, fir[u] = e_cnt++; }
int N, f[MAX_N << 2], g[MAX_N << 2], h[MAX_N << 2];
int siz[MAX_N], cnt[MAX_N];
void dfs(int x, int fa) {
siz[x] = 1;
for (int i = fir[x]; ~i; i = e[i].next) {
int v = e[i].to; if (v == fa) continue;
dfs(v, x), siz[x] += siz[v];
cnt[siz[v]]++;
}
cnt[N - siz[x]]++;
}
int fac[MAX_N], ifc[MAX_N];
int C(int n, int m) {
if (n < 0 || m < 0 || n < m) return 0;
else return 1ll * fac[n] * ifc[n - m] % Mod * ifc[m] % Mod;
}
int main () {
clearGraph();
N = gi();
for (int i = 1; i < N; i++) {
int u = gi(), v = gi();
Add_Edge(u, v), Add_Edge(v, u);
}
fac[0] = 1; for (int i = 1; i <= N; i++) fac[i] = 1ll * fac[i - 1] * i % Mod;
ifc[N] = fpow(fac[N], Mod - 2);
for (int i = N - 1; ~i; i--) ifc[i] = 1ll * ifc[i + 1] * (i + 1) % Mod;
dfs(1, 0);
for (int i = 1; i <= N; i++) g[i] = 1ll * cnt[i] * fac[i] % Mod;
for (int i = 0; i <= N; i++) h[i] = ifc[i];
reverse(&g[0], &g[N + 1]);
FFT_prepare(N << 1);
NTT(g, 1), NTT(h, 1);
for (int i = 0; i < Limit; i++) f[i] = 1ll * g[i] * h[i] % Mod;
NTT(f, 0);
for (int i = 1; i <= N; i++) {
int ans = 1ll * N * C(N, i) % Mod;
ans = (ans - 1ll * ifc[i] * f[N - i] % Mod + Mod) % Mod;
printf("%d\n", ans);
}
return 0;
}