【AGC005F】Many Easy Problems

题面

洛谷

题解

这当前处理的点集大小为(k)，那么考虑将每个点的贡献拆开来算，那么如果这(K)个点都在以(x)为根的一棵子树内，这个点就没有贡献

令(size_x)表示(x)子树的大小，有

[f(k)={N choose k} - sum_{x=1}^Nsum_{(x,v)}{ size_v choose K} ]

令(cnt_i)表示无根树大小为(i)的子树个数，那么有，

[f(k)=sum_{i=k}^N cnt_i imes frac{i!}{k!(i-k)!} ]

其中(frac {1}{k!})可以最后再算。

令(g(i)=cnt_i imes i!)，(h(i)=frac {1}{i!})，注意这个式子两字母之差相等，考虑将(g)翻转，

有

[egin{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} ]

然后(mathcal{NTT})就可以了。

代码

#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
", ans); 
	} 
    return 0; 
}