BZOJ5093 [Lydsy1711月赛]图的价值 【第二类斯特林数 + NTT】

题目链接

BZOJ5093

题解

点之间是没有区别的，所以我们可以计算出一个点的所有贡献，然后乘上\(n\)

一个点可能向剩余的\(n - 1\)个点连边，那么就有

\[ans = 2^{{n - 1 \choose 2}}n \sum\limits_{i = 0}^{n - 1} {n - 1 \choose i} i^k
\]

显然要求

\[\sum\limits_{i = 0}^{n} {n \choose i} i^k
\]

然后我就不知道怎么做了。。

翻翻题解

有这样一个结论：

\[n^k = \sum\limits_{i = 0}^{k} \begin{Bmatrix} k \\ i \end{Bmatrix} {n \choose i} i!
\]

那么就有

\[\begin{aligned}
\sum\limits_{i = 0}^{n} {n \choose i} i^k &= \sum\limits_{i = 0}^{n} {n \choose i} \sum\limits_{j = 0}^{i} \begin{Bmatrix} k \\ j \end{Bmatrix} {i \choose j}j! \\
&= \sum\limits_{j = 0}^{n}\begin{Bmatrix} k \\ j \end{Bmatrix} j! \sum\limits_{i = j}^{n} {n \choose i} {i \choose j} \\
&= \sum\limits_{j = 0}^{n}\begin{Bmatrix} k \\ j \end{Bmatrix} j! {n \choose j} 2^{n - j} \\
\end{aligned}
\]

解释一下最后一步

\[\sum\limits_{i = j}^{n} {n \choose i} {i \choose j}
\]

直观来看是从\(n\)中取出\(i\)个，然后从\(i\)中取出\(j\)个

实际上等价于从\(n\)中取出\(j\)个，剩余随便取

最后只需要求出第二类斯特林数，用第二类斯特林反演即可

\[\begin{aligned}
\begin{Bmatrix} n \\ m \end{Bmatrix} &= \frac{1}{m!} \sum\limits_{i = 0}^{m} (-1)^{i}{m \choose i}(m - i)^{n} \\
&= \sum\limits_{i = 0}^{m} \frac{(-1)^{i}}{i!} \times \frac{(m - i)^{n}}{(m - i)!} \\
\end{aligned}
\]

\(NTT\)即可

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<map>
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define mp(a,b) make_pair<int,int>(a,b)
#define cls(s) memset(s,0,sizeof(s))
#define cp pair<int,int>
#define LL long long int
using namespace std;
const int maxn = 800005,maxm = 100005,INF = 1000000000;
inline int read(){
	int out = 0,flag = 1; char c = getchar();
	while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}
	while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
	return out * flag;
}
const int G = 3,P = 998244353;
inline int qpow(int a,LL b){
	int re = 1;
	for (; b; b >>= 1,a = 1ll * a * a % P)
		if (b & 1) re = 1ll * re * a % P;
	return re;
}
int R[maxn];
void NTT(int* a,int n,int f){
	for (int i = 0; i < n; i++) if (i < R[i]) swap(a[i],a[R[i]]);
	for (int i = 1; i < n; i <<= 1){
		int gn = qpow(G,(P - 1) / (i << 1));
		for (int j = 0; j < n; j += (i << 1)){
			int g = 1,x,y;
			for (int k = 0; k < i; k++,g = 1ll * g * gn % P){
				x = a[j + k],y = 1ll * g * a[j + k + i] % P;
				a[j + k] = (x + y) % P,a[j + k + i] = ((x - y) % P + P) % P;
			}
		}
	}
	if (f == 1) return;
	int nv = qpow(n,P - 2); reverse(a + 1,a + n);
	for (int i = 0; i < n; i++) a[i] = 1ll * a[i] * nv % P;
}
int fac[maxn],inv[maxn],fv[maxn],C[maxn];
int S[maxn],B[maxn],N,K;
void init(){
	fac[0] = fac[1] = inv[0] = inv[1] = fv[0] = fv[1] = 1;
	for (int i = 2; i <= K; i++){
		fac[i] = 1ll * fac[i - 1] * i % P;
		inv[i] = 1ll * (P - P / i) * inv[P % i] % P;
		fv[i] = 1ll * fv[i - 1] * inv[i] % P;
	}
	C[0] = 1; int E = min(N - 1,K);
	for (int i = 1; i <= E; i++) C[i] = 1ll * C[i - 1] * (N - i) % P * inv[i] % P;
}
void work(){
	int n = 1,L = 0;
	for (int i = 0; i <= K; i++){
		S[i] = (((i & 1) ? -1 : 1) * fv[i] + P) % P;
		B[i] = 1ll * qpow(i,K) * fv[i] % P;
	}
	while (n <= (K << 1)) n <<= 1,L++;
	for (int i = 1; i < n; i++) R[i] = (R[i >> 1] >> 1) | ((i & 1) << (L - 1));
	NTT(S,n,1); NTT(B,n,1);
	for (int i = 0; i < n; i++) S[i] = 1ll * S[i] * B[i] % P;
	NTT(S,n,-1);
	//REP(i,10) printf("%d ",S[i]); puts("");
	int ans = 0;
	for (int i = 0; i < N; i++){
		if (i > K) break;
		ans = (ans + 1ll * S[i] * fac[i] % P * C[i] % P * qpow(2,N - 1 - i) % P) % P;
	}
	ans = 1ll * ans * N % P * qpow(2,1ll * (N - 1) * (N - 2) / 2) % P;
	printf("%d\n",ans);
}
int main(){
	N = read(); K = read();
	if (N == 1){puts("0"); return 0;}
	if (N == 2){puts("2"); return 0;}
	init();
	work();
	return 0;
}