射命丸文的筆記

題目連結：luogu P4233

題目大意

問你有每個 n 個點的有哈密頓迴路的競賽圖的期望哈密頓迴路數。
其中哈密頓迴路就是從一個點出發不重複的經過每個點最終回到自己的路徑。
競賽圖就是任意兩點之間都有一條有向邊。

思路

考慮求出所有圖的哈密頓迴路總數和有多少個圖有哈密頓迴路。

前者比較好求，你就選一套路徑 \(\dfrac{n!}{n}=(n-1)!\)，然後其它邊就可以隨意了 \(2^{(\frac{n(n-1)}{2}-n)}\)。

那接著問題就是後者，考慮一個容斥 DP：
\(f_n=2^{\frac{n(n-1)}{2}}-\sum\limits_{i=1}^{n-1}f_i\binom{n}{i}2^{\frac{(n-i)(n-i-1)}{2}}\)

然後你考慮優化：
發現在 \(i=n\) 的時候 \(\binom{n}{i}2^{\frac{(n-i)(n-i-1)}{2}}\) 這個東西是 \(1\)，那我們完全可以把右邊的減移到左邊，從而變成這個：
\(\sum\limits_{i=1}^{n}f_i\binom{n}{i}2^{\frac{(n-i)(n-i-1)}{2}}=2^{\frac{n(n-1)}{2}}\)

然後發現我們設 \(F=\sum\limits_{i}\dfrac{f_i}{i!}x^i,G=\sum\limits_{i}\dfrac{2^{\frac{i(i-1)}{2}}}{i!}x^i\)

程式碼

#include<cstdio>
#include<cstring>
#include<algorithm>
#define ll long long
#define mo 998244353

#define clear(f, n) memset(f, 0, (n) * sizeof(ll))
#define cpy(f, g, n) memcpy(f, g, (n) * sizeof(ll))

using namespace std;

int n, an[800001], limit, l_size;
ll f[800001], G, Gv, g[800001];
ll w[800001], r[800001], tmp[800001];
ll jc[800001], inv[800001];

ll ksm(ll x, ll y) {
	ll re = 1;
	while (y) {
		if (y & 1) re = re * x % mo;
		x = x * x % mo;
		y >>= 1;
	}
	return re;
}

void get_an() {
	for (int i = 0; i < limit; i++)
		an[i] = (an[i >> 1] >> 1) | ((i & 1) << (l_size - 1));
}

void NTT(ll *f, ll op) {
	get_an();
	for (int i = 0; i < limit; i++)
		if (i < an[i]) swap(f[i], f[an[i]]);
	
	for (int mid = 1; mid < limit; mid <<= 1) {
		ll Wn = ksm(op == 1 ? G : Gv, (mo - 1) / (mid << 1));
		for (int R = (mid << 1), j = 0; j < limit; j += R) {
			ll w = 1;
			for (int k = 0; k < mid; k++, w = w * Wn % mo) {
				ll x = f[j + k], y = w * f[j + mid + k] % mo;
				f[j + k] = (x + y) % mo;
				f[j + mid + k] = (x - y + mo) % mo;
			}
		}
	}
	
	if (op == -1) {
		ll liv = ksm(limit, mo - 2);
		for (int i = 0; i < limit; i++)
			f[i] = f[i] * liv % mo;
	}
}

void px(ll *x, ll *y) {
	for (int i = 0; i < limit; i++)
		x[i] = x[i] * y[i] % mo;
}

void cheng(ll *x, int n, ll *y, int m) {
	limit = 1; l_size = 0;
	while (limit < n + m + 1) {
		limit <<= 1; l_size++;
	}
	NTT(x, 1); NTT(y, 1);
	px(x, y); NTT(x, -1);
}

void invp(ll *F, int n) {
	w[0] = ksm(F[0], mo - 2);
	l_size = 0;
	for (int len = 2; (len >> 1) <= n; len <<= 1) {
		limit = len; l_size++;
		for (int i = 0; i < (len >> 1); i++) r[i] = w[i];
		cpy(tmp, F, len);
		NTT(tmp, 1); NTT(r, 1);
		px(r, tmp); NTT(r, -1);
		clear(r, (len >> 1));
		cpy(tmp, w, len);
		NTT(tmp, 1); NTT(r, 1);
		px(r, tmp); NTT(r, -1);
		
		for (int i = len >> 1; i < len; i++)
			w[i] = (w[i] * 2 - r[i] + mo) % mo;
	}
	cpy(F, w, n);
	clear(tmp, n); clear(w, n); clear(r, n);
}

int main() {
	G = 3; Gv = ksm(G, mo - 2);
	
	scanf("%d", &n);
	jc[0] = 1; for (int i = 1; i <= n; i++) jc[i] = jc[i - 1] * i % mo;
	inv[0] = inv[1] = 1; for (int i = 2; i <= n; i++) inv[i] = inv[mo % i] * (mo - mo / i) % mo;
	for (int i = 1; i <= n; i++) inv[i] = inv[i - 1] * inv[i] % mo;
	for (int i = 0; i <= n; i++) {
		f[i] = ksm(2, 1ll * i * (i - 1) / 2) * inv[i] % mo;
	}
	
	cpy(g, f, n + 1); g[0] = (g[0] - 1 + mo) % mo;
	invp(f, n + 1);
	cheng(f, n + 1, g, n + 1);
	
	for (int i = 1; i <= n; i++) {
		if (i == 1) {
			printf("1\n"); continue;
		}
		if (i == 2) {
			printf("-1\n"); continue;
		}
		f[i] = f[i] * jc[i] % mo;
		ll all = ksm(2, 1ll * i * (i - 1) / 2 - i) * jc[i - 1] % mo;
		printf("%lld\n", all * ksm(f[i], mo - 2) % mo);
	}
	
	return 0;
}