JZOJ6676. 【2020.06.01省选模拟】查拉图斯特拉如是说 （第二类斯特林数+多项式多点求值）

Description:

https://gmoj.net/senior/#main/show/6676

题解：

考虑用第二类斯特林数把(i^k)搞掉：

(Ans=2^n*f(0)（这一项提出来好，后面不写了）+sum_{k=1}^m f(k)*sum_{i=0}^ninom{n}{i}*i^k)
(=sum_{k=1}^m f(k)*sum_{i=0}^ninom{n}{i}*sum_{j=1}^m inom{i}{j}*s[k][j]*j!)
(=sum_{k=1}^m f(k)*sum_{j=1}^m s[k][j]*j!*sum_{i=0}^ninom{n}{i}*inom{i}{j})
(=sum_{k=1}^m f(k)*sum_{j=1}^m s[k][j]*j!*inom{n}{j}*2^{n-j})

获得了(O(m^2))的简单做法。

继续搞：

设(G[j]=j!*inom{n}{j}*2^{n-j})
(=sum_{k=1}^m f(k)*sum_{j=1}^m s[k][j]*G[j])

斯特林数有一个容斥的求法：
(s[n][m]=frac{1}{m!} sum_{i=1}^{m(这里填infty也行)}~~~~ i^n*inom{m}{i}*(-1)^{m-i})

(原式=sum_{j=1}^m G(j)*frac{1}{j!}*sum_{k=1}^m sum_{i=1}^m i^k*inom{j}{i}*(-1)^{j-i})
(=sum_{j=1}^m G(j)*frac{1}{j!} * sum_{i=1}^minom{j}{i}*(-1)^{j-i}*sum_{k=1}^m f(k)*i^k)

最后一个循环多点求值，再做一遍NTT即可。

Code：

#include<bits/stdc++.h>
#define fo(i, x, y) for(int i = x, _b = y; i <= _b; i ++)
#define ff(i, x, y) for(int i = x, _b = y; i <  _b; i ++)
#define fd(i, x, y) for(int i = x, _b = y; i >= _b; i --)
#define ll long long
#define pp printf
#define hh pp("
")
using namespace std;

const int mo = 998244353;

ll ksm(ll x, ll y) {
	ll s = 1;
	for(; y; y /= 2, x = x * x % mo)
		if(y & 1) s = s * x % mo;
	return s;
}

#define V vector<ll>
#define si size()
#define re resize

namespace ntt {
	const int nm = 1 << 18;
	ll w[nm], a[nm], b[nm];
	int r[nm];
	void build() {
		for(int i = 1; i < nm; i *= 2) {
			w[i] = 1; ll v = ksm(3, (mo - 1) / 2 / i);
			ff(j, 1, i) w[i + j] = w[i + j - 1] * v % mo;
		}
	}
	void dft(ll *a, int n, int f) {
		ff(i, 0, n) {
			r[i] = r[i / 2] / 2 + (i & 1) * (n / 2);
			if(i < r[i]) swap(a[i], a[r[i]]);
		} ll v;
		for(int i = 1; i < n; i *= 2) for(int j = 0; j < n; j += 2 * i) ff(k, 0, i) {
			v = a[i + j + k] * w[i + k], a[i + j + k] = (a[j + k] - v) % mo, a[j + k] = (a[j + k] + v) % mo;
		}
		if(f == -1) {
			reverse(a + 1, a + n);
			v = ksm(n, mo - 2);
			ff(i, 0, n) a[i] = (a[i] + mo) * v % mo;
		}
	}
	V operator * (V p, V q) {
		int n0 = p.si + q.si - 1, n = 1;
		while(n < n0) n *= 2;
		ff(i, 0, n) a[i] = b[i] = 0;
		ff(i, 0, p.si) a[i] = p[i];
		ff(i, 0, q.si) b[i] = q[i];
		dft(a, n, 1); dft(b, n, 1);
		ff(i, 0, n) a[i] = a[i] * b[i] % mo;
		dft(a, n, -1);
		p.re(n0);
		ff(i, 0, n0) p[i] = a[i];
		return p;
	}
	void dft(V &p, int f) {
		int n = p.si;
		ff(i, 0, n) a[i] = p[i];
		dft(a, n, f);
		ff(i, 0, n) p[i] = a[i];
	}
}

using ntt :: operator *;
using ntt :: dft;

V operator - (V a, V b) {
	a.re(max(a.si, b.si));
	ff(i, 0, a.si) a[i] = (a[i] - b[i]) % mo;
	return a;
}

V qni(V a) {
	int n0 = a.si;
	V b; b.re(1); b[0] = ksm(a[0], mo - 2);
	for(int n = 2; n < n0 * 2; n *= 2) {
		V c = a; c.re(n); c.re(2 * n);
		b.re(2 * n);
		dft(c, 1); dft(b, 1);
		ff(i, 0, 2 * n) b[i] = (2 * b[i] - c[i] * b[i] % mo * b[i]) % mo;
		dft(b, -1); b.re(n);
	}
	b.re(n0); return b;
}

V qmo(V a, V b) {
	int n = a.si - 1, m = b.si - 1;
	if(n < m) return a;
	V a0 = a, b0 = b;
	reverse(a.begin(), a.end());
	reverse(b.begin(), b.end());
	b.re(a.si);
	V c = a * qni(b);
	c.re(n - m + 1);
	reverse(c.begin(), c.end());
	V d = a0 - b0 * c;
	d.re(m);
	return d;
}

const int N = 1e5 + 5;

int n, m; ll f[N];
ll fac[N], nf[N], h[N];

void build(int n) {
	fac[0] = 1; fo(i, 1, n) fac[i] = fac[i - 1] * i % mo;
	nf[n] = ksm(fac[n], mo - 2); fd(i, n, 1) nf[i - 1] = nf[i] * i % mo;
}

void build_h() {
	h[0] = 1;
	fo(i, 1, m) h[i] = h[i - 1] * (n - i + 1) % mo;
	fo(i, 0, m) h[i] = h[i] * nf[i] % mo;
}

ll C(int n, int m) {
	if(n < m) return 0;
	return fac[n] * nf[n - m] % mo * nf[m] % mo;
}

ll g[N];

V t[N * 4];

#define i0 i + i
#define i1 i + i + 1
void dg(int i, int x, int y) {
	if(x == y) {
		t[i].re(2);
		t[i][0] = -x; t[i][1] = 1;
		return;
	}
	int m = x + y >> 1;
	dg(i0, x, m); dg(i1, m + 1, y);
	t[i] = t[i0] * t[i1];
}

ll p[N];

void fz(V a, int i, int x, int y) {
	a = qmo(a, t[i]);
	if(y - x + 1 < 128)  {
		fo(j, x, y) {
			ll s = 0, v = 1;
			ff(k, 0, a.si)  {
				s = (s + v * a[k]) % mo;
				v = v * j % mo;
			}
			p[j] = s;
		}
		return;
	}
	if(x == y) {
		p[x] = a[0];
		return;
	}
	int m = x + y >> 1;
	fz(a, i0, x, m); fz(a, i1, m + 1, y);
}

int main() {
	ntt :: build();
	freopen("number.in", "r", stdin);
	freopen("number.out", "w", stdout);
	scanf("%d %d", &n, &m);
	fo(i, 0, m) scanf("%lld", &f[i]);
	build(m);
	build_h();
	ll ans = f[0] * ksm(2, n) % mo;
	fo(j, 1, m) g[j] = fac[j] * ksm(2, n - j) % mo * h[j] % mo;
	V a, b; a.clear(); b.clear();
	a.re(m + 1); b.re(m + 1);
	dg(1, 1, m);
	V c; c.re(m + 1);
	fo(i, 1, m) c[i] = f[i];
	fz(c, 1, 1, m);
	fo(i, 1, m) {
		a[i] = nf[i] * p[i] % mo;
	}
	fo(i, 0, m) b[i] = nf[i] * (i % 2 ? -1 : 1);
	a = a * b;
	fo(i, 1, m) ans = (ans + a[i] * g[i]) % mo;
	ans = (ans % mo + mo) % mo;
	pp("%lld
", ans);
}