2021ICPC济南站 L. Strange Series 题解

2021ICPC济南站 L. Strange Series 题解

L. Strange Series

题意 ($2\mathrm{s}$)

给定一个$n$次多项式$P(x)=a_0+a_{1}x+\cdots +a_n{x}^n$,其中$a_i\ge 0(i=0,\cdots ,n)$.考察级数$S=\sum _{m=0}^{\infty }\frac{P(m)}{m!}$.可以证明$S$是良好定义的,且$S=p\mathrm{e}$,其中$p$是一个非负整数,$\mathrm{e}=\sum _{m=0}^{\infty }\frac{1}{m!}\approx 2.71828$.求$p$.

有$t(1\le t\le 100)$组测试数据.每组测试数据输入一个整数$n(0\le n\le 1\mathrm{e}5)$.第二行输入$(n+1)$个整数$a_0,a_1,\cdots ,a_n(0\le a_i<998244353)$.数据保证至多有$4$组测试数据满足$n>1000$.

对每组测试数据,输出$p$对$998244353$取模的结果.

思路

[引理] $\sum _{m=0}^{\infty }\frac{m^n}{m!}{x}^m={\mathrm{e}}^x\cdot Q(x)$,其中$Q(x)$是关于$x$的$n$次多项式.

[证] 用数学归纳法证明.不妨设$n-1$时结论成立.

$\sum _{m=0}^{\infty }\frac{1}{m!}\cdot m^n{x}^m=x\sum _{m=0}^{\infty }\frac{1}{m!}\cdot m^{n-1}\cdot m{x}^{m-1}=x\sum _{m=0}^{\infty }\frac{1}{m!}\cdot m^{n-1}\cdot \frac{\mathrm{d}}{\mathrm{d}x}(x^m)=x\cdot \frac{\mathrm{d}}{\mathrm{d}x}\sum _{m=0}^{\infty }\frac{1}{m!}\cdot m^{n-1}{x}^m$

​ $=x\cdot \frac{\mathrm{d}}{\mathrm{d}x}[{\mathrm{e}}^{x}Q(x)]$,其中$Q(x)$是关于$x$的$(n-1)$次多项式.

注意到对${\mathrm{e}}^{x}Q(x)$求导后多项式部分的最高次项仍为$x^{n-1}$,乘上前面的$x$即证.

$P(x)=a_0+a_{1}x+\cdots +a_n{x}^n(a_i\ge 0),S=\sum _{m=0}^{\infty }\frac{P(m)}{m!}=p\mathrm{e}$,则$p=\frac{1}{\mathrm{e}}S=\frac{1}{\mathrm{e}}\underset{s\to \infty }{\lim }\sum _{m=0}^s\frac{P(m)}{m!}$.

因$\frac{P(m)}{m!}=\frac{1}{m!}(a_0+a_{1}m+\cdots +a_n{m}^n)=\sum _{i=0}^n{a}_i\frac{k^i}{k!}$,

​ 则$p=\frac{1}{\mathrm{e}}\underset{s\to \infty }{\lim }\sum _{m=0}^s\sum _{i=0}^n{a}_i\frac{m^i}{m!}=\sum _{i=0}^n{a}_i\left(\frac{1}{\mathrm{e}}\underset{s\to \infty }{\lim }\sum _{m=0}^s\frac{m^i}{m!}\right)=\sum _{i=0}^n{a}_i\left[\frac{1}{\mathrm{e}}\sum _{m=0}^{\infty }\frac{m^i}{m!}\right]$.

由引理知:$\sum _{m=0}^{\infty }\frac{m^i}{m!}{x}^m={\mathrm{e}}^x\cdot Q(x)$,其中$Q(x)$是关于$x$的$i$次多项式,进而$p=\sum _{i=0}^n{a}_i\cdot Q(1)$,同时也证明了$p\in \mathbb{N}$.

下面讨论如何求$Q(x)$的系数.

$i=0$时,$\sum _{m=0}^{\infty }\frac{1}{m!}{x}^m={\mathrm{e}}^x\cdot 1$,此时$Q(x)=1,Q(1)=1$.

$i=1$时,$\sum _{m=1}^{\infty }\frac{m}{m!}{x}^m={\mathrm{e}}^x\cdot x$,此时$Q(x)=x,Q(1)=1$.

$i=2$时,$\sum _{m=1}^{\infty }\frac{m^2}{m!}{x}^m={\mathrm{e}}^x\cdot x(x+1)$,此时$Q(x)=x(x+1),Q(1)=2$.

$i=3$时,$\sum _{m=1}^{\infty }\frac{m^3}{m!}{x}^m={\mathrm{e}}^x\cdot x(x^2+3x+1)$,此时$Q(x)=x(x^2+3x+1),Q(1)=5$.

$i=4$时,$\sum _{m=1}^{\infty }\frac{m^4}{m!}{x}^m={\mathrm{e}}^x\cdot x(x^3+6x^2+7x+1)$,此时$Q(x)=x(x^3+6x^2+7x+1),Q(1)=15$.

$i=5$时,$\sum _{m=1}^{\infty }\frac{m^5}{m!}{x}^m={\mathrm{e}}^x\cdot x(x^4+10x^3+25x^2+15x+1)$,

​ 此时$Q(x)=x(x^4+10x^3+25x^2+15x+1),Q(1)=52$.

易联想到Bell数:$B_0=1,B_1=1,B_2=2,B_3=5,B_4=15,B_5=52,\cdots$,它可用递推求得.

注意到$n$最大为$1\mathrm{e}5$,暴力递推出Bell数会TLE.考虑用Bell数的生成函数求Bell数.

Bell数的定义:将$n$个不同的元素的集合划分为任意多个非空子集的方案数.

设$n$个元素的方案数为$f_n$.固定最后一个元素$x$,枚举最后一个元素所在的集合的元素个数$i\in [1,n]$,从剩下的$(n-1)$个元素中选$(i-1)$个元素与$x$在同一集合中,其他$(n-i)$个元素任意划分,故$f_n=\sum _{i=1}^n{C}_{n-1}^{i-1}\cdot f_{n-i}$,调整下标得$f_{n+1}=\sum _{i=0}^n{C}_n^i\cdot f_{n-i}$.

设$f$的EGF为$F(x)$,则$\frac{f_{n+1}}{(n+1)!}=\sum _{i=0}^n\frac{f_{n-i}}{(n-i)!}\cdot \frac{1}{i!(n+1)}$,进而$(n+1)\cdot \frac{f_{n+1}}{(n+1)!}=\sum _{i=0}^n\frac{f_{n-i}}{(n-i)!}\cdot \frac{1}{i!}$,

​ 即$[x^n]F’(x)=\sum _{i=0}^n[x^{n-i}]F(x)\cdot [x^i]{\mathrm{e}}^x$,亦即$F^{\prime }(x)=F(x){\mathrm{e}}^x$.

令$y=F(x)$,则$\frac{\mathrm{d}y}{\mathrm{d}x}={\mathrm{e}}^{x}y$,移项得$\frac{\mathrm{d}y}{y}={\mathrm{e}}^x\mathrm{d}x$,两边积分得$\ln y={\mathrm{e}}^x+C$,故$y=\exp ({\mathrm{e}}^x+C)$.

$n=1$时只有一种划分方案,代入得$C=-1$,故$F(x)=\exp ({\mathrm{e}}^x-1)$.多项式exp求即可,时间复杂度$O(n\log n)$.

代码

const int MAXN = 6e5 + 5;
const int MOD = 998244353;

int rev[MAXN];  // rev[x]表示二进制数x镜像后的结果

void NTT(int* y, int lim, int op) {  // op=1时为NTT,op=-1时为INTT
	for (int i = 0; i < lim; i++)  // 做一遍BRP,将系数位置调整为递归后的位置
		if (rev[i] < i) swap(y[i], y[rev[i]]);

	for (int l = 2; l <= lim; l <<= 1) {  // 模拟合并过程,l为当前区间长度
		int g = qpow(3, (MOD - 1) / l, MOD);
		for (int i = 0; i < lim; i += l) {
			int cur = 1;
			int k = l >> 1;
			for (int j = 0; j < k; j++, cur = (ll)cur * g % MOD) {
				int tmp = (ll)y[i + j + k] * cur % MOD;
				y[i + j + k] = (y[i + j] - tmp + MOD) % MOD;  // 奇次项
				y[i + j] = ((ll)y[i + j] + tmp) % MOD;  // 偶次项
			}
		}
	}

	if (op == -1) {  // INTT
		reverse(y + 1, y + lim);
		int inv = qpow(lim, MOD - 2, MOD);
		for (int i = 0; i < lim; i++) y[i] = (ll)y[i] * inv % MOD;
	}
}

void PolyMul(int* a, int* b, int n, int m) {  // n次多项式a乘m次多项式b,结果放在a中
	int lim = 1, len = 0;
	while (lim < (n + m << 1)) lim <<= 1, len++;
	for (int i = 1; i < lim; i++) rev[i] = (rev[i >> 1] >> 1 | ((i & 1) << (len - 1)));

	NTT(a, lim, 1), NTT(b, lim, 1);
	for (int i = 0; i < lim; i++) a[i] = (ll)a[i] * b[i] % MOD;
	NTT(a, lim, -1);
}

int n;
int f[MAXN], derf[MAXN], invf[MAXN];  // f、f的导数、f的逆
int bell[MAXN];

int tmp[MAXN];  // 递归过程中临时存放系数的数组

void PolyInv(int deg, int* a, int* res) {
	if (deg == 1) {
		res[0] = qpow(a[0], MOD - 2, MOD);
		return;
	}

	PolyInv((deg + 1) >> 1, a, res);  // 递归求解

	int lim = 1, len = 0;
	while (lim < (deg << 1)) lim <<= 1, len++;

	for (int i = 1; i < lim; i++) rev[i] = (rev[i >> 1] >> 1 | ((i & 1) << (len - 1)));

	for (int i = 0; i < lim; i++) tmp[i] = a[i] * (i < deg);  // i≥deg的项在模意义下为0

	NTT(tmp, lim, 1), NTT(res, lim, 1);
	for (int i = 0; i < lim; i++)
		res[i] = (ll)(2 - (ll)tmp[i] * res[i] % MOD + MOD) % MOD * res[i] % MOD;
	NTT(res, lim, -1);

	for (int i = deg; i < lim; i++) res[i] = 0;  // i≥deg的项在模意义下为0
}

void PolyDer(int deg, int* a, int* res) {  // 对deg次多项式a求导,结果放在res中
	for (int i = 1; i < deg; i++) res[i - 1] = (ll)i * a[i] % MOD;
	res[deg - 1] = 0;
}

void PolyInt(int deg, int* a, int* res) {  // 对deg次多项式a求不定积分(C=0),结果放在res中
	res[0] = 0;
	for (int i = 1; i < deg; i++) res[i] = (ll)a[i - 1] * qpow(i, MOD - 2, MOD) % MOD;
}

void PolyLn(int n, int* a, int* res) {  // 对n次多项式a求ln,结果放在res中
	memset(derf, 0, so(derf));
	memset(invf, 0, so(invf));

	int deg = 1;
	while (deg <= n) deg <<= 1;

	PolyDer(deg, a, derf);  // a的导数
	PolyInv(deg, a, invf);  // a的逆
	PolyMul(derf, invf, deg, deg);
	PolyInt(deg, derf, res);
}

int lnres[MAXN];  // res的ln
int tmpa[MAXN];  // 递归过程中临时存放系数的数组

void PolyExp(int deg, int* a, int* res) {  // 对deg次多项式a求exp,结果放在res中
	if (deg == 1) {
		res[0] = 1;
		return;
	}

	PolyExp(deg + 1 >> 1, a, res);

	PolyLn(deg, res, lnres);

	int lim = 1, len = 0;
	while (lim < (deg << 1)) lim <<= 1, len++;
	for (int i = 1; i < lim; i++) rev[i] = (rev[i >> 1] >> 1 | ((i & 1) << (len - 1)));

	for (int i = 0; i < deg; i++) tmpa[i] = a[i];
	for (int i = deg; i < lim; i++) tmpa[i] = lnres[i] = 0;
	for (int i = 0; i < lim; i++) tmpa[i] = (((ll)tmpa[i] - lnres[i]) % MOD + MOD) % MOD;
	tmpa[0]++;

	NTT(tmpa, lim, 1), NTT(res, lim, 1);
	for (int i = 0; i < lim; i++) res[i] = (ll)res[i] * tmpa[i] % MOD;
	NTT(res, lim, -1);
	for (int i = deg; i < lim; i++) res[i] = 0;
}

int fac[MAXN], ifac[MAXN];

void init() {  // 预处理Bell数
	fac[0] = 1;
	for (int i = 1; i < MAXN; i++) fac[i] = (ll)fac[i - 1] * i % MOD;

	ifac[MAXN - 1] = qpow(fac[MAXN - 1], MOD - 2, MOD);
	for (int i = MAXN - 1; i; i--) ifac[i - 1] = (ll)ifac[i] * i % MOD;

	n = 1e5 + 1;
	for (int i = 1; i < n; i++) f[i] = ifac[i];

	PolyExp(n, f, bell);

	for (int i = 0; i < n; i++) bell[i] = (ll)bell[i] * fac[i] % MOD;
}

int main() {
	init();  // 预处理Bell数
	
	CaseT{
		int n; cin >> n;
		int ans = 0;
		for (int i = 0; i <= n; i++) {
			int a; cin >> a;
			ans = ((ll)ans + (ll)a * bell[i]) % MOD;
		}
		cout << ans << endl;
	}
}

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