[2021ICPC济南 L] Strange Series (Bell 数 多项式exp)
题意
T T T 组输入，给定一个 n n n 次多项式 f ( x ) = a 0 + a 1 x + ⋯ + a n x n f(x) = a_0 + a_1x + \cdots + a_nx ^ n f(x)=a0+a1x+⋯+anxn，定义 S = ∑ i = 0 ∞ f ( i ) i ! S = \sum\limits_{i = 0} ^ {\infty} \dfrac{f(i)}{i!} S=i=0∑∞i!f(i)，可以证明 S S S 一定是 e e e 的倍数，即 S = p × e S = p \times e S=p×e，求 p p p 对 998 244 353 998\,244\,353 998244353 取模。
1 ≤ T ≤ 100 , 0 ≤ n ≤ 1 0 5 , 0 ≤ a i < 998 244 353 1 \le T \le 100, 0 \le n \le 10 ^ 5,0 \le a_i < 998\,244\,353 1≤T≤100,0≤n≤105,0≤ai<998244353
分析：
首先将
f
(
x
)
f(x)
f(x) 代入
S
S
S 得
∑
i
=
0
∞
1
i
!
∑
j
=
0
n
a
j
×
i
j
\sum_{i = 0} ^ {\infty}\frac{1}{i!} \sum_{j = 0} ^ {n}a_j \times i ^ j
i=0∑∞i!1j=0∑naj×ij
看到自然数幂想到展开
i
k
=
∑
j
=
0
k
{
k
j
}
i
j
‾
i ^ k = \sum\limits_{j = 0} ^ {k} {k \brace j} i ^{\underline j}
ik=j=0∑k{jk}ij，代入得
∑
i
=
0
∞
1
i
!
∑
j
=
0
n
a
j
∑
k
=
0
j
{
j
k
}
i
k
‾
\sum_{i = 0} ^ {\infty} \frac{1}{i!} \sum_{j = 0} ^ {n}a_j \sum_{k = 0} ^ {j} {j \brace k} i ^ {\underline k}
i=0∑∞i!1j=0∑najk=0∑j{kj}ik
交换求和次序，先对
i
i
i 求和
∑
j
=
0
n
a
j
∑
k
=
0
j
{
j
k
}
∑
i
=
0
∞
i
k
‾
i
!
\sum_{j = 0} ^ {n}a_j \sum_{k = 0} ^ {j} {j \brace k} \sum_{i = 0} ^ {\infty} \frac{i ^ {\underline k}}{i!}
j=0∑najk=0∑j{kj}i=0∑∞i!ik
把下降幂消掉
∑
j
=
0
n
a
j
∑
k
=
0
j
{
j
k
}
∑
i
=
k
∞
1
(
i
−
k
)
!
\sum_{j = 0} ^ {n}a_j \sum_{k = 0} ^ {j} {j \brace k} \sum_{i = k} ^ {\infty} \frac{1}{(i-k)!}
j=0∑najk=0∑j{kj}i=k∑∞(i−k)!1
做变换
(
i
−
k
)
→
i
(i - k) \rightarrow i
(i−k)→i
∑
j
=
0
n
a
j
∑
k
=
0
j
{
j
k
}
∑
i
=
0
∞
1
i
!
\sum_{j = 0} ^ {n}a_j \sum_{k = 0} ^ {j} {j \brace k} \sum_{i = 0} ^ {\infty} \frac{1}{i!}
j=0∑najk=0∑j{kj}i=0∑∞i!1
由于
e
=
∑
i
=
0
∞
1
i
!
e = \sum\limits_{i = 0} ^ {\infty} \dfrac{1}{i!}
e=i=0∑∞i!1，所以原式为
e
e
e 的倍数得证，那么式子变为
∑
j
=
0
n
a
j
∑
k
=
0
j
{
j
k
}
×
e
\sum_{j = 0} ^ {n}a_j \sum_{k = 0} ^ {j} {j \brace k} \times e
j=0∑najk=0∑j{kj}×e
事实上
Bell
n
=
∑
i
=
0
n
{
n
i
}
\text{Bell}_{n} = \sum\limits_{i = 0} ^ {n} {n \brace i}
Belln=i=0∑n{in}，其中
Bell
n
\text{Bell}_{n}
Belln 为第
n
n
n 项贝尔数，代表
n
n
n 个元素的集合划分为任意非空子集的方案数，所以答案就为
∑
i
=
0
n
a
i
×
Bell
i
\sum_{i = 0} ^ {n} a_i \times \text{Bell}_{i}
i=0∑nai×Belli
考虑快速求解贝尔数，设贝尔数的
EGF
\textbf{EGF}
EGF 为
B
(
x
)
=
∑
i
=
0
∞
F
(
x
)
i
i
!
B(x) = \sum\limits_{i = 0} ^ {\infty} \dfrac{F(x) ^ i}{i!}
B(x)=i=0∑∞i!F(x)i，其中
F
(
x
)
=
∑
i
=
1
∞
x
i
i
!
=
e
x
−
1
F(x) = \sum\limits_{i = 1} ^ {\infty}\dfrac{x ^ i}{i!} = e ^ x - 1
F(x)=i=1∑∞i!xi=ex−1，那么
B
(
x
)
=
∑
i
=
0
∞
(
e
x
−
1
)
i
i
!
=
e
e
x
−
1
B(x) = \sum\limits_{i = 0} ^ {\infty} \dfrac{(e ^ x - 1) ^ i}{i!} = e ^ {e ^ {x} - 1}
B(x)=i=0∑∞i!(ex−1)i=eex−1，直接多项式
exp
\text{exp}
exp 就好了。
代码：
#include <bits/stdc++.h>
using namespace std;
using i64 = long long;
constexpr int mod = 998244353;
int norm(int x) {
if (x < 0) {
x += mod;
}
if (x >= mod) {
x -= mod;
}
return x;
}
template<class T>
T power(T a, int b) {
T res = 1;
for (; b; b /= 2, a *= a) {
if (b % 2) {
res *= a;
}
}
return res;
}
struct Z {
int x;
Z(int x = 0) : x(norm(x)) {}
int val() const {
return x;
}
Z operator-() const {
return Z(norm(mod - x));
}
Z inv() const {
assert(x != 0);
return power(*this, mod - 2);
}
Z &operator*=(const Z &rhs) {
x = i64(x) * rhs.x % mod;
return *this;
}
Z &operator+=(const Z &rhs) {
x = norm(x + rhs.x);
return *this;
}
Z &operator-=(const Z &rhs) {
x = norm(x - rhs.x);
return *this;
}
Z &operator/=(const Z &rhs) {
return *this *= rhs.inv();
}
friend Z operator*(const Z &lhs, const Z &rhs) {
Z res = lhs;
res *= rhs;
return res;
}
friend Z operator+(const Z &lhs, const Z &rhs) {
Z res = lhs;
res += rhs;
return res;
}
friend Z operator-(const Z &lhs, const Z &rhs) {
Z res = lhs;
res -= rhs;
return res;
}
friend Z operator/(const Z &lhs, const Z &rhs) {
Z res = lhs;
res /= rhs;
return res;
}
friend istream &operator>>(istream &is, Z &a) {
i64 v;
is >> v;
a = Z(v);
return is;
}
friend ostream &operator<<(ostream &os, const Z &a) {
return os << a.val();
}
};
vector<int> rev;
vector<Z> roots{0, 1};
void dft(vector<Z> &a) {
int n = a.size();
if (int(rev.size()) != n) {
int k = __builtin_ctz(n) - 1;
rev.resize(n);
for (int i = 0; i < n; i ++) {
rev[i] = rev[i >> 1] >> 1 | (i & 1) << k;
}
}
for (int i = 0; i < n; i ++) {
if (rev[i] < i) {
swap(a[i], a[rev[i]]);
}
}
if (int(roots.size()) < n) {
int k = __builtin_ctz(roots.size());
roots.resize(n);
while ((1 << k) < n) {
Z e = power(Z(3), (mod - 1) >> (k + 1));
for (int i = 1 << (k - 1); i < (1 << k); i ++) {
roots[i << 1] = roots[i];
roots[i << 1 | 1] = roots[i] * e;
}
k ++;
}
}
for (int k = 1; k < n; k *= 2) {
for (int i = 0; i < n; i += 2 * k) {
for (int j = 0; j < k; j ++) {
Z u = a[i + j], v = a[i + j + k] * roots[k + j];
a[i + j] = u + v, a[i + j + k] = u - v;
}
}
}
}
void idft(vector<Z> &a) {
int n = a.size();
reverse(a.begin() + 1, a.end());
dft(a);
Z inv = (1 - mod) / n;
for (int i = 0; i < n; i ++) {
a[i] *= inv;
}
}
struct Poly {
vector<Z> a;
Poly() {}
Poly(const vector<Z> &a) : a(a) {}
Poly(const initializer_list<Z> &a) : a(a) {}
int size() const {
return a.size();
}
void resize(int n) {
a.resize(n);
}
Z operator[](int idx) const {
if (idx < size()) {
return a[idx];
} else {
return 0;
}
}
Z &operator[](int idx) {
return a[idx];
}
Poly mulxk(int k) const {
auto b = a;
b.insert(b.begin(), k, 0);
return Poly(b);
}
Poly modxk(int k) const {
k = min(k, size());
return Poly(vector<Z>(a.begin(), a.begin() + k));
}
Poly divxk(int k) const {
if (size() <= k) {
return Poly();
}
return Poly(vector<Z>(a.begin() + k, a.end()));
}
friend Poly operator+(const Poly &a, const Poly &b) {
vector<Z> res(max(a.size(), b.size()));
for (int i = 0; i < int(res.size()); i ++) {
res[i] = a[i] + b[i];
}
return Poly(res);
}
friend Poly operator-(const Poly &a, const Poly &b) {
vector<Z> res(max(a.size(), b.size()));
for (int i = 0; i < int(res.size()); i ++) {
res[i] = a[i] - b[i];
}
return Poly(res);
}
friend Poly operator*(Poly a, Poly b) {
if (a.size() == 0 || b.size() == 0) {
return Poly();
}
int sz = 1, tot = a.size() + b.size() - 1;
while (sz < tot) {
sz *= 2;
}
a.a.resize(sz);
b.a.resize(sz);
dft(a.a);
dft(b.a);
for (int i = 0; i < sz; i ++) {
a.a[i] = a[i] * b[i];
}
idft(a.a);
a.resize(tot);
return a;
}
friend Poly operator*(Z a, Poly b) {
for (int i = 0; i < int(b.size()); i ++) {
b[i] *= a;
}
return b;
}
friend Poly operator*(Poly a, Z b) {
for (int i = 0; i < int(a.size()); i ++) {
a[i] *= b;
}
return a;
}
Poly &operator+=(Poly b) {
return (*this) = (*this) + b;
}
Poly &operator-=(Poly b) {
return (*this) = (*this) - b;
}
Poly &operator*=(Poly b) {
return (*this) = (*this) * b;
}
Poly deriv() const {
if (a.empty()) {
return Poly();
}
vector<Z> res(size() - 1);
for (int i = 0; i < size() - 1; i ++) {
res[i] = (i + 1) * a[i + 1];
}
return Poly(res);
}
Poly integr() const {
vector<Z> res(size() + 1);
for (int i = 0; i < size(); i ++) {
res[i + 1] = a[i] / (i + 1);
}
return Poly(res);
}
Poly inv(int m) const {
Poly x{a[0].inv()};
int k = 1;
while (k < m) {
k *= 2;
x = (x * (Poly{2} - modxk(k) * x)).modxk(k);
}
return x.modxk(m);
}
Poly log(int m) const {
return (deriv() * inv(m)).integr().modxk(m);
}
Poly exp(int m) const {
Poly x{1};
int k = 1;
while (k < m) {
k *= 2;
x = (x * (Poly{1} - x.log(k) + modxk(k))).modxk(k);
}
return x.modxk(m);
}
Poly pow(int k, int m) const {
int i = 0;
while (i < size() && a[i].val() == 0) {
i ++;
}
if (i == size() || 1LL * i * k >= m) {
return Poly(vector<Z>(m));
}
Z v = a[i];
auto f = divxk(i) * v.inv();
return (f.log(m - i * k) * k).exp(m - i * k).mulxk(i * k) * power(v, k);
}
Poly sqrt(int m) const {
Poly x{1};
int k = 1;
while (k < m) {
k *= 2;
x = (x + (modxk(k) * x.inv(k)).modxk(k)) * ((mod + 1) / 2);
}
return x.modxk(m);
}
Poly mulT(Poly b) const {
if (b.size() == 0) {
return Poly();
}
int n = b.size();
reverse(b.a.begin(), b.a.end());
return ((*this) * b).divxk(n - 1);
}
};
vector<Z> fact, infact, f;
Poly bell;
void init(int n) {
fact.resize(n + 1), infact.resize(n + 1), f.resize(n + 1);
fact[0] = infact[0] = 1;
for (int i = 1; i <= n; i ++) {
fact[i] = fact[i - 1] * i;
}
infact[n] = fact[n].inv();
for (int i = n; i; i --) {
infact[i - 1] = infact[i] * i;
}
for (int i = 1; i <= n; i ++) {
f[i] = infact[i];
}
bell = Poly(f).exp(n + 1);
for (int i = 1; i <= n; i ++) {
bell[i] *= fact[i];
}
}
void solve() {
int n;
cin >> n;
Z res;
for (int i = 0; i <= n; i ++) {
int x;
cin >> x;
res += bell[i] * x;
}
cout << res << "\n";
}
signed main() {
init(1e5);
cin.tie(0) -> sync_with_stdio(0);
int T;
cin >> T;
while (T --) {
solve();
}
}