ntt

constexpr int P = 167772161;
vector<int> rev, roots{0, 1};
int power(int a, int b) {
int res = 1;
for (; b; b >>= 1, a = 1LL * a * a % P)
if (b & 1)
res = 1LL * res * a % P;
return res;
}
void dft(vector<int> &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) {
int e = power(3, (P - 1) >> (k + 1));
for (int i = 1 << (k - 1); i < (1 << k); ++i) {
roots[2 * i] = roots[i];
roots[2 * i + 1] = 1LL * roots[i] * e % P;
}
++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) {
int u = a[i + j];
int v = 1LL * a[i + j + k] * roots[k + j] % P;
int x = u + v;
if (x >= P)
x -= P;
a[i + j] = x;
x = u - v;
if (x < 0)
x += P;
a[i + j + k] = x;
}
}
}
}
void idft(vector<int> &a) {
int n = a.size();
reverse(a.begin() + 1, a.end());
dft(a);
int inv = power(n, P - 2);
for (int i = 0; i < n; ++i)
a[i] = 1LL * a[i] * inv % P;
}
struct Poly {
vector<int> a;
Poly() {}
Poly(int a0) {
if (a0)
a = {a0};
}
Poly(const vector<int> &a1) : a(a1) {
while (!a.empty() && !a.back())
a.pop_back();
}
int size() const {
return a.size();
}
int operator[](int idx) const {
if (idx < 0 || idx >= size())
return 0;
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<int>(a.begin(), a.begin() + k));
}
Poly divxk(int k) const {
if (size() <= k)
return Poly();
return Poly(vector<int>(a.begin() + k, a.end()));
}
friend Poly operator+(const Poly a, const Poly &b) {
vector<int> res(max(a.size(), b.size()));
for (int i = 0; i < int(res.size()); ++i) {
res[i] = a[i] + b[i];
if (res[i] >= P)
res[i] -= P;
}
return Poly(res);
}
friend Poly operator-(const Poly a, const Poly &b) {
vector<int> res(max(a.size(), b.size()));
for (int i = 0; i < int(res.size()); ++i) {
res[i] = a[i] - b[i];
if (res[i] < 0)
res[i] += P;
}
return Poly(res);
}
friend Poly operator*(Poly a, Poly b) {
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] = 1LL * a[i] * b[i] % P;
idft(a.a);
return Poly(a.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<int> res(size() - 1);
for (int i = 0; i < size() - 1; ++i)
res[i] = 1LL * (i + 1) * a[i + 1] % P;
return Poly(res);
}
Poly integr() const {
if (a.empty())
return Poly();
vector<int> res(size() + 1);
for (int i = 0; i < size(); ++i)
res[i + 1] = 1LL * a[i] * power(i + 1, P - 2) % P;
return Poly(res);
}
Poly inv(int m) const {
Poly x(power(a[0], P - 2));
int k = 1;
while (k < m) {
k *= 2;
x = (x * (2 - modxk(k) * x)).modxk(k);
}
return x.modxk(m);
}
Poly pow(i64 m, int n) const {
m %= P;
Poly p2 = this -> log(n);
for(int i=0;i<n;i++){
p2.a[i] = 1LL*p2.a[i]*m % P;
}
return p2.exp(n);
}
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 * (1 - x.log(k) + modxk(k))).modxk(k);
}
return x.modxk(m);
}
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)) * ((P + 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<int> eval(vector<int> x) const {
if (size() == 0)
return vector<int>(x.size(), 0);
const int n = max(int(x.size()), size());
vector<Poly> q(4 * n);
vector<int> ans(x.size());
x.resize(n);
function<void(int, int, int)> build = [&](int p, int l, int r) {
if (r - l == 1) {
q[p] = vector<int>{1, (P - x[l]) % P};
} else {
int m = (l + r) / 2;
build(2 * p, l, m);
build(2 * p + 1, m, r);
q[p] = q[2 * p] * q[2 * p + 1];
}
};
build(1, 0, n);
function<void(int, int, int, const Poly &)> work = [&](int p, int l, int r, const Poly &num) {
if (r - l == 1) {
if (l < int(ans.size()))
ans[l] = num[0];
} else {
int m = (l + r) / 2;
work(2 * p, l, m, num.mulT(q[2 * p + 1]).modxk(m - l));
work(2 * p + 1, m, r, num.mulT(q[2 * p]).modxk(r - m));
}
};
work(1, 0, n, mulT(q[1].inv(n)));
return ans;
}
};

例如:求一行斯特林数

void solve() {
int n;
cin >> n;
vector<Z> a(n + 1), b(n + 1), f(n + 1, 1);
for (int i = 1; i <= n; i++) {
f[i] = f[i - 1] * i;
}
for (int i = 0; i <= n; i++) {
a[i] = power(i, n) / f[i];
b[i] = power(Z(-1), i) / f[i];
}
Poly p1, p2;
for (int i = 0; i <= n; i++) {
p1.a.emplace_back(a[i].x);
p2.a.emplace_back(b[i].x);
}
auto c = p1 * p2;
for (int i = 0; i <= n; i++) {
cout << c[i] << " \n"[i == n];
}
}

小心等比数列公比为 1 的情况,要单独讨论!!! 把 \sum 的上限尽量搞成 n,便于交换 \sum 的顺序 ntt 的作用就是把所有卷积形式的 \sum 预处理成一个答案数组,可以 O(1) 访问,比如: image-20241107111450310

(f * g)(j) 就是积多项式 xjx^j 的系数!可以 O(1) 获取!