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 的上限尽量搞成 n,便于交换 的顺序
ntt 的作用就是把所有卷积形式的 预处理成一个答案数组,可以 O(1) 访问,比如:

(f * g)(j) 就是积多项式 的系数!可以 O(1) 获取!