[2021CCPC 威海G] Shinyruo and KFC (下降幂多项式乘法+下降幂转普通幂+多项式多点求值)

题意

给定 n n n 个正整数 a 1 , a 2 , ⋯ , a n a_1,a_2,\cdots,a_n a1,a2,⋯,an，并给定正整数 m m m，对于每个 k ∈ [ 1 , m ] k \in [1, m] k∈[1,m]，计算 ∏ i = 1 n ( k a i ) \prod\limits_{i = 1} ^ {n} \dbinom{k}{a_i} i=1∏n(aik)

对 998 244 353 998\,244\,353 998244353 取模。

( 1 ≤ n , m ≤ 5 × 1 0 4 , ∑ i = 1 n a i ≤ 1 0 5 ) (1 \le n, m \le 5 \times 10 ^ 4, \sum\limits_{i = 1} ^ {n}a_i \le 10 ^ 5) (1≤n,m≤5×104,i=1∑nai≤105)

分析：

考虑拆组合数
∏ i = 1 n ( k a i ) = ∏ i = 1 n k ! a i ! × ( k − a i ) ! = 1 ∏ i = 1 n a i ! × ∏ i = 1 n k a i ‾ \prod_{i = 1} ^ {n}\binom{k}{a_i}=\prod_{i = 1} ^ {n}\frac{k!}{a_i! \times (k - a_i)!} \\ = \frac{1}{\prod\limits_{i = 1} ^ {n}a_i!} \times \prod_{i = 1} ^ {n} k ^ {\underline {a_i}} i=1∏n(aik)=i=1∏nai!×(k−ai)!k!=i=1∏nai!1×i=1∏nkai
所以可以把 k a i ‾ k ^ {\underline{a_i}} kai 看作一个下降幂多项式，那么使用分治下降幂多项式乘法可以求出 ∏ i = 1 n k a i ‾ \prod\limits_{i = 1} ^ {n} k ^ {\underline {a_i}} i=1∏nkai，再转为普通幂多项式，再对 ( 1 , 2 , ⋯ , m ) (1, 2, \cdots,m) (1,2,⋯,m) 使用多项式多点求值即可求出答案，时间复杂度 O ( n log ⁡ 2 n ) O(n\log ^ 2 n) O(nlog2n)

代码：

#include <bits/stdc++.h>
using namespace std;
using i64 = long long;
constexpr int mod = 998244353;
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;
}
template<int mod>
struct ModInt {int x;ModInt() : x(0) {}ModInt(i64 y) : x(y >= 0 ? y % mod : (mod - (-y) % mod) % mod) {}ModInt &operator+=(const ModInt &p) {if ((x += p.x) >= mod) x -= mod;return *this;}ModInt &operator-=(const ModInt &p) {if ((x += mod - p.x) >= mod) x -= mod;return *this;}ModInt &operator*=(const ModInt &p) {x = (int)(1LL * x * p.x % mod);return *this;}ModInt &operator/=(const ModInt &p) {*this *= p.inv();return *this;}ModInt operator-() const {return ModInt(-x);}ModInt operator+(const ModInt &p) const {return ModInt(*this) += p;}ModInt operator-(const ModInt &p) const {return ModInt(*this) -= p;}ModInt operator*(const ModInt &p) const {return ModInt(*this) *= p;}ModInt operator/(const ModInt &p) const {return ModInt(*this) /= p;}bool operator==(const ModInt &p) const {return x == p.x;}bool operator!=(const ModInt &p) const {return x != p.x;}ModInt inv() const {int a = x, b = mod, u = 1, v = 0, t;while (b > 0) {t = a / b;swap(a -= t * b, b);swap(u -= t * v, v);}return ModInt(u);}ModInt pow(i64 n) const {ModInt res(1), mul(x);while (n > 0) {if (n & 1) res *= mul;mul *= mul;n >>= 1;}return res;}friend ostream &operator<<(ostream &os, const ModInt &p) {return os << p.x;}friend istream &operator>>(istream &is, ModInt &a) {i64 t;is >> t;a = ModInt<mod>(t);return (is);}int val() const {return x;}static constexpr int val_mod() {return mod;}
};
using Z = ModInt<mod>;
vector<Z> fact, infact;
void init(int n) {fact.resize(n + 1), infact.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;}
}
Z C(int n, int m) {if (n < 0 || m < 0 || n < m) return Z(0);return fact[n] * infact[n - m] * infact[m];
}
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] = a[i + 1] * (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> eval(vector<Z> x) const {if (size() == 0) {return vector<Z>(x.size(), 0);}const int n = max(int(x.size()), size());vector<Poly> q(n << 2);vector<Z> ans(x.size());x.resize(n);function<void(int, int, int)> build = [&](int p, int l, int r) {if (r - l == 1) {q[p] = Poly{1, -x[l]};} else {int m = l + r >> 1;build(p << 1, l, m);build(p << 1 | 1, m, r);q[p] = q[p << 1] * q[p << 1 | 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(p << 1, l, m, num.mulT(q[p << 1 | 1]).modxk(m - l));work(p << 1 | 1, m, r, num.mulT(q[p << 1]).modxk(r - m));}};work(1, 0, n, mulT(q[1].inv(n)));return ans;}Poly inter(const Poly &y) const {vector<Poly> Q(a.size() << 2), P(a.size() << 2);function<void(int, int, int)> dfs1 = [&](int p, int l, int r) {int m = l + r >> 1;if (l == r) {Q[p].a.push_back(-a[m]);Q[p].a.push_back(Z(1));return;}dfs1(p << 1, l, m), dfs1(p << 1 | 1, m + 1, r);Q[p] = Q[p << 1] * Q[p << 1 | 1];};dfs1(1, 0, a.size() - 1);Poly f;f.a.resize((int)(Q[1].size()) - 1);for (int i = 0; i + 1 < Q[1].size(); i ++) {f[i] = Q[1][i + 1] * (i + 1);}Poly g = f.eval(a);function<void(int, int, int)> dfs2 = [&](int p, int l, int r) {int m = l + r >> 1;if (l == r) {P[p].a.push_back(y[m] * power(g[m], mod - 2));return;}dfs2(p << 1, l, m), dfs2(p << 1 | 1, m + 1, r);P[p].a.resize(r - l + 1);Poly A = P[p << 1] * Q[p << 1 | 1];Poly B = P[p << 1 | 1] * Q[p << 1];for (int i = 0; i <= r - l; i ++) {P[p][i] = A[i] + B[i];}};dfs2(1, 0, a.size() - 1);return P[1];}
};
Poly toFPP(vector<Z> &a) {int n = a.size();vector<Z> b(n);iota(b.begin(), b.end(), 0);auto F = Poly(a).eval(b);vector<Z> f(n), g(n);for (int i = 0, sign = 1; i < n; i ++, sign *= -1) {f[i] = F[i] * infact[i];g[i] = Z(sign) * infact[i];}return Poly(f) * Poly(g);
}
Poly toOP(vector<Z> &a) {int n = a.size();vector<Z> g(n);for (int i = 0; i < n; i ++) {g[i] = infact[i];}auto F = Poly(a) * Poly(g);for (int i = 0; i < n; i ++) {F[i] *= fact[i];}vector<Z> p(n);iota(p.begin(), p.end(), 0);return Poly(p).inter(F);
}
Poly FPPMul(Poly a, Poly b) {int n = a.size() + b.size() - 1;Poly p;p.resize(n);for (int i = 0; i < n; i ++) {p[i] = infact[i];}a *= p, b *= p;for (int i = 0; i < n; i ++) {a[i] *= b[i] * fact[i];}for (int i = 1; i < n; i += 2) {p[i] = -p[i];}a *= p;a.resize(n);return a;
}
signed main() {init(2e5);cin.tie(0) -> sync_with_stdio(0);int n, m;cin >> n >> m;Z inv = 1;vector<int> a(n + 1);vector<vector<Z>> num(n + 1);for (int i = 1; i <= n; i ++) {cin >> a[i];inv *= infact[a[i]];num[i].resize(a[i] + 1);num[i][a[i]] = 1;}function<Poly(int, int)> dc = [&](int l, int r) {if (l == r) {return Poly(num[l]);}int mid = l + r >> 1;return FPPMul(dc(l, mid), dc(mid + 1, r));};vector<Z> q(m + 1);iota(q.begin(), q.end(), 0);auto ans = dc(1, n).a;auto res = toOP(ans).eval(q);for (int i = 1; i <= m; i ++) {cout << res[i] * inv << "\n";}
}

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[2021CCPC 威海G] Shinyruo and KFC (下降幂多项式乘法+下降幂转普通幂+多项式多点求值)

代码：

[2021CCPC 威海G] Shinyruo and KFC (下降幂多项式乘法+下降幂转普通幂+多项式多点求值)相关推荐

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