[2022 牛客多校4 C] Easy Counting Problem (生成函数 NTT)

题意

给定一个正整数 $ww$ 及 $ww$ 个数 $c_0,c_1,\cdots ,c_{w-1}c\_0,\; c\_1,\; \backslash cdots,c\_\{w\; -\; 1\}$

$qq$ 组询问，每次询问给定一个正整数 $nn$，计算有多少个长度为 $nn$ 的字符串满足：

• 每个字符只能取数字 $0\sim w-10\; \backslash sim\; w\; -\; 1$
• 数字 $ii$ 至少出现 $c_{i}c\_i$ 次

对 $998244353998\; \backslash ,\; 244\; \backslash ,\; 353$ 取模。

$2\le w\le 10,1\le c_i\le 5\times 10^4,{\sum }_{i=0}^{w-1}{c}_i\le 5\times 10^{4}2\; \backslash le\; w\; \backslash le\; 10,\; 1\; \backslash le\; c\_i\; \backslash le\; 5\; \backslash times\; 10\; ^\; 4,\; \backslash sum\backslash limits\_\{i\; =\; 0\}\; ^\; \{w\; -\; 1\}c\_i\; \backslash le\; 5\; \backslash times\; 10\; ^\; 4$

$1\le q\le 300,1\le n\le 10^{7}1\; \backslash le\; q\; \backslash le\; 300,\; 1\; \backslash le\; n\; \backslash le\; 10\; ^\; 7$

分析：

首先我们可以写出每个数字 $ii$ 的 $\mathbf{\text{EGF}}\backslash textbf\{EGF\}$

那么每个数字的 $\mathbf{\text{EGF}}\backslash textbf\{EGF\}$ 做乘积表示满足条件的所有长度的字符串的方案数

可以把和式用前缀和相减拆一下 ${\sum }_{j=c_i}^{\mathrm{\infty }}{\displaystyle \frac{x^j}{j!}}={\sum }_{j=0}^{\mathrm{\infty }}{\displaystyle \frac{x^j}{j!}}-{\sum }_{j=0}^{c_i-1}{\displaystyle \frac{x^j}{j!}}\backslash sum\backslash limits\_\{j\; =\; c\_i\}\; ^\; \{\backslash infty\}\; \backslash dfrac\{x\; ^\; j\}\{j!\}\; =\; \backslash sum\backslash limits\_\{j\; =\; 0\}\; ^\; \{\backslash infty\}\; \backslash dfrac\{x\; ^\; j\}\{j!\}\; -\; \backslash sum\backslash limits\_\{j\; =\; 0\}\; ^\; \{c\_i\; -\; 1\}\; \backslash dfrac\{x\; ^\; j\}\{j!\}$，发现第一项为 $e^{x}e\; ^\; x$，故答案为

由于 $w\le 10w\; \backslash le\; 10$，所以考虑暴力展开式子，做换元 $e^x\to ye\; ^\; x\; \backslash rightarrow\; y$

在展开式子的过程中，假设当前的多项式为 $f=A_0+A_{1}y+A_2{y}^2+A_3{y}^3+\cdots f\; =\; A\_0\; +\; A\_1\; y\; +\; A\_2y\; ^\; 2\; +\; A\_3y^3\; +\; \backslash cdots$，那么新遇到一个多项式 $(y+g_i)(y\; +\; g\_i)$ 其中 $g_i=-{\sum }_{j=0}^{c_i-1}{\displaystyle \frac{x^j}{j!}}g\_i\; =\; -\backslash sum\backslash limits\_\{j\; =\; 0\}\; ^\; \{c\_i\; -\; 1\}\; \backslash dfrac\{x\; ^\; j\}\{j!\}$， 则结果变为 $f\ast y+f\ast g_{i}f\; *\; y\; +\; f\; *\; g\_i$ ($\ast *$ 表示多项式卷积)，前一项为 $A_{0}y+A_1{y}^2+A_2{y}^3+A_3{y}^4+\cdots A\_0y\; +\; A\_1\; y\; ^\; 2\; +\; A\_2y\; ^\; 3\; +\; A\_3y^4\; +\; \backslash cdots$，那么后一项是 $ff$ 的每一项系数与 $g_{i}g\_i$ 的多项式卷积，为 $A_0\ast g_i+(A_1\ast g_i)y+(A_2\ast g_i)y^2+(A_3\ast g_i)y^3+\cdots A\_0\; *\; g\_i\; +\; (A\_1\; *\; g\_i)\; y\; +\; (A\_2\; *\; g\_i)y\; ^\; 2\; +\; (A\_3\; *\; g\_i)y^3\; +\; \backslash cdots$，那么答案就为

这样就预处理好了总答案，现考虑回答每组询问，我们知道最后的答案是形如 ${\sum }_{i=0}^{w-1}{e}^{ix}{F}_i(x)\backslash sum\backslash limits\_\{i\; =\; 0\}\; ^\; \{w\; -\; 1\}\; e\; ^\; \{ix\}\; F\_i(x)$ 的多项式，我们需要知道每一项的第 $nn$ 项系数，由于 ${\sum }_{i=0}^{w-1}{c}_i\le 5\times 10^4\backslash sum\backslash limits\_\{i\; =\; 0\}\; ^\; \{w\; -\; 1\}c\_i\; \backslash le\; 5\; \backslash times\; 10\; ^\; 4$，我们可以在询问里对于每个 $ii$ 直接枚举 $F_i(x)F\_i(x)$ 的项数，设当前枚举到了第 $jj$ 项，那么需要在 $e^{ix}e\; ^\; \{ix\}$ 中取出第 $n-jn\; -\; j$ 项，也就是 $e^{ix}=1+{\displaystyle \frac{(ix{)}^1}{1!}}+{\displaystyle \frac{(ix{)}^2}{2!}}+{\displaystyle \frac{(ix{)}^3}{3!}}+\cdots e\; ^\; \{ix\}\; =\; 1\; +\; \backslash dfrac\{(ix)\; ^\; 1\}\{1!\}\; +\; \backslash dfrac\{(ix)\; ^\; 2\}\{2!\}\; +\; \backslash dfrac\{(ix)\; ^\; 3\}\{3!\}\; +\; \backslash cdots$ 的第 $n-jn\; -\; j$ 项，为 ${\displaystyle \frac{i^{n-j}}{(n-j)!}}\backslash dfrac\{i\; ^\; \{n\; -\; j\}\}\{(n\; -\; j)!\}$

那么答案就为

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;
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;
    }
}
signed main() {
    init(1e7);
    cin.tie(0) -> sync_with_stdio(0);
    int w;
    cin >> w;
    vector<Poly> ans(w + 1);
    ans[0] = {1};
    for (int i = 1; i <= w; i ++) {
        int c;
        cin >> c;
        vector<Z> g(c);
        for (int j = 0; j < c; j ++) {
            g[j] = -infact[j];
        }
        for (int j = i; j; j --) {
            ans[j] = ans[j] * Poly(g) + ans[j - 1];
        }
        ans[0] = ans[0] * Poly(g);
    }
    int m;
    cin >> m;
    while (m --) {
        int n;
        cin >> n;
        Z res;
        for (int i = 0; i <= w; i ++) {
            int v = min(ans[i].size() - 1, n);
            Z Pow = power(Z(i), n - v);
            for (int j = v; ~j; j --) {
                res += ans[i][j] * Pow * infact[n - j];
                Pow *= i;
            }
        }
        cout << res * fact[n] << "\n";
    }
}