Luogu P4705 玩游戏

题目描述

https://www.luogu.com.cn/problem/P4705

简要题意：现在有两个序列 $a$ 和 $b$，长度分别为 $n$ 和 $m$，现在随机从 $a$ 中选一个数 $x$，然后再随机从 $b$ 中选一个数
$y$，求对于 $k\in [1,t]$，$(x+y)^k$ 的期望是多少，答案对 $998244353$ 取模

$a,b,t\le 10^5$

Solution

我们将 $(x+y)^k$ 拆开，然后得到 $ans_k=\sum_{x=1}^n\sum_{y=1}^m\sum_{i=0}^k\binom{k}{i}a_x^ib_y^{k-i}$，稍微交换一下求和顺序，$ans_k=\sum_{i=0}^n\binom{k}{i}(\sum_{i=1}^na_x^i)(\sum_{j=1}^mb_y^{k-i})$，我们令 $A_k$ 表示 $\sum_{i=1}^na_i^k$，如果我们能预处理 $A_0$ 到 $A_1$，那么 $ans$ 我们可以直接卷积

我们现在考虑如何求 $A_k$，我们令 $A_k$ 的生成函数为 $F(x)$，容易得到 $F(x)=\sum_{i=1}^n\frac{1}{1-a_ix}$，发现对于分数的求和并没有什么特别好的做法，一个比较简单的做法是类似于分治+NTT的做法，我们直接维护分子和分母这两个多项式，然后左右两边乘的时候暴力通分即可

我们接下来考虑一个比较神奇的做法，我们考虑将这个 $\sum$ 变成 $\prod$，实现这个的最好做法是 $\ln$，但原始并没有 $\ln$，所以我们造一个 $\ln$，我们知道 $\ln’(1-a_ix)=\frac{-a_i}{1-a_ix}$，我们令 $G(x)=\sum_{i=1}^n\ln’(1-a_ix)$，那么我们有 $F(x)=-xG(x)+n$

而 $G(x)=\sum_{i=1}^n\ln’(1-a_ix)=\ln’(\prod_{i=1}^n1-a_ix)$，这个直接分治+NTT加多项式 $\ln$ 即可

时间复杂度 $O(n\log^2 n)$

#include <iostream>
#include <vector>
#include <algorithm>
#define maxn 100010
#define ll long long
using namespace std;

const int p = 998244353;
ll pow_mod(ll x, ll n) {
    ll s = 1;
    for (; n; n >>= 1, x = x * x % p)
        if (n & 1) s = s * x % p;
    return s; 
}

#define Poly vector<int>
#define len(A) ((int) A.size())
namespace Pol {
    inline int add(int a, int b) { return (a += b) >= p ? a -= p : a; }
    inline int mul(int a, int b) { return 1ll * a * b % p; } 
    Poly operator - (const int &v, const Poly &a) {
        Poly res(a); 
        for (int i = 0; i < len(res); ++i) res[i] = p - res[i]; 
        res[0] = add(res[0], v); return res;
    }
    Poly operator - (const Poly &a, const int &v) {
        Poly res(a); res[0] = add(res[0], p - v); return res;
    }
    Poly operator * (const Poly &a, const int &v) {
        Poly res(a);
        for (int i = 0; i < len(res) ; ++i) res[i] = mul(res[i], v); 
        return res;
    }
 
    const int N = 4200000;
    int P[N];
    void init_P(int n) {
        int l = 0; while ((1 << l) < n) ++l;
        for (int i = 0; i < n; ++i) P[i] = (P[i >> 1] >> 1) | ((i & 1) << l - 1);
    }
    void NTT(Poly &a, int type) {
        static int w[N]; ll G = 3, Gi = pow_mod(G, p - 2); int n = len(a);
        for (int i = 0; i < n; ++i) if (i < P[i]) swap(a[i], a[P[i]]);
        for (int i = 2, m = 1; i <= n; m = i, i *= 2) {
            ll wn = pow_mod(type > 0 ? G : Gi, (p - 1) / i);
            w[0] = 1; for (int j = 1; j < m; ++j) w[j] = wn * w[j - 1] % p;
            for (int j = 0; j < n; j += i)
                for (int k = 0; k < m; ++k) {
                    int t1 = a[j + k], t2 = 1ll * a[j + k + m] * w[k] % p;
                    a[j + k] = add(t1, t2);
                    a[j + k + m] = add(t1, p - t2);
                }
        }
        if (type < 0) {
            int inv = pow_mod(n, p - 2);
            for (int i = 0; i < n; ++i) a[i] = mul(a[i], inv);
        }
    }
    Poly operator * (const Poly &A, const Poly &B) {
        int n = 1, n1 = len(A), n2 = len(B); while (n < n1 + n2 - 1) n <<= 1; init_P(n);
        Poly a(n), b(n);
        for (int i = 0; i < n1; ++i) a[i] = add(A[i], p);
        for (int i = 0; i < n2; ++i) b[i] = add(B[i], p);
        NTT(a, 1); NTT(b, 1);
        for (int i = 0; i < n; ++i) a[i] = mul(a[i], b[i]);
        NTT(a, -1); return a; 
    }
    Poly Der(const Poly &a) {
        Poly res(a);
        for (int i = 0; i < len(a) - 1; ++i) res[i] = mul(i + 1, res[i + 1]);
        res[len(a) - 1] = 0; return res;
    }
    Poly Int(const Poly &a) {
        static int inv[N];
        if (!inv[1]) {
            inv[1] = 1; 
            for (int i = 2; i < N; ++i) inv[i] = mul(p - p / i, inv[p % i]);
        }
        Poly res(a); res.resize(len(a) + 1);
        for (int i = len(a); i; --i) res[i] = mul(res[i - 1], inv[i]);
        res[0] = 0; return res;
    }
    Poly Inv(const Poly &a) {
        Poly res(1, pow_mod(a[0], p - 2)); 
        int n = 1; while (n < len(a)) n <<= 1; 
        for (int k = 2; k <= n; k <<= 1) {
            int L = 2 * k; init_P(L); Poly t(L);
            copy_n(a.begin(), min(k, len(a)), t.begin()); 
            t.resize(L); res.resize(L);
            NTT(res, 1); NTT(t, 1);
            for (int i = 0; i < L; ++i) res[i] = mul(res[i], add(2, p - mul(t[i], res[i])));
            NTT(res, -1); res.resize(k);
        } res.resize(len(a)); return res;
    }
    pair<Poly, Poly> Divide(const Poly &a, const Poly &b) {
        int n = len(a), m = len(b); 
        Poly t1(a.rbegin(), a.rbegin() + n - m + 1), t2(b.rbegin(), b.rend()); t2.resize(n - m + 1);
        Poly Q = Inv(t2) * t1; Q.resize(n - m + 1); reverse(Q.begin(), Q.end());
        Poly R = Q * b; R.resize(m - 1); for (int i = 0; i < len(R); ++i) R[i] = add(a[i], p - R[i]);
        return make_pair(Q, R); 
    }
    Poly Ln(const Poly &a) {
        Poly res = Int(Der(a) * Inv(a));
        res.resize(len(a)); return res;
    }
    Poly Exp(const Poly &a) {
        Poly res(1, 1);
        int n = 1; while (n < len(a)) n <<= 1;
        for (int k = 2; k <= n; k <<= 1) {
            Poly t(res.begin(), res.end()); t.resize(k); t = Ln(t);
            for (int i = 0; i < min(len(a), k); ++i) t[i] = add(a[i], p - t[i]); t[0] = add(t[0], 1);
            res = res * t; res.resize(k);
        } res.resize(len(a)); return res;
    }
    Poly Sqrt(const Poly &a) { // a[0] = 1
        Poly res(1, 1); ll inv2 = pow_mod(2, p - 2); 
        int n = 1; while (n < len(a)) n <<= 1;
        for (int k = 2; k <= n; k <<= 1) {
            Poly t(res.begin(), res.end()), ta(a.begin(), a.begin() + min(len(a), k));
            t.resize(k); t = Inv(t) * ta; 
            res.resize(k); for (int i = 0; i < k; ++i) res[i] = mul(add(res[i], t[i]), inv2);
        } res.resize(len(a)); return res;
    }
    Poly Pow(const Poly &a, int k) { // a[0] = 1
        return Exp(Ln(a) * k);
    }
    Poly Pow(const Poly &a, int k, int kk) { 
        int n = len(a), t = n, m, v, inv, powv; Poly res(n);
        for (int i = n - 1; ~i; --i) if (a[i]) t = i, v = a[i];
        if (k && t >= (n + k - 1) / k) return res;
        if (t == n) { if (!k) res[0] = 1; return res; }
        m = n - t * k; res.resize(m);
        inv = pow_mod(v, p - 2); powv = pow_mod(v, kk); 
        for (int i = 0; i < m; ++i) res[i] = mul(a[i + (k > 0) * t], inv);
        res = Exp(Ln(res) * k); res.resize(n); 
        for (int i = m - 1; ~i; --i) {
            ll tmp = mul(res[i], powv);
            res[i] = 0, res[i + t * k] = tmp;
        }
        return res;
    }
}  // namespace Pol

int n, m, k, a[maxn], b[maxn];

Poly solve(int *a, int l, int r) {
    if (l == r) {
        Poly res; res.push_back(1); res.push_back(p - a[l]);
        return res; 
    } int m = l + r >> 1;
    return Pol::operator*(solve(a, l, m), solve(a, m + 1, r));
}

ll fac[maxn], inv[maxn];
void init_C(int n) {
    fac[0] = 1; for (int i = 1; i <= n; ++i) fac[i] = fac[i - 1] * i % p;
    inv[n] = pow_mod(fac[n], p - 2); for (int i = n - 1; ~i; --i) inv[i] = inv[i + 1] * (i + 1) % p; 
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr); cout.tie(nullptr);

    cin >> n >> m; 
    for (int i = 1; i <= n; ++i) cin >> a[i];
    for (int i = 1; i <= m; ++i) cin >> b[i];
    cin >> k; init_C(k); Poly A = solve(a, 1, n), B = solve(b, 1, m);
    A.resize(k + 1); B.resize(k + 1); 
    A = Pol::Der(Pol::Ln(A)); B = Pol::Der(Pol::Ln(B));
    for (int i = k; i; --i) A[i] = -A[i - 1], B[i] = -B[i - 1]; A[0] = n; B[0] = m; 
    for (int i = 0; i < k + 1; ++i) A[i] = A[i] * inv[i] % p, B[i] = B[i] * inv[i] % p; 
    Poly res = Pol::operator*(A, B); ll invnm = pow_mod(1ll * n * m % p, p - 2); 
    for (int i = 1; i <= k; ++i) cout << (res[i] * invnm % p * fac[i] % p + p) % p << "\n";
    return 0;
}