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P5162 WD与积木

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P5162 WD与积木

$n$ 个有标号的物品，放入若干个集合。集合内没有顺序，集合之间有顺序。对于所有方案，求集合个数的期望。

考虑如果确定了一个向量 $b _ i$ 表示第 $i$ 个集合有 $b _ i$ 个数，那么有方案数 $\frac {n!} {\prod b _ i !}$ 。考虑 EGF ，对于一个集合，就有 $F(x) = \sum _ {i = 1} \frac {x ^ i}{i!} = e ^ x - 1$ 。

考虑枚举有多少个集合，那么答案为

$$G(n) = \sum _ {k = 1} [x ^ n] F(x) ^ k = [x ^ n] \sum _ {k = 1} F(x) ^ k = [x ^ n]\frac {F(x)} {1 - F(x)} = [x ^ n](\frac 1 {2 - e ^ x} - 1)$$

。再考虑计算所有方案的和，即

$$H(n) =\sum _ {k = 1} k[x ^ n]F(x) ^ k = [x ^ n] \sum _ {k = 1} kF(x) ^ k = [x ^ n] \frac {F(x)} {(1 - F(x)) ^ 2} = \frac {e ^ x - 1} {(2 - e ^ x) ^ 2}$$

最终答案为 $\frac {H(n)} {G(n)}$ 。

查看代码 
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#include <cstdio>
#include <algorithm>
using namespace std;
template <class Type>
void read (Type &x)
{
    char c;
    bool flag = false;
    while ((c = getchar()) < '0' || c > '9')
        flag |= c == '-';
    x = c - '0';
    while ((c = getchar()) >= '0' && c <= '9')
        x = (x << 3) + (x << 1) + c - '0';
    if (flag) x = ~x + 1;
}
template <class Type, class ...Rest>
void read (Type &x, Rest &...y) { read(x); read(y...); }
template <class Type>
void write (Type x)
{
    if (x < 0) putchar('-'), x = ~x + 1;
    if (x > 9) write(x / 10);
    putchar('0' + x % 10);
}
typedef long long LL;
const int N = 1e6 + 10, mod = 998244353, inv2 = mod + 1 >> 1;
int rev[N];
int binpow (int b, int k = mod - 2)
{
    int res = 1;
    for (; k; k >>= 1, b = (LL)b * b % mod)
        if (k & 1) res = (LL)res * b % mod;
    return res;
}
void ntt (int *x, int bit, int op)
{
    int tot = 1 << bit;
    for (int i = 1; i < tot; ++i)
        if ((rev[i] = rev[i >> 1] >> 1 | (i & 1) << bit - 1) > i)
            swap(x[rev[i]], x[i]);
    for (int mid = 1; mid < tot; mid <<= 1)
    {
        int w1 = binpow(3, (mod - 1) / (mid << 1));
        if (!~op) w1 = binpow(w1);
        for (int i = 0; i < tot; i += mid << 1)
            for (int j = 0, k = 1; j < mid; ++j, k = (LL)k * w1 % mod)
            {
                int p = x[i | j], q = (LL)k * x[i | j | mid] % mod;
                x[i | j] = (p + q) % mod, x[i | j | mid] = (p - q) % mod;
            }
    }
    if (~op) return;
    int itot = binpow(tot);
    for (int i = 0; i < tot; ++i)
        x[i] = (LL)x[i] * itot % mod;
}
void PolyMul (int n, int *f, int m, int *g, int nm, int *res)
{
    int bit = 0;
    while (1 << bit < n + m - 1) ++bit;
    int tot = 1 << bit;
    for (int i = n; i < tot; ++i) f[i] = 0;
    for (int i = m; i < tot; ++i) g[i] = 0;
    ntt(f, bit, 1), ntt(g, bit, 1);
    for (int i = 0; i < tot; ++i)
        res[i] = (LL)f[i] * g[i] % mod;
    ntt(res, bit, -1);
    for (int i = nm; i < tot; ++i) res[i] = 0;
}
void _PolyMul (int n, int *f, int m, int *g, int nm, int *res)
{
    int bit = 0;
    while (1 << bit < n + n + m - 2) ++bit;
    int tot = 1 << bit;
    for (int i = n; i < tot; ++i) f[i] = 0;
    for (int i = m; i < tot; ++i) g[i] = 0;
    ntt(f, bit, 1), ntt(g, bit, 1);
    for (int i = 0; i < tot; ++i)
        res[i] = (LL)f[i] * f[i] % mod * g[i] % mod;
    ntt(res, bit, -1);
    for (int i = nm; i < tot; ++i) res[i] = 0;
}
void PolyInv(int n, int *x, int *g)
{
    if (n == 1) return void(g[0] = binpow(x[0]));
    int m = n + 1 >> 1;
    int bit = 0;
    while (1 << bit < n + m + m - 2) ++bit;
    int tot = 1 << bit;
    PolyInv(m, x, g);
    for (int i = m; i < tot; ++i) g[i] = 0;
    static int A[N];
    for (int i = 0; i < n; ++i) A[i] = x[i];
    for (int i = n; i < tot; ++i) A[i] = 0;
    ntt(g, bit, 1), ntt(A, bit, 1);
    for (int i = 0; i < tot; ++i)
        g[i] = (2 - (LL)g[i] * A[i]) % mod * g[i] % mod;
    ntt(g, bit, -1);
    for (int i = n; i < tot; ++i) g[i] = 0;
}
int fact[N], ifact[N];
int main ()
{
    const int n = 1e5 + 1;
    fact[0] = 1;
    for (int i = 1; i <= n; ++i)
        fact[i] = (LL)fact[i - 1] * i % mod;
    ifact[n] = binpow(fact[n]);
    for (int i = n; i; --i)
        ifact[i - 1] = (LL)ifact[i] * i % mod;
    static int A[N], B[N], C[N];
    for (int i = 0; i < n; ++i) B[i] = -ifact[i];
    (B[0] += 2) %= mod;
    PolyInv(n, B, A);
    for (int i = 0; i < n; ++i)
        B[i] = A[i], C[i] = ifact[i];
    (--C[0]) %= mod;
    _PolyMul(n, B, n, C, n, B);
    int T; read(T);
    for (int k; T; --T, puts(""))
        read(k), write(((LL)B[k] * binpow(A[k]) % mod + mod) % mod);
    return 0;
}
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