P5824 十二重计数法

P5824 十二重计数法

球不同，盒不同

$\text I$

每个球都是独立的，各有 $m$ 个选择。答案为 $m ^ n$ 。

$\text {II}$

每选择一个盒子后，可以选择的方案少一个，即 $m ^ {\underline n}$ 。

$\text {III}$

记 $f (x)$ 表示钦定 $x$ 个盒子为空的方案数，$g(x)$ 表示恰有 $x$ 个盒子为空的答案。那么有 $f(x) = (m - x) ^ n = \sum _ {i = x} \binom i x g(i)$ 。由二项式反演，$g(x) = \sum _ {i = x} (-1) ^ {i - x}\binom i x f(i) $ ，那么 $g(0) = \sum _ {i = 0} (-1) ^ i \binom i x (m - i) ^ n$ 。

球不同，盒相同

$\text {IV}$

枚举有多少个集合为空，集合非空时即为第二类斯特林数，答案为 $\sum _ {i = 1} ^ m {n \brace i}$ 。

$\text V$

对于 $n > m$ ，答案为 $0$ 。对于 $n < m$ ，所有盒子相同，方案为 $1$ 。

$\text {VI}$

即第二类斯特林数 ${n \brace m}$ 。

球相同，盒不同

$\text {VII}$

插板法，答案为 $binom {n + m - 1} {m - 1}$ 。

$\text {VIII}$

$m$ 个盒子里面选择 $n$ 个不为空，答案为 $\binom m n $ 。

$\text {IX}$

插板法，答案为 $\binom {n - 1} {m - 1}$ 。

球相同，盒相同

$\text X$

直接统计是困难的。考虑按照这样的办法统计答案：每一次操作选择前 $i \in [1, m]$ 个盒子使球的个数 +1 。这样只用 $m$ 种操作即可刻画出答案。那么问题变化为使用体积为 $[1, m]$ 的物品装满体积为 $n$ 的背包的方案数，那么和 付公主的背包 是一样的了。

$\text {XI}$

对于 $n > m$ ，答案为 $0$ 。对于 $n < m$ ，所有盒子相同，方案为 $1$ 。

$\text {XII}$

将每个盒子放一个球后，和 $\text X$ 是一样的。

查看代码

#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')
        c == '-' && (flag = true);
    x = c - '0';
    while ((c = getchar()) >= '0' && c <= '9')
        x = (x << 1) + (x << 3) + c - '0';
    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)
{
    x < 0 && (putchar('-'), x = ~x + 1);
    x > 9 && (write(x / 10), 0);
    putchar('0' + x % 10);
}
typedef long long LL;
const int N = 1e6 + 10, mod = 998244353;
int n, m, rev[N], inv[N], fact[N], ifact[N], sn[N], pm[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 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;
}
void PolyDerivate(int n, int *x, int *g)
{
    for (int i = 1; i < n; ++i)
        g[i - 1] = (LL)x[i] * i % mod;
    g[n - 1] = 0;
}
void PolyIntegrate(int n, int *x, int *g)
{
    for (int i = 1; i < n; ++i)
        g[i] = (LL)x[i - 1] * binpow(i) % mod;
    g[0] = 0;
}
void PolyLn(int n, int *x, int *g)
{
    static int A[N], B[N];
    PolyDerivate(n, x, A);
    PolyInv(n, x, B);
    PolyMul(n, A, n, B, n, A);
    PolyIntegrate(n, A, g);
}
void PolyExp(int n, int *x, int *g)
{
    if (n == 1) return void(g[0] = 1);
    int m = n + 1 >> 1;
    PolyExp(m, x, g);
    for (int i = m; i < n; ++i) g[i] = 0;
    static int A[N];
    PolyLn(n, g, A);
    for (int i = 0; i < n; ++i)
        A[i] = (x[i] - A[i]) % mod;
    ++A[0];
    PolyMul(n, A, m, g, n, g);
}
void init ()
{
    inv[1] = 1;
    for (int i = 2; i < N; ++i)
        inv[i] = -(LL)inv[mod % i] * (mod / i) % mod;
    fact[0] = ifact[0] = 1;
    for (int i = 1; i < N; ++i)
    {
        fact[i] = (LL)fact[i - 1] * i % mod;
        ifact[i] = (LL)ifact[i - 1] * inv[i] % mod;
    }
}
int C (int a, int b) { return a < b ? 0 : (LL)fact[a] * ifact[b] % mod * ifact[a - b] % mod; }
int main ()
{
    init();
    read(n, m);
    {
        static int A[N], B[N];
        for (int i = 0; i <= n; i++)
            A[i] = (LL)binpow(i, n) * ifact[i] % mod;
        for (int i = 0; i <= n; i++)
            B[i] = ifact[i] * (i & 1 ? -1 : 1);
        PolyMul(n + 1, A, n + 1, B, n + 1, sn);
    }
    {
        static int A[N];
        for (int i = 1; i <= m; ++i)
            for (int j = 1; i * j <= n; ++j)
                (A[i * j] += inv[j]) %= mod;
        PolyExp(n + 1, A, pm);
    }
    write(binpow(m, n)), puts("");
    write(n <= m ? ((LL)fact[m] * ifact[m - n] % mod + mod) % mod : 0), puts("");
    {
        int res = 0;
        for (int i = 0; i <= m; ++i)
            (res += (i & 1 ? -1ll : 1ll) * C(m, i) * binpow(m - i, n) % mod) %= mod;
        write((res + mod) % mod), puts("");
    }
    {
        int res = 0;
        for (int i = 1; i <= min(n, m); ++i)
            (res += sn[i]) %= mod;
        write((res + mod) % mod), puts("");
    }
    write(n <= m), puts("");
    write((sn[m] + mod) % mod), puts("");
    write((C(n + m - 1, m - 1) + mod) % mod), puts("");
    write((C(m, n) + mod) % mod), puts("");
    write((C(n - 1, m - 1) + mod) % mod), puts("");
    write((pm[n] + mod) % mod), puts("");
    write(n <= m), puts("");
    write(n <= m ? 0 : (pm[n - m] + mod) % mod), puts("");
    return 0;
}