P5408 第一类斯特林数·行

P5408 第一类斯特林数·行

根据递推公式，第一类斯特林数的生成函数 $ \sum _ {i = 0} ^ n {n \brack i} k ^ i = k ^ {\overline n}$ 。考虑如何快速计算上升幂，考虑倍增，$k ^ {\overline {2n}} = k ^ {\overline n} (k + n) ^ {\overline n} $ 。

记 $f _ i$ 表示多项式的第 $i$ 次项系数。对于多项式平移，有：

$$
\begin {aligned}
& f(x + c) \\
= & \sum _ {i = 0} ^ n f _ i (x + c) ^ i \\
= & \sum _ {i = 0} ^ n f _ i \sum _ {j = 0} ^ i \binom i j x _ j c ^ {i - j} \\
= & \sum _ {i = 0} ^ n x ^ i \sum _ {j = i} ^ n f _ j \binom j i c _ {j - i} \\
= & \sum _ {i = 0} ^ n \frac {x ^ i} {i!} \sum _ {j = i} ^ n f _ j j! \frac {c _ {j - i}} {(j - i)!}
\end {aligned}
$$

可以做卷积。

复杂度 $T(n) = T(\frac n 2) + O(n \log n) = O(n \log n)$ 。

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')
        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 = 1 << 20, mod = 167772161;
int fact[N], ifact[N], 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 init ()
{
    fact[0] = 1;
    for (int i = 1; i < N; ++i)
        fact[i] = (LL)fact[i - 1] * i % mod;
    ifact[N - 1] = binpow(fact[N - 1]);
    for (int i = N - 1; i; --i)
        ifact[i - 1] = (LL)ifact[i] * i % mod;
}
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] = ((LL)p + q) % mod, x[i | j | mid] = ((LL)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 *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);
}
void PolyMove (int *x, int len, int k)
{
    static int A[N], B[N];
    for (int i = 0, t = 1; i < len; ++i, t = (LL)t * k % mod)
        A[len - 1 - i] = (LL)t * ifact[i] % mod;
    for (int i = 0; i < len; ++i)
        B[i] = (LL)fact[i] * x[i] % mod;
    PolyMul(len, A, len, B, x);
    for (int i = 0; i < len; ++i)
        x[i] = (LL)x[len - 1 + i] * ifact[i] % mod;
}
void solve (int n, int *f)
{
    if (!n) return void(f[0] = 1);
    int m = n >> 1;
    solve(m, f);
    static int g[N];
    for (int i = 0; i <= m; ++i) g[i] = f[i];
    PolyMove(g, m + 1, m);
    PolyMul(m + 1, f, m + 1, g, f);
    if (n & 1) for (int i = n; ~i; --i)
        f[i] = ((LL)(n - 1) * f[i] + (i ? f[i - 1] : 0)) % mod;
}
int main ()
{
    init();
    int n; read(n);
    static int res[N];
    solve(n, res);
    for (int i = 0; i <= n; ++i)
        write((res[i] + mod) % mod), putchar(' ');
    return 0;
}