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P5850 calc加强版

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P5850 calc加强版

两个重要的展开式：

$$
e ^ {f(x)} = \sum _ {i = 1} \frac {f(x) ^ i} {i!}
$$

$$
-\ln (1 - f(x)) = \sum _ {i = 1} \frac {f(x) ^ i} i
$$

现在如果不考虑选择的 $n$ 个数的顺序那么有（如果考虑，只需要乘 $n!$ 即可）：

$$
\begin {aligned}
ANS & = \prod _ {i = 1} ^ m (1 + ix) \\
& = e ^ {\sum _ {i = 1} ^ m \ln (1 + ix)}
\end {aligned}
$$

根据 $\ln$ 的展开式：

$$
\begin {aligned}
& \ln (1 + ix)\\
= & -(-\ln (1 - (- ix))) \\
= & -\sum _ {k = 1} \frac {(-ix) ^ k} k \\
= & \sum _ {k = 1} \frac {(-1) ^ {k + 1} i ^ k} k x ^ k
\end {aligned}
$$

那么

$$
\begin {aligned}
& \sum _ {i = 1} ^ m \ln (1 + ix) \\
= & \sum _ {i = 1} ^ m \sum _ {k = 1} \frac {(-1) ^ {k + 1} i ^ k} k x ^ k \\
= & \sum _ {k = 1} \frac {(-1) ^ {k + 1} \sum _ {i = 1} ^ m i ^ k} k x ^ k
\end {aligned}
$$

考虑如何快速计算 $\sum _ {i = 0} ^ m i ^ k$ 。考虑构造 EGF ：

$$
\begin {aligned}
& \sum _ {k = 0} \sum _ {i = 0} ^ m i ^ k \frac {x ^ k} {k!} \\
= & \sum _ {i = 0} ^ m\sum _ {k = 0} \frac {(ix) ^ k} {k!} \\
= & \sum _ {i = 0} ^ m e ^ {ix} \\
= & \frac {e ^ {(m + 1)x} - 1}{e ^ x - 1}
\end {aligned}
$$

注意一些细节，对于 $k > 0$ ，$0 ^ k = 0$ ，因此这里多算的 $i = 0$ 没有影响。而 $e ^ 0 - 1 = 0$ ，这意味着分子的常数项为 $0$ ，无法直接求逆，但是注意到分母的常数项也为 $0$ ，因此直接同时除以 $x$ 。

卡常。$e ^ x$ 和 $e ^ {(m + 1)x} $ 需要手动展开算。

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 = 3e6 + 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 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);
}
int main ()
{
    static int n, m, A[N], B[N], C[N];
    read(m, n); ++n;
    for (int i = 1, t = 1; i <= n; t = (LL)t * ++i % mod)
    {
        A[i - 1] = (LL)binpow(t) * binpow(m + 1, i) % mod;
        B[i - 1] = binpow(t);
    }        
    PolyInv(n, B, C), PolyMul(n, A, n, C, n, A);
    A[0] = 0;
    for (int i = 1, t = 1; i < n; t = (LL)t * ++i % mod)
        A[i] = (LL)A[i] * t % mod * binpow(i) * (i & 1 ? 1 : -1) % mod;
    PolyExp(n, A, B);
    for (int i = 1, t = 1; i < n; t = (LL)t * ++i % mod)
        write(((LL)t * B[i] % mod + mod) % mod), puts("");
    return 0;
}
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