Chirp Z 变换 模板

给定一个 $n$ 项多项式 $P(x)$ 以及 $c, m$，计算 $P(c^0),P(c^1),\dots,P(c^{m-1})$。

注意到 $ki = \binom {i + k} 2 - \binom i 2 - \binom k 2$ ，那么有

$$
\begin {aligned}
P(c ^ k) & = \sum _ {i = 0} ^ {n - 1} [x ^ i]P (c ^ k) ^ i \\
& = \sum _ {i = 0} ^ {n - 1} [x ^ i]Pc ^ {ki} \\
& = \sum _ {i = 0} ^ {n - 1} [x ^ i]Pc ^ {\binom {i + k} 2 - \binom i 2 - \binom k 2} \\
& = c ^ { - \binom k 2}\sum _ {i = 0} ^ {n - 1} [x ^ i]Pc ^ {\binom {i + k} 2 - \binom i 2} \\
\end {aligned}
$$

可以写作关于 $i$ 和关于 $i + k$ 的多项式的卷积。

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 = 5e6 + 10, mod = 998244353;
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;
}
int main ()
{
    static int n, c, m, A[N], B[N];
    read(n, c, m);
    auto calc = [](int x, int op)
    {
        x = (LL)x * (x - 1) / 2 % (mod - 1);
        return binpow(c, ~op ? x : mod - 1 - x);
    };
    for (int i = 0; i < n; ++i)
        read(A[i]), A[i] = (LL)A[i] * calc(i, -1) % mod;
    for (int i = 0; i < n + m; ++i)
        B[n + m - 1 - i] = calc(i, 1);
    PolyMul(n, A, n + m, B, n + m, A);
    for (int i = 0; i < m; ++i, putchar(' '))
        write(((LL)A[n + m - 1 - i] * calc(i, -1) % mod + mod) % mod);
    return 0;
}