P5667 拉格朗日插值2

P5667 拉格朗日插值2

上拉插。

$$\begin{array}{r}f(i+m)=\sum _{j=0}^{n}f(j)\prod _{k\ne j}\frac{(i+m-k)}{(j-k)}\\ & =\prod _{j=m+i-n}^{m+i}j\sum _{j=0}^n\frac{f(j)}{(m+i-j)(-1{)}^{n-j}j!(n-j)!}\end{array}$$

其中 $\frac{f(j)}{(-1{)}^{n-j}j!(n-j)!}$ 是关于 $j$ 的项，$\frac{1}{(m+i-j)}$ 是关于 $i-j$ 的项，直接卷积即可。注意到 $i-j$ 可能小于 $0$ ，所以向右平移 $n$ 个长度。

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], fact[N], ifact[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 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;
}
int main ()
{
    init();
    static int A[N], B[N];
    read(n, m);
    for (int i = 0; i <= n; ++i)
    {
        read(A[i]);
        A[i] = (n - i & 1 ? -1ll : 1ll) * ifact[i] * ifact[n - i] % mod * A[i] % mod;
    }
    for (int i = 0; i <= 2 * n; ++i) B[i] = binpow(m - n + i);
    PolyMul(n + 1, A, 2 * n + 1, B, 2 * n + 1, A);
    int k = 1;
    for (int i = 0; i <= n; ++i) k = (LL)k * (m - i) % mod;
    for (int i = 0; i <= n; ++i)
    {
        write(((LL)A[i + n] * k % mod + mod) % mod), putchar(' ');
        k = (LL)(m + i + 1) * binpow(m + i - n) % mod * k % mod;
    }
    return 0;
}