P4721 【模板】分治 FFT

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给定序列$g_{1\dots n - 1}$，求序列 $f_{0\dots n - 1}$。

其中 $f_i=\sum_{j=1}^if_{i-j}g_jf_i$
​边界为 $f_0=1$ 。

答案对$998244353$ 取模。

方法一

$f=f\times g+1,f=(1-g)^{-1}$

方法二

样例

3 1 2

$f_0$会对所有产生贡献，以此类推

1 3 1 2

1 3 1+9 2+3

1 3 10 5+30

假设我们现在已经做完了$[𝑙,𝑚𝑖𝑑]$，考虑对右边的影响。

$f(i)=\sum_{j=l}^{mid}f(j)g(i-j)$
那么我们可以把$𝑓$的$[𝑙,𝑚𝑖𝑑]$项拿出来，其他项置$0$，在把这个和$𝑔$的$[0,𝑟−𝑙]$卷起来就可以得到影响，然后加上去就好了。

  • 为了避免每次分治$n\log n$
  • $A[i - l] = f[i]$,最后得到$f[i]+=C[i-l]$
分治做法 
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#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define pii pair<int, int>
#define mk make_pair
const int N = 1e6 + 10;
const int mod = 998244353;

int read()
{
    int x = 0, f = 1;
    char c = getchar();
    while (c < '0' || c > '9')
    {
        if (c == '-')
            f = -1;
        c = getchar();
    }
    while (c >= '0' && c <= '9')
        x = (x << 1) + (x << 3) + c - '0', c = getchar();
    return x * f;
}

int qpow(int a, int p, int mod)
{
    int res = 1;
    while (p)
    {
        if (p & 1)
            res = 1ll * res * a % mod;
        a = 1ll * a * a % mod;
        p >>= 1;
    }
    return res;
}
int inc(int x, int y, int mo)
{
    if (x + y >= mo)
        return (x + y - mo);
    else
        return (x + y);
}

int del(int x, int y, int mo)
{
    if (x - y < 0)
        return (x - y + mo);
    else
        return (x - y);
}

const int P = 998244353, gi = 3;
const double pi = acos(-1.0);
int rev[N];
void NTT(int *A, int n, int inv)
{
    for (int i = 0; i < n; i++)
        if (i < rev[i])
            swap(A[i], A[rev[i]]);
    for (int l = 1; l < n; l <<= 1)
    {
        int tt = qpow(gi, (P - 1) / (l << 1), P);
        int temp = (inv == 1 ? tt : qpow(tt, P - 2, P));
        for (int i = 0; i < n; i += (l << 1))
        {
            int omega = 1;
            for (int j = 0; j < l; j++, omega = 1ll * omega * temp % P)
            {
                int x = A[i + j], y = 1ll * omega * A[i + j + l] % P;
                A[i + j] = inc(x, y, P);
                A[i + j + l] = del(x, y, P);
            }
        }
    }
    int invv = qpow(n, P - 2, P);
    if (inv == -1)
        for (int i = 0; i < n; i++)
            A[i] = 1ll * A[i] * invv % P;
}
int NTTinit(int n, int m)
{
    int ML = 1, bit = 0;
    while (ML < n + m)
        ML <<= 1, bit++;
    for (int i = 0; i < ML; i++)
        rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (bit - 1));
    return ML;
}
void NTTX(int *a, int n, int *b, int m, int *ans)
{
    int ML = NTTinit(n, m);
    NTT(a, ML, 1);
    NTT(b, ML, 1);
    for (int i = 0; i < ML; i++)
        ans[i] = 1ll * a[i] * b[i] % P;
    NTT(ans, ML, -1);
}

int A[N], B[N], C[N];

void CDQNTT(int *f, int *g, int l, int r)
{
    if (l == r)
        return;
    int mid = (l + r) >> 1;
    CDQNTT(f, g, l, mid);
    int ML = NTTinit(r - l + 1, r - l + 1);
    for (int i = 0; i < ML; i++)
        A[i] = B[i] = C[i] = 0;
    for (int i = l; i <= mid; ++i)
        A[i - l] = f[i];
    for (int i = 0; i <= r - l; ++i)
        B[i] = g[i];
    NTTX(A, r - l + 1, B, r - l + 1, C);
    for (int i = mid + 1; i <= r; ++i)
        f[i] = inc(f[i], C[i - l], mod);
    CDQNTT(f, g, mid + 1, r);
}
int f[N], g[N];
int main()
{
    int n = read();
    for (int i = 1; i < n; i++)
        g[i] = read();
    f[0] = 1;
    CDQNTT(f, g, 0, n);
    for (int i = 0; i < n; i++)
        printf("%d ", f[i]);
}