P4238 多项式求逆

发表于 | 阅读数

$f(x)\times g(x)\equiv1\mod x^n$

多项式求逆。

首先$(1+x+x^2)$的逆是$(1-x)$，满足$a_0=1,a_{i\leq n}=0$

$n=1$时即为逆元$\frac{1}{a_0}$,显然$n=3$也适用与$n=2$。

考虑已经知道$g_{n/2}(x)\times f(x)\equiv1\mod x^{\frac{n}{2}}$

$g_{n/2}(x)-g_n(x),a_{i\leq \frac{n}{2}}=0$,卷上自己必定都是$0$。

同乘$f(x)$

  • $n/2$向上取整，保证了都会满足。反正多了无所谓。
  • 每次$NTT$，只保留$x_{i\leq n}$的项数。
代码 
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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 C[N], D[N];
void finv(int *a, int *b, int len)
{
    if (len == 1)
    {
        b[0] = qpow(a[0], mod - 2, mod);
        return;
    }
    //cout << len << endl;
    finv(a, b, (len + 1) >> 1);
    int ML = 1, bit = 0;
    while (ML <= (len << 1))
        ML <<= 1, bit++;
    for (int i = 0; i < ML; i++)
        rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (bit - 1));
    memset(C, 0, sizeof(C));
    memset(D, 0, sizeof(D));
    //cout << ML << endl;
    for (int i = 0; i < len; i++)
        C[i] = a[i];
    for (int i = 0; i < len; i++)
        D[i] = b[i];
    NTT(C, ML, 1);
    NTT(D, ML, 1);
    for (int i = 0; i < ML; i++)
        D[i] = 1LL * (2 - 1ll * C[i] * D[i] % mod + mod) % mod * D[i] % mod;
    NTT(D, ML, -1);
    for (int i = 0; i < len; i++)
        b[i] = D[i];
}
int a[N], b[N];
int main()
{
    int n = read();
    for (int i = 0; i < n; i++)
        a[i] = read();
    finv(a, b, n);
    for (int i = 0; i < n; i++)
        printf("%d ", b[i]);
}