【模板】多项式除法

问题描述

给定一个 $n$次多项式 $F(x)$ 和一个 $m$ 次多项式 $G(x)$，请求出多项式 $Q(x)$，$R(x)$，满足以下条件：

• $Q(x)$ 次数为 $n-m$，$R(x)$ 次数小于 $m$
• $F(x) = Q(x) * G(x) + R(x)$

所有运算在模998244353意义下进行

详见洛谷 P4512

分析

具体来说，设多项式$A$为$n$次多项式，考虑一种操作$R$，使得

$displaystyle A_R(x)=x^n A(frac{1}{x})$

稍微想象一下，可以发现$A_R[i]=A[n-i]$($[i]$表示多项式的第$i$次系数)。

这个操作可以$O(n)$完成。

然后开始化式子。

$$F(x)=Q(x) * G(x)+R(x)$$

$$displaystyle F(frac{1}{x})=Q(frac{1}{x}) * G(frac{1}{x})+R(frac{1}{x})$$

$$displaystyle x^n F(frac{1}{x})=x^{n-m} Q(frac{1}{x}) * x^m G(frac{1}{x})+x^{n-m+1} * x^{m-1} R(frac{1}{x})$$

$$displaystyle F_R(x)=Q_R(x)*G_R(x)+x^{n-m+1} * R_R(x)$$

$$displaystyle F_R(x) equiv Q_R(x)*G_R(x)pmod {x^{n-m+1}}$$

$$displaystyle Q_R(x) equiv F_R(x)*G_R^{-1}(x)pmod {x^{n-m+1}}$$

求一遍$G_R$的逆，然后就可以利用多项式乘法求出$Q$。然后

$$R(x)=F(x)-G(x)*Q(x)$$

直接计算即可。系数翻转可以用自带的 $reverse$ 函数，逆元最好迭代求解。

总的时间复杂度$O(nlogn)$。

代码：

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
const int N = 1<<20;

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;
}
namespace Polynomial
{
    const ll P = 998244353, g = 3, gi = 332748118;
    static int rev[N];
    int lim, bit;
    ll add(ll a, ll b)
    {
        return (a += b) >= P ? a - P : a;
    }
    ll qpow(ll a, ll b)
    {
        ll prod = 1;
        while(b)
        {
            if(b & 1) prod = prod * a % P;
            a = a * a % P;
            b >>= 1;
        }
        return (prod + P) % P;
    }
    void calrev() {
        for(int i = 1; i < lim; i++)
            rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (bit - 1));
    }
    void NTT(ll *A, int inv)
    {
        for(int i = 0; i < lim; i++)
            if(i < rev[i]) swap(A[i], A[rev[i]]);
        for(int mid = 1; mid < lim; mid <<= 1)
        {
            ll tmp = qpow(inv == 1 ? g : gi, (P - 1) / (mid << 1));
            for(int j = 0; j < lim; j += (mid << 1))
            {
                ll omega = 1;
                for(int k = 0; k < mid; k++, omega = (omega * tmp) % P) {
                    int x = A[j + k], y = omega * A[j + k + mid] % P;
                    A[j + k] = (x + y) % P;
                    A[j + k + mid] = (x - y + P) % P;
                }
            }
        }
        if(inv == 1) return;
        int invn = qpow(lim, P - 2);
        for(int i = 0; i < lim; i++)
            A[i] = A[i] * invn % P;
    }
    static ll x[N], y[N];
    void mul(ll *a, ll *b)
    {
        memset(x, 0, sizeof x);
        memset(y, 0, sizeof y);
        for(int i = 0; i < (lim >> 1); i++)
            x[i] = a[i], y[i] = b[i];
        NTT(x, 1), NTT(y, 1);
        for(int i = 0; i < lim; i++)
            x[i] = x[i] * y[i] % P;
        NTT(x, -1);
        for(int i = 0; i < lim; i++)
            a[i] = x[i];
    }
    static ll c[2][N];
    void Inv(ll *a, int n)
    {
        int p = 0;
        memset(c, 0, sizeof c);
        c[0][0] = qpow(a[0], P - 2);
        lim = 2, bit = 1;
        while(lim <= (n << 1))
        {
            lim <<= 1, bit++;
            calrev();
            p ^= 1;
            memset(c[p], 0, sizeof c[p]);
            for(int i = 0; i <= lim; i++)
                c[p][i] = add(c[p^1][i], c[p^1][i]);
            mul(c[p^1], c[p^1]);
            mul(c[p^1], a);
            for(int i = 0; i <= lim; i++)
                c[p][i] = add(c[p][i], P - c[p^1][i]);
        }
        for(int i = 0; i < lim; i++)
            a[i] = c[p][i];
    }
}
using namespace Polynomial;
int n,  m;
ll F[N], G[N], Q[N], R[N], Gr[N];
int main()
{
    n = read(), m = read();
    for(int i = 0; i <= n; i++)
        F[i] = read(), Q[n - i] = F[i];
    for(int i = 0; i <= m; i++)
        G[i] = read(), Gr[m - i] = G[i];
    for(int i = n - m + 2; i <= m; i++)
        Gr[i] = 0;
    Inv(Gr, n - m + 1);    //Gr=Gr的逆
    mul(Q, Gr);            //Q=Q*Gr
    reverse(Q, Q + n - m + 1);   //Q=reverse(Q)
    for(int i = n - m + 1; i <= n; i++)
        Q[i] = 0;
    for(int i = 0; i <= n - m; i++)
        printf("%lld ", Q[i]);
    printf("
");
    lim = 1, bit = 0;
    while(lim <= (n << 2))
        lim <<= 1, bit++;
    calrev();
    mul(Q, G);
    for(int i = 0; i < m; i++)
        printf("%lld ", add(F[i], P - Q[i]));   //R=F - Q*G
    return 0;
}

代码转载自：https://www.luogu.org/blog/AKIOIorz/p4512-mu-ban-duo-xiang-shi-chu-fa