Luogu 4512 【模板】多项式除法

高级操作，感觉非常神仙。

题目中的字母太难懂了，重新定义一下。

$$A(x) = B(x) * C(x) + D(x)$$

其中，$A(x)$的次数是$n$，$B(x)$的次数是$m$，$A, B$都已知，要求$C$的次数是$n - m$，$D$的次数小于$m$。

定义一种操作:

如果$A$的次数为$n$，那么

$$A_R(x) = x^nA(\frac{1}{x})$$

其实就是把一个多项式各项系数翻转过来。

比如$A(x) = x^3 + 2x^2 + 3x + 5$，有

$$A_R(x) = 5x^3 + 3x^2 + 2x + 1$$

有了这个操作之后就可以变魔术了。

发现这个$D(x)$很难受，想办法把它搞掉。

$$A(x) = B(x) * C(x) + D(x)$$

$$A(\frac{1}{x}) = B(\frac{1}{x})C(\frac{1}{x}) + D(\frac{1}{x})$$

两边乘上$x^n$，

$$x^nA(\frac{1}{x}) = x^mB(\frac{1}{x})x^{n - m}C(\frac{1}{x}) + x^{n - m + 1} * x^{m - 1}D(\frac{1}{x})$$

$$A_R(x) = B_R(x)C_R(x) + x^{n - m + 1} D_R(x)$$

两边模上$x^{n - m + 1}$，

$$A_R(x) \equiv B_R(x)C_R(x) (\mod x^{n - m + 1}) $$

其实就是求逆了呀。

求出$C(x)$之后只要重新算一遍多项式乘法减掉就可以算出$D(x)$了。

时间复杂度$O(nlogn)$。

Code：

#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
using namespace std;
typedef long long ll;
typedef vector <ll> poly;

const int N = 1e5 + ;

namespace Poly {
    const int L =  << ;
    const ll P = 998244353LL;

    int lim, pos[L];

    inline void deb(poly c) {
        for (int i = ; i < (int)c.size(); i++)
            printf("%lld%c", c[i], " \n"[i == (int)c.size() - ]);
    }

    template <typename T>
    inline void inc(T &x, T y) {
        x += y;
        if (x >= P) x -= P;
    }

    template <typename T>
    inline void sub(T &x, T y) {
        x -= y;
        if (x < ) x += P;
    }

    inline ll fpow(ll x, ll y) {
        ll res = ;
        for (; y > ; y >>= ) {
            if (y & ) res = res * x % P;
            x = x * x % P;
        }
        return res;
    }

    inline void prework(int len) {
        int l = ;
        for (lim = ; lim < len; lim <<= , ++l);
        for (int i = ; i < lim; i++)
            pos[i] = (pos[i >> ] >> ) | ((i & ) << (l - ));
    }

    inline void ntt(poly &c, int opt) {
        c.resize(lim, );
        for (int i = ; i < lim; i++)
            if (i < pos[i]) swap(c[i], c[pos[i]]);
        for (int i = ; i < lim; i <<= )  {
            ll wn = fpow(, (P - ) / (i << ));
            if (opt == -) wn = fpow(wn, P - );
            for (int len = i << , j = ; j < lim; j += len) {
                ll w = ;
                for (int k = ; k < i; k++, w = w * wn % P) {
                    ll x = c[j + k], y = w * c[j + k + i] % P;
                    c[j + k] = (x + y) % P, c[j + k + i] = (x - y + P) % P;
                }
            }
        }

        if (opt == -) {
            ll inv = fpow(lim, P - );
            for (int i = ; i < lim; i++) c[i] = c[i] * inv % P;
        }
    }

    inline poly mul(const poly x, const poly y) {
        poly u = x, v = y, res;
        prework(x.size() + y.size() - );
        ntt(u, ), ntt(v, );
        for (int i = ; i < lim; i++) res.push_back(u[i] * v[i] % P);
        ntt(res, -);
        res.resize(x.size() + y.size() - );
        return res;
    }

    poly getInv(poly x, int len) {
        x.resize(len, );
        if (len == ) {
            poly res;
            res.push_back(fpow(x[], P - ));
            return res;
        }

        poly y = getInv(x, (len + ) >> );

        prework(len << );
        poly u = x, v = y, res;
        ntt(u, ), ntt(v, );
        for (int i = ; i < lim; i++) res.push_back(v[i] * ((2LL - u[i] * v[i] % P + P) % P) % P);
        ntt(res, -);

        res.resize(len, );
        return res;
    }

    inline poly getDiv(poly x, poly y) {
        poly u = x, v = y, res;

        reverse(u.begin(), u.end());
        reverse(v.begin(), v.end());
//        deb(u), deb(v);

        int len = x.size() - y.size() + ;
        u.resize(len, ), v.resize(len, );

//        deb(u), deb(v);

        res = getInv(v, len);

//        deb(res);

        res = mul(u, res);
        res.resize(len);
//        deb(u);

        reverse(res.begin(), res.end());

        return res;
    }

    inline poly getRest(poly x, poly y) {
        poly u = x, v = y, res = getDiv(x, y);

        v = mul(v, res);
        res = u;
        int len = max(u.size(), v.size());
        res.resize(len, ), v.resize(len, );
        for (int i = ; i < len; i++) res[i] = (res[i] - v[i] + P) % P;

        for (; !res.empty() && !res[res.size() - ]; res.pop_back());
        return res;
    }

}

using Poly :: getDiv;
using Poly :: getRest;

template <typename T>
inline void read(T &X) {
    X = ; char ch = ; T op = ;
    for (; ch > ''|| ch < ''; ch = getchar())
        if (ch == '-') op = -;
    for (; ch >= '' && ch <= ''; ch = getchar())
        X = (X << ) + (X << ) + ch - ;
    X *= op;
}

int main() {
    int n, m;
    read(n), read(m);
    ++n, ++m;

    poly x, y;
    x.resize(n, ), y.resize(m, );
    for (int i = ; i < n; i++) read(x[i]);
    for (int i = ; i < m; i++) read(y[i]);

    poly u = getDiv(x, y), v = getRest(x, y);
    for (int i = ; i < (int)u.size(); i++)
        printf("%lld%c", u[i], " \n"[i == (int)u.size() - ]);
    for (int i = ; i < (int)v.size(); i++)
        printf("%lld%c", v[i], " \n"[i == (int)v.size() - ]);

    return ;
}