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 + 5;

namespace Poly {
    const int L = 1 << 18;
    const ll P = 998244353LL;
    
    int lim, pos[L];
    
    inline void deb(poly c) {
        for (int i = 0; i < (int)c.size(); i++)
            printf("%lld%c", c[i], " 
"[i == (int)c.size() - 1]);
    }
    
    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 < 0) x += P;
    }
    
    inline ll fpow(ll x, ll y) {
        ll res = 1;
        for (; y > 0; y >>= 1) {
            if (y & 1) res = res * x % P;
            x = x * x % P;
        }
        return res;
    }
    
    inline void prework(int len) {
        int l = 0;
        for (lim = 1; lim < len; lim <<= 1, ++l);
        for (int i = 0; i < lim; i++)
            pos[i] = (pos[i >> 1] >> 1) | ((i & 1) << (l - 1));
    }
    
    inline void ntt(poly &c, int opt) {
        c.resize(lim, 0);
        for (int i = 0; i < lim; i++)
            if (i < pos[i]) swap(c[i], c[pos[i]]);
        for (int i = 1; i < lim; i <<= 1)  {
            ll wn = fpow(3, (P - 1) / (i << 1));
            if (opt == -1) wn = fpow(wn, P - 2);
            for (int len = i << 1, j = 0; j < lim; j += len) {
                ll w = 1;
                for (int k = 0; 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 == -1) {
            ll inv = fpow(lim, P - 2);
            for (int i = 0; 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() - 1);
        ntt(u, 1), ntt(v, 1);
        for (int i = 0; i < lim; i++) res.push_back(u[i] * v[i] % P);
        ntt(res, -1);
        res.resize(x.size() + y.size() - 1);
        return res;
    }
    
    poly getInv(poly x, int len) {
        x.resize(len, 0);
        if (len == 1) {
            poly res;
            res.push_back(fpow(x[0], P - 2));
            return res;
        }
        
        poly y = getInv(x, (len + 1) >> 1);
        
        prework(len << 1);
        poly u = x, v = y, res;
        ntt(u, 1), ntt(v, 1);
        for (int i = 0; i < lim; i++) res.push_back(v[i] * ((2LL - u[i] * v[i] % P + P) % P) % P);
        ntt(res, -1);
        
        res.resize(len, 0);
        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() + 1;
        u.resize(len, 0), v.resize(len, 0);
        
//        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, 0), v.resize(len, 0);
        for (int i = 0; i < len; i++) res[i] = (res[i] - v[i] + P) % P;
        
        for (; !res.empty() && !res[res.size() - 1]; res.pop_back());
        return res;
    }
    
}

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

template <typename T>
inline void read(T &X) {
    X = 0; char ch = 0; T op = 1;
    for (; ch > '9'|| ch < '0'; ch = getchar())
        if (ch == '-') op = -1;
    for (; ch >= '0' && ch <= '9'; ch = getchar())
        X = (X << 3) + (X << 1) + ch - 48;
    X *= op;
}

int main() {
    int n, m;
    read(n), read(m);
    ++n, ++m;
    
    poly x, y;
    x.resize(n, 0), y.resize(m, 0);
    for (int i = 0; i < n; i++) read(x[i]);
    for (int i = 0; i < m; i++) read(y[i]);
    
    poly u = getDiv(x, y), v = getRest(x, y);
    for (int i = 0; i < (int)u.size(); i++)
        printf("%lld%c", u[i], " 
"[i == (int)u.size() - 1]);
    for (int i = 0; i < (int)v.size(); i++) 
        printf("%lld%c", v[i], " 
"[i == (int)v.size() - 1]);
    
    return 0;
}