luoguP4512 【模板】多项式除法

求 $F(x)=Q(x) imes G(x)+R(x)$  中的 $Q(x),R(x)$

 

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

 

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

 

带入 $frac{1}{x}$，再乘以 $x^{n}$ 其实就是将系数翻转了

 

令 $F_{R}$ 表示将 $F$ 翻转

 

$F_{R}(x)=Q_{R}(x)G_{R}(x)+x^{n-m+1}R_{R}(x)$

 

$F_{R}(x)equiv Q_{R}(x)G_{R}(x)+x^{n-m+1}R_{R}(x)($mod $x^{n-m+1})$

 

$F_{R}(x)equiv Q_{R}(x) imes G_{R}(x)$ (mod $x^{n-m+1}$) 

 

$Q_{R}(x)equiv F_{R}(x) imes G_{R}^{-1}(x)$(mod $x^{n-m+1}$) 

 

这里一定要注意，对 $G_{R}$ 求逆时模的是 $x^{n-m+1}$，所以要先将 $G_{R}$ 的长度定为 $n-m+1$

$R(x)=F(x)-G(x) imes Q(x)$   

// luogu-judger-enable-o2
#include <cstdio>
#include <string>
#include <algorithm>    
#include <cstring>
#include <vector>     
#define setIO(s) freopen(s".in","r",stdin)   
typedef long long ll;
const int maxn=2100005;  
const ll mod=998244353; 
using namespace std;                   
inline ll qpow(ll base,ll k) {
    ll tmp=1;      
    for(;k;k>>=1,base=base*base%mod)if(k&1) tmp=tmp*base%mod;   
    return tmp;     
}       
inline ll inv(ll a) { return qpow(a, mod-2); }     
inline void NTT(ll *a,int len,int flag) {
    for(int i=0,k=0;i<len;++i) {
        if(i>k) swap(a[i],a[k]);     
        for(int j=len>>1;(k^=j)<j;j>>=1);  
    }  
    for(int mid=1;mid<len;mid<<=1) {
        ll wn=qpow(3, (mod-1)/(mid<<1)),x,y;  
        if(flag==-1) wn=qpow(wn,mod-2);  
        for(int i=0;i<len;i+=(mid<<1)) {     
            ll w=1;
            for(int j=0;j<mid;++j) {        
                x=a[i+j],y=w*a[i+j+mid]%mod;   
                a[i+j]=(x+y)%mod, a[i+j+mid]=(x-y+mod)%mod;   
                w=w*wn%mod;    
            }
        }  
    }
    if(flag==-1) {
        int re=qpow(len,mod-2); 
        for(int i=0;i<len;++i) a[i]=a[i]*re%mod;   
    }
}
ll A[maxn],B[maxn];           
struct poly {
    vector<ll>a; 
    int len; 
    poly(){}                
    inline void clear() { len=0; a.clear(); }   
    inline void rev() {reverse(a.begin(), a.end()); }              
    inline void push(int x) { a.push_back(x),++len; }   
    inline void resize(int x) { len=x; a.resize(x); }                  
    void getinv(poly &b,int n) {
        if(n==1) { b.clear(); b.push(inv(a[0]));  return; } 
        getinv(b,n>>1);     
        int t=n<<1,lim=min(len,n);
        for(int i=0;i<lim;++i) A[i]=a[i];
        for(int i=lim;i<t;++i) A[i]=0;
        for(int i=0;i<b.len;++i) B[i]=b.a[i];
        for(int i=b.len;i<t;++i) B[i]=0;  
        NTT(A,t,1),NTT(B,t,1);  
        for(int i=0;i<t;++i)  A[i]=(2-A[i]*B[i]%mod+mod)*B[i]%mod; 
        NTT(A,t,-1);             
        b.clear();          
        for(int i=0;i<n;++i) b.push(A[i]);   
    }    
    poly Inv() {
        int n=1;
        while(n<=len)n<<=1;  
        poly b;           
        b.clear(), getinv(b,n);          
        return b;                         
    }           
    poly operator * (const poly &b) const {
        int n=1;
        while(n<=len+b.len) n<<=1;  
        for(int i=0;i<len;++i) A[i]=a[i];
        for(int i=len;i<n;++i) A[i]=0;
        for(int i=0;i<b.len;++i) B[i]=b.a[i];
        for(int i=b.len;i<n;++i) B[i]=0;
        NTT(A,n,1), NTT(B,n,1);
        for(int i=0;i<n;++i) A[i]=A[i]*B[i]%mod;    
        NTT(A,n,-1);    
        poly c;
        c.clear();
        for(int i=0;i<len+b.len-1;++i) c.push(A[i]);    
        return c;        
    }     
    poly operator + (const poly &b) const {
        poly c; 
        c.clear();    
        for(int i=0;i<len;++i) c.push(a[i]); 
        for(int i=0;i<b.len;++i) {
            if(i<len) c.a[i]=(c.a[i]+b.a[i])%mod;    
            else c.push(b.a[i]); 
        }
        return c;     
    }
    poly operator - (const poly &b) const {
        poly c; 
        c.clear();   
        for(int i=0;i<len;++i) c.push(a[i]); 
        for(int i=0;i<b.len;++i) {
            if(i<len) c.a[i]=(c.a[i]-b.a[i]+mod)%mod;  
            else c.push((mod-b.a[i])%mod);  
        }
        return c;  
    }
    friend poly operator / (poly f,poly g) {   
        poly Q;       
        int l=f.len-g.len+1;
        f.rev(), g.rev(), g.resize(l), f.resize(l);                  
        g=g.Inv(), Q=f*g, Q.resize(l),Q.rev();           
        return Q;   
    }
    friend poly operator % (poly f,poly g) {
        poly u=f-(f/g)*g;    
        u.resize(g.len-1);
        return u;     
    }              
}po[4];        
inline void inv() {
    int n,x;
    scanf("%d",&n), po[0].clear();
    for(int i=0;i<n;++i) scanf("%d",&x), po[0].push(x);   
    po[1]=po[0].Inv();
    for(int i=0;i<po[1].len;++i) printf("%lld ",po[1].a[i]);  
}         
inline void mult() {
    int n,m,x;
    scanf("%d%d",&n,&m);
    for(int i=0;i<=n;++i) scanf("%d",&x), po[0].push(x);
    for(int i=0;i<=m;++i) scanf("%d",&x), po[1].push(x);   
    po[1]=po[0]*po[1]; 
    for(int i=0;i<po[1].len;++i) printf("%lld ",po[1].a[i]);
}                           
inline void divide() {
    int n,m,x;
    scanf("%d%d",&n,&m);                           
    for(int i=0;i<=n;++i) scanf("%d",&x), po[0].push(x);   
    for(int i=0;i<=m;++i) scanf("%d",&x), po[1].push(x);  
    po[2]=po[0]/po[1];           
    for(int i=0;i<po[2].len;++i) printf("%lld ",po[2].a[i]); 
    printf("
"); 
    po[2]=po[0]%po[1];
    for(int i=0;i<po[2].len;++i) printf("%lld ",po[2].a[i]); 
}
int main() {
    // setIO("input");  
    divide();
    return 0; 
}