Polynomial inverse polynomial division is the foundation, if you will not inverse polynomial, please see here
Problem: Given two polynomials $ F (x) $ (the number is n), $ G (x) $ (the number is m), the sum of two polynomials $ Q (x) $ and $ R (x) $, satisfying $ F (x) = G (x) Q (x) + R (x) $, all operations carried out in the sense die 998 244 353
Pushed out of the equation:
$F(x)=G(x)Q(x)+R(x)$
With $ \ frac {1} {x} $ alternative $ x $, to give:
$F(\frac{1}{x})=G(\frac{1}{x})Q(\frac{1}{x})+R(\frac{1}{x})$
Both sides by a $ x ^ {n} $, to give:
$x^{n}F(\frac{1}{x})=x^{m}G(\frac{1}{x})x^{n-m}Q(\frac{1}{x})+x^{n}R(\frac{1}{x})$
The expression $ F (x) = G (x) Q (x) + R (x) $ analyzed, it can be seen if $ F (x) $ is the number of n, (x) $ $ G is m times , the $ Q (x) $ is the number of $ nm $, $ R (x) $ does not exceed the number of m-1
Then take the inverse of the first argument and then by a maximum number of times a structure equivalent to the original polynomial coefficients of the polynomial is just the opposite!
也即若$F(x)=\sum_{i=0}^{n}a_{i}x^{i}$，则$x^{n}F(\frac{1}{x})=\sum_{i=0}^{n}a_{n-i}x^{i}$
We note $ x ^ {n} F (\ frac {1} {x}) = \ sum_ {i = 0} ^ {n} a_ {ni} x ^ {i} = F ^ {T} (x) $
Then the above equation can be simplified into $ F ^ {T} (x) = G ^ {T} (x) Q ^ {T} (x) + R ^ {T} (x) $
Can be found, $ R ^ {T} (x) $ before the polynomial $ (n-m + 1) coefficients are items $ 0!
So if we at $ mod $ $ x ^ {n-m + 1} $ sense, this equation can be derived immediately:
$F^{T}(x)\equiv G^{T}(x)Q^{T}(x)$($mod$ $x^{n-m+1}$)
So transposition that was:
$Q^{T}(x)\equiv \frac{F^{T}(x)}{G^{T}(x)}$($mod$ $x^{n-m+1}$)
This determined the $ Q (x) $, then based on the original expression, we get:
$R(x)=F(x)-G(x)Q(x)$
$ R (x) $ even out
#include <cstdio> #include <cmath> #include <cstring> #include <cstdlib> #include <iostream> #include <algorithm> #include <queue> #include <stack> #define ll long long using namespace std; const ll mode=998244353; int to[(1<<20)+5]; int n,m; ll FF[100005],GG[100005],F[100005],G[100005],Q[100005],GF[100005]; struct node { ll g[100005]; int len; }las,now; ll pow_mul(ll x,ll y) { ll ret=1; while(y) { if(y&1)ret=ret*x%mode; x=x*x%mode,y>>=1; } return ret; } void NTT(ll *a,int len,int k) { for(int i=0;i<len;i++)if(i<to[i])swap(a[i],a[to[i]]); for(int i=1;i<len;i<<=1) { ll w0=pow_mul(3,(mode-1)/(i<<1)); for(int j=0;j<len;j+=(i<<1)) { ll w=1; for(int o=0;o<i;o++,w=w*w0%mode) { ll w1=a[j+o],w2=a[j+o+i]*w%mode; a[j+o]=(w1+w2)%mode,a[j+o+i]=((w1-w2)%mode+mode)%mode; } } } if(k==-1) { ll inv=pow_mul(len,mode-2); for(int i=1;i<(len>>1);i++)swap(a[i],a[len-i]); for(int i=0;i<len;i++)a[i]=a[i]*inv%mode; } } ll a[(1<<20)+5],b[(1<<20)+5],c[(1<<20)+5]; void solve(int dep) { if(dep==1) { now.g[0]=pow_mul(G[0],mode-2); now.len=1; las=now; return; } int nxt=(dep+1)/2; solve(nxt); int lim=1,l=0; while(lim<=2*dep)lim<<=1,l++; for(int i=0;i<lim;i++)to[i]=((to[i>>1]>>1)|((i&1)<<(l-1))); for(int i=0;i<lim;i++)a[i]=b[i]=0; for(int i=0;i<dep;i++)a[i]=G[i]; for(int i=0;i<nxt;i++)b[i]=las.g[i]; NTT(a,lim,1),NTT(b,lim,1); for(int i=0;i<lim;i++)c[i]=a[i]*b[i]%mode*b[i]%mode; NTT(c,lim,-1); now.len=dep; for(int i=0;i<dep;i++)now.g[i]=((2*las.g[i]-c[i])%mode+mode)%mode; las=now; } void mul(ll *A,ll *B,int len) { int lim=1,l=0; while(lim<=2*len)lim<<=1,l++; for(int i=1;i<lim;i++)to[i]=((to[i>>1]>>1)|((i&1)<<(l-1))); for(int i=0;i<lim;i++)a[i]=b[i]=0; for(int i=0;i<len;i++)a[i]=A[i],b[i]=B[i]; NTT(a,lim,1),NTT(b,lim,1); for(int i=0;i<lim;i++)c[i]=a[i]*b[i]%mode; NTT(c,lim,-1); } int main() { scanf("%d%d",&n,&m); for(int i=0;i<=n;i++)scanf("%lld",&FF[i]),F[n-i]=FF[i]; for(int i=0;i<=m;i++)scanf("%lld",&GG[i]),G[m-i]=GG[i]; for(int i=n-m+2;i<=m;i++)G[i]=0; for(int i=n-m+1;i<=n;i++)F[i]=0; solve(n-m+1); for(int i=0;i<=n-m;i++)GF[i]=now.g[i]; mul(F,GF,n-m+1); for(int i=0;i<=n-m;i++)Q[i]=c[i]; for(int i=0;i<=n-m;i++)Q[i]=c[n-m-i]; for(int i=0;i<=n-m;i++)printf("%lld ",Q[i]); printf("\n"); mul(Q,GG,max(m+1,n-m+1)); for(int i=0;i<m;i++)printf("%lld ",((FF[i]-c[i])%mode+mode)%mode); printf("\n"); return 0; }