luoguP4512 【模板】多项式除法
求 $F(x)=Q(x)\times G(x)+R(x)$ 中的 $Q(x),R(x)$
$F(\frac{1}{x})=Q(\frac{1}{x})\times G(\frac{1}{x}) + R(\frac{1}{x})$
$x^{n}F(\frac{1}{x})=x^{n-m}Q(\frac{1}{x})x^{m}G(\frac{1}{x})+x^{n-m+1}x^{m-1}R(\frac{1}{x})$
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 $x^{n-m+1})$
$F_{R}(x)\equiv Q_{R}(x)\times G_{R}(x)$ (mod $x^{n-m+1}$)
$Q_{R}(x)\equiv F_{R}(x)\times G_{R}^{-1}(x)$(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 $n-m+1$
$R(x)=F(x)-G(x)\times Q(x)$
// 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("\n"); 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; }