洛谷P5158 【模板】多项式快速插值
题面
前置芝士
题解
首先根据拉格朗日插值公式我们可以暴力\(O(n^2)\)插出这个多项式，然而这显然是\(gg\)的
那么看看怎么优化，先来看一看拉格朗日插值的公式
\[f(x)=\sum_{i = 1}^{n} y_i \prod_{i \not = j} \frac{x - x_j}{x_i - x_j}
\]
转化一下
\[f(x)=\sum_{i = 1}^{n}{ y_i\over \prod_{i \not = j}{x_i - x_j}} \prod_{i \not = j}(x - x_j)
\]
来考虑一下\({ y_i\over \prod_{i \not = j}{x_i - x_j}}\)这个东西，上面是个常数，那么只考虑下面。如果我们设多项式\(g(x)=\prod_{i=1}^n (x-x_i)\)，那么下面那个东西就是\({g(x_i)\over (x-x_i)}\)
这分子分母全为\(0\)我怎么求啊……
根据我也不知道是啥的洛必达法则，如果
\[\lim_{x\to a}f(x)=0,\lim_{x\to a}g(x)=0
\]
则有
\[\lim_{x\to a}{f(x)\over g(x)}=\lim_{x\to a}{f'(x)\over g'(x)}
\]
那么我们代入之后可以发现\({g(x_i)\over (x-x_i)}=g'(x_i)\)
先分治\(NTT\)算出\(g\)，然后多点求值把每个点处的值算出来就好了
那么接下来我们就考虑分治，设\(f_{l,r}\)表示\((x_l,y_l),(x_r,y_r)\)这些点算出来的答案，则有
\[\begin{aligned}
f_{l,r}
&=\sum_{i = l}^{r}{ y_i\over g'(x_i)} \prod_{j=l,i \not = j}^r(x - x_j)\\
&=\prod_{j=mid+1}^r(x - x_j)\sum_{i = l}^{mid}{ y_i\over g'(x_i)} \prod_{j=l,i \not = j}^{mid}(x - x_j)+\prod_{j=l}^{mid}(x - x_j)\sum_{i = mid+1}^{r}{ y_i\over g'(x_i)} \prod_{j=mid+1,i \not = j}^{r}(x - x_j)\\
&=\prod_{j=mid+1}^r(x - x_j)f_{l,mid}+\prod_{j=l}^{mid}(x - x_j)f_{mid+1,r}\\
\end{aligned}
\]
然后没有然后了
复杂度为\(O(n\log^2n)\)
//minamoto
#include<bits/stdc++.h>
#define R register
#define fp(i,a,b) for(R int i=(a),I=(b)+1;i<I;++i)
#define fd(i,a,b) for(R int i=(a),I=(b)-1;i>I;--i)
#define go(u) for(int i=head[u],v=e[i].v;i;i=e[i].nx,v=e[i].v)
using namespace std;
char buf[1<<21],*p1=buf,*p2=buf;
inline char getc(){return p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++;}
int read(){
R int res,f=1;R char ch;
while((ch=getc())>'9'||ch<'0')(ch=='-')&&(f=-1);
for(res=ch-'0';(ch=getc())>='0'&&ch<='9';res=res*10+ch-'0');
return res*f;
}
char sr[1<<21],z[20];int C=-1,Z=0;
inline void Ot(){fwrite(sr,1,C+1,stdout),C=-1;}
void print(R int x){
if(C>1<<20)Ot();if(x<0)sr[++C]='-',x=-x;
while(z[++Z]=x%10+48,x/=10);
while(sr[++C]=z[Z],--Z);sr[++C]=' ';
}
const int N=(1<<18)+5,P=998244353;
inline int add(R int x,R int y){return x+y>=P?x+y-P:x+y;}
inline int dec(R int x,R int y){return x-y<0?x-y+P:x-y;}
inline int mul(R int x,R int y){return 1ll*x*y-1ll*x*y/P*P;}
int ksm(R int x,R int y){
R int res=1;
for(;y;y>>=1,x=mul(x,x))(y&1)?res=mul(res,x):0;
return res;
}
int r[19][N],w[2][N],lg[N],inv[19];
void Pre(){
fp(d,1,18){
fp(i,1,(1<<d)-1)r[d][i]=(r[d][i>>1]>>1)|((i&1)<<(d-1));
lg[1<<d]=d,inv[d]=ksm(1<<d,P-2);
}
for(R int t=(P-1)>>1,i=1,x,y;i<262144;i<<=1,t>>=1){
x=ksm(3,t),y=ksm(332748118,t),w[0][i]=w[1][i]=1;
fp(k,1,i-1)
w[1][k+i]=mul(w[1][k+i-1],x),
w[0][k+i]=mul(w[0][k+i-1],y);
}
}
int lim,d,n,m;
inline void init(R int len){lim=1,d=0;while(lim<len)lim<<=1,++d;}
void NTT(int *A,int ty){
fp(i,0,lim-1)if(i<r[d][i])swap(A[i],A[r[d][i]]);
for(R int mid=1;mid<lim;mid<<=1)
for(R int j=0,t;j<lim;j+=(mid<<1))
fp(k,0,mid-1)
A[j+k+mid]=dec(A[j+k],t=mul(w[ty][mid+k],A[j+k+mid])),
A[j+k]=add(A[j+k],t);
if(!ty)fp(i,0,lim-1)A[i]=mul(A[i],inv[d]);
}
void Inv(int *a,int *b,int len){
if(len==1)return b[0]=ksm(a[0],P-2),void();
Inv(a,b,len>>1),lim=(len<<1),d=lg[lim];
static int A[N],B[N];
fp(i,0,len-1)A[i]=a[i],B[i]=b[i];fp(i,len,lim-1)A[i]=B[i]=0;
NTT(A,1),NTT(B,1);
fp(i,0,lim-1)A[i]=mul(A[i],mul(B[i],B[i]));
NTT(A,0);
fp(i,0,len-1)b[i]=dec(add(b[i],b[i]),A[i]);
fp(i,len,lim-1)b[i]=0;
}
struct node{
node *lc,*rc;vector<int>vec;int deg;
void Mod(const int *a,int *r,int n){
static int A[N],B[N],D[N];
int len=1;while(len<=n-deg)len<<=1;
fp(i,0,n)A[i]=a[n-i];fp(i,0,deg)B[i]=vec[deg-i];
fp(i,n-deg+1,len-1)B[i]=0;
Inv(B,D,len);
lim=(len<<1),d=lg[lim];
fp(i,n-deg+1,lim-1)A[i]=D[i]=0;
NTT(A,1),NTT(D,1);
fp(i,0,lim-1)A[i]=mul(A[i],D[i]);
NTT(A,0);
reverse(A,A+n-deg+1);
init(n+1);
fp(i,n-deg+1,lim-1)A[i]=0;
fp(i,0,deg)B[i]=vec[i];fp(i,deg+1,lim-1)B[i]=0;
NTT(A,1),NTT(B,1);
fp(i,0,lim-1)A[i]=mul(A[i],B[i]);
NTT(A,0);
fp(i,0,deg-1)r[i]=dec(a[i],A[i]);
}
void Mul(){
static int A[N],B[N];deg=lc->deg+rc->deg,vec.resize(deg+1),init(deg+1);
fp(i,0,lc->deg)A[i]=lc->vec[i];fp(i,lc->deg+1,lim-1)A[i]=0;
fp(i,0,rc->deg)B[i]=rc->vec[i];fp(i,rc->deg+1,lim-1)B[i]=0;
NTT(A,1),NTT(B,1);
fp(i,0,lim-1)A[i]=mul(A[i],B[i]);
NTT(A,0);
fp(i,0,deg)vec[i]=A[i];
}
}pool[N],*rt;
struct QAQ{
QAQ *lc,*rc;vector<int>vec;int l,r;
void Mul(node* p){
static int A[N],B[N],C[N],D[N];
int mid=(l+r)>>1;init(r-l+1+1);
fp(i,0,mid-l+1)A[i]=lc->vec[i],C[i]=p->lc->vec[i];
fp(i,mid-l+2,lim-1)A[i]=C[i]=0;
fp(i,0,r-mid)B[i]=rc->vec[i],D[i]=p->rc->vec[i];
fp(i,r-mid+1,lim-1)B[i]=D[i]=0;
NTT(A,1),NTT(B,1),NTT(C,1),NTT(D,1);
fp(i,0,lim-1)A[i]=add(mul(A[i],D[i]),mul(B[i],C[i]));
NTT(A,0);vec.resize(r-l+2);
fp(i,0,r-l+1)vec[i]=A[i];
}
}o[N],*qwq;
int a[N],tot,cnt;
inline node* newnode(){return &pool[tot++];}
inline QAQ* newQAQ(){return &o[cnt++];}
void solve(node* &p,int l,int r){
p=newnode();
if(l==r)return p->deg=1,p->vec.resize(2),p->vec[0]=P-a[l],p->vec[1]=1,void();
int mid=(l+r)>>1;
solve(p->lc,l,mid),solve(p->rc,mid+1,r);
p->Mul();
}
int b[25],f[N];
void calc(node* p,int l,int r,const int *A){
if(r-l<=512){
fp(i,l,r){
int x=a[i],c1,c2,c3,c4,now=A[r-l];
b[0]=1;fp(j,1,16)b[j]=mul(b[j-1],x);
for(R int j=r-l-1;j-15>=0;j-=16){
c1=(1ll*now*b[16]+1ll*A[j]*b[15]+1ll*A[j-1]*b[14]+1ll*A[j-2]*b[13])%P,
c2=(1ll*A[j-3]*b[12]+1ll*A[j-4]*b[11]+1ll*A[j-5]*b[10]+1ll*A[j-6]*b[9])%P,
c3=(1ll*A[j-7]*b[8]+1ll*A[j-8]*b[7]+1ll*A[j-9]*b[6]+1ll*A[j-10]*b[5])%P,
c4=(1ll*A[j-11]*b[4]+1ll*A[j-12]*b[3]+1ll*A[j-13]*b[2]+1ll*A[j-14]*b[1])%P,
now=(0ll+c1+c2+c3+c4+A[j-15])%P;
}
fd(j,(r-l)%16-1,0)now=(1ll*now*x+A[j])%P;
f[i]=now;
}
return;
}
int mid=(l+r)>>1,b[p->deg+1];
p->lc->Mod(A,b,p->deg-1),calc(p->lc,l,mid,b);
p->rc->Mod(A,b,p->deg-1),calc(p->rc,mid+1,r,b);
}
int x[N],y[N],A[N];
void loli(QAQ* &qwq,node *p,int l,int r){
qwq=newQAQ(),qwq->l=l,qwq->r=r;
if(l==r)return qwq->vec.resize(2),qwq->vec[0]=mul(y[l],ksm(f[l],P-2)),qwq->vec[1]=0,void();
int mid=(l+r)>>1;
loli(qwq->lc,p->lc,l,mid),loli(qwq->rc,p->rc,mid+1,r);
qwq->Mul(p);
}
int main(){
// freopen("testdata.in","r",stdin);
n=read(),Pre();
fp(i,1,n)x[i]=a[i]=read(),y[i]=read();
solve(rt,1,n);
fp(i,1,n)A[i-1]=mul(rt->vec[i],i);A[n]=0;
calc(rt,1,n,A);
loli(qwq,rt,1,n);
fp(i,0,n-1)print(qwq->vec[i]);
return Ot(),0;
}
深深地明白自己的弱小
