多项式快速插值
前言
給定一堆點(x1,y1),(x2,y2)???(xn,yn)(x_1,y_1),(x_2,y_2)···(x_n,y_n)(x1?,y1?),(x2?,y2?)???(xn?,yn?),求一個n?1n-1n?1次多項式A(x)A(x)A(x)滿足A(xi)=yiA(x_i)=y_iA(xi?)=yi?
做法
先考慮拉格朗日插值
對于每個iii求出一個n?1n-1n?1次多項式Gi(x)G_i(x)Gi?(x)滿足Gi(xi)=yiG_i(x_i)=y_iGi?(xi?)=yi?,并且Gi(xj)=0(j≠i)G_i(x_j)=0(j\neq i)Gi?(xj?)=0(j??=i)
這樣一來L(x)=∑i=1nGi(x)L(x)=\sum_{i=1}^nG_i(x)L(x)=i=1∑n?Gi?(x)
即
L(x)=∑i=0n(yi∏j=1,i≠jnx?xjxi?xj)L(x)=\sum_{i=0}^n\left(y_i\prod_{j=1,i\neq j}^n\frac{x-x_j}{x_i-x_j}\right)L(x)=i=0∑n????yi?j=1,i??=j∏n?xi??xj?x?xj?????
也可以換個形式(好像是重心法??)
L(x)=(∏j=1n(x?xj))(∑i=0nyi(x?xi)∏j=1,i≠jn(xi?xj))L(x)=\left(\prod_{j=1}^n(x-x_j)\right)\left(\sum_{i=0}^n\frac{y_i}{(x-x_i)\prod_{j=1,i\neq j}^n(x_i-x_j)}\right)L(x)=(j=1∏n?(x?xj?))(i=0∑n?(x?xi?)∏j=1,i??=jn?(xi??xj?)yi??)
本題中,我們將L(x)=∑i=0n(yi∏j=1,i≠jnx?xjxi?xj)L(x)=\sum_{i=0}^n\left(y_i\prod_{j=1,i\neq j}^n\frac{x-x_j}{x_i-x_j}\right)L(x)=i=0∑n????yi?j=1,i??=j∏n?xi??xj?x?xj?????
轉化成L(x)=∑i=0n(yi∏j=1,i≠jn(xi?xj)∏j=1,i≠jn(x?xj))L(x)=\sum_{i=0}^n\left(\frac{y_i}{\prod_{j=1,i\neq j}^n(x_i-x_j)}\prod_{j=1,i\neq j}^n(x-x_j)\right)L(x)=i=0∑n????∏j=1,i??=jn?(xi??xj?)yi??j=1,i??=j∏n?(x?xj?)???
先求yi∏j=1,i≠jn(xi?xj)\frac{y_i}{\prod_{j=1,i\neq j}^n(x_i-x_j)}∏j=1,i??=jn?(xi??xj?)yi??,我們發現分子是一個常數,所以轉化成求所有∏j=1,i≠jn(xi?xj)\prod_{j=1,i\neq j}^n(x_i-x_j)j=1,i??=j∏n?(xi??xj?)
我們設R(x)=∏j=1n(x?xj)R(x)=\prod_{j=1}^n(x-x_j)R(x)=∏j=1n?(x?xj?)
我們發現∏j=1,i≠jn(xi?xj)=lim?x→xiR(x)x?xi\prod_{j=1,i\neq j}^n(x_i-x_j)=\lim_{x\rightarrow x_i} \frac{R(x)}{x-x_i}j=1,i??=j∏n?(xi??xj?)=x→xi?lim?x?xi?R(x)?
我們發現lim?x→xiR(x)=0\lim_{x\rightarrow x_i}R(x)=0x→xi?lim?R(x)=0并且lim?x→xix?xi=0\lim_{x\rightarrow x_i}x-x_i=0x→xi?lim?x?xi?=0
根據洛必達法則
我們對分子分母分別求導,得到
∏j=1,i≠jn(xi?xj)=lim?x→xiR(x)x?xi=lim?x→xiR′(x)(x?xi)′=lim?x→xi(R′(x))=R′(xi)\prod_{j=1,i\neq j}^n(x_i-x_j)=\lim_{x\rightarrow x_i} \frac{R(x)}{x-x_i}=\lim_{x\rightarrow x_i} \frac{R'(x)}{(x-x_i)'}=\lim_{x\rightarrow x_i}(R'(x))=R'(x_i)j=1,i??=j∏n?(xi??xj?)=x→xi?lim?x?xi?R(x)?=x→xi?lim?(x?xi?)′R′(x)?=x→xi?lim?(R′(x))=R′(xi?)
所以我們只要求出R(x)R(x)R(x)并對其求導,然后進行多項式多點求值即可完成
然后求出值后用每個yiy_iyi?除,即可
設vi=yi∏j=1,i≠jn(xi?xj)v_i=\frac{y_i}{\prod_{j=1,i\neq j}^n(x_i-x_j)}vi?=∏j=1,i??=jn?(xi??xj?)yi??
我們要求的變成
L(x)=∑i=0n(vi∏j=1,i≠jn(x?xj))L(x)=\sum_{i=0}^n\left(v_i\prod_{j=1,i\neq j}^n(x-x_j)\right)L(x)=i=0∑n????vi?j=1,i??=j∏n?(x?xj?)???
用分治套NTTNTTNTT即可計算
總復雜度O(nlog2n)\mathcal O(nlog^2n)O(nlog2n)
代碼
#include<cstdio> #include<cctype> #include<cstring> #include<cstdlib> #include<vector> namespace fast_IO {const int IN_LEN=10000000,OUT_LEN=10000000;char ibuf[IN_LEN],obuf[OUT_LEN],*ih=ibuf+IN_LEN,*oh=obuf,*lastin=ibuf+IN_LEN,*lastout=obuf+OUT_LEN-1;inline char getchar_(){return (ih==lastin)&&(lastin=(ih=ibuf)+fread(ibuf,1,IN_LEN,stdin),ih==lastin)?EOF:*ih++;}inline void putchar_(const char x){if(oh==lastout)fwrite(obuf,1,oh-obuf,stdout),oh=obuf;*oh++=x;}inline void flush(){fwrite(obuf,1,oh-obuf,stdout);} } using namespace fast_IO; #define getchar() getchar_() #define putchar(x) putchar_((x)) typedef long long ll; #define rg register template <typename T> inline T max(const T a,const T b){return a>b?a:b;} template <typename T> inline T min(const T a,const T b){return a<b?a:b;} template <typename T> inline T mind(T&a,const T b){a=a<b?a:b;} template <typename T> inline T maxd(T&a,const T b){a=a>b?a:b;} template <typename T> inline T abs(const T a){return a>0?a:-a;} template <typename T> inline void swap(T&a,T&b){T c=a;a=b;b=c;} template <typename T> inline void swap(T*a,T*b){T c=a;a=b;b=c;} template <typename T> inline T gcd(const T a,const T b){if(!b)return a;return gcd(b,a%b);} template <typename T> inline T square(const T x){return x*x;}; template <typename T> inline void read(T&x) {char cu=getchar();x=0;bool fla=0;while(!isdigit(cu)){if(cu=='-')fla=1;cu=getchar();}while(isdigit(cu))x=x*10+cu-'0',cu=getchar();if(fla)x=-x; } template <typename T> void printe(const T x) {if(x>=10)printe(x/10);putchar(x%10+'0'); } template <typename T> inline void print(const T x) {if(x<0)putchar('-'),printe(-x);else printe(x); } const int maxn=2097152,mod=998244353; inline int Md(const int x){return x>=mod?x-mod:x;} template<typename T> inline int pow(int x,T y) {rg int res=1;x%=mod;for(;y;y>>=1,x=(ll)x*x%mod)if(y&1)res=(ll)res*x%mod;return res; } namespace Poly///namespace of Poly { int W_[maxn],ha[maxn],hb[maxn],Inv[maxn]; inline void init(const int x) {rg int tim=0,lenth=1;while(lenth<x)lenth<<=1,tim++;for(rg int i=1;i<lenth;i<<=1){const int WW=pow(3,(mod-1)/(i*2));W_[i]=1;for(rg int j=i+1,k=i<<1;j<k;j++)W_[j]=(ll)W_[j-1]*WW%mod;}Inv[0]=Inv[1]=1;for(rg int i=2;i<x;i++)Inv[i]=(ll)(mod-mod/i)*Inv[mod%i]%mod; } int L; inline void DFT(int*A)//prepare:init L {for(rg int i=0,j=0;i<L;i++){if(i>j)swap(A[i],A[j]);for(rg int k=L>>1;(j^=k)<k;k>>=1);}for(rg int i=1;i<L;i<<=1)for(rg int j=0,J=i<<1;j<L;j+=J)for(rg int k=0;k<i;k++){const int x=A[j+k],y=(ll)A[j+k+i]*W_[i+k]%mod;A[j+k]=Md(x+y),A[j+k+i]=Md(mod+x-y);} } void IDFT(int*A) {for(rg int i=1;i<L-i;i++)swap(A[i],A[L-i]);DFT(A); } inline int Quadratic_residue(const int a) {if(a==0)return 0;int b=(rand()<<14^rand())%mod;while(pow(b,(mod-1)>>1)!=mod-1)b=(rand()<<14^rand())%mod;int s=mod-1,t=0,x,inv=pow(a,mod-2),f=1;while(!(s&1))s>>=1,t++,f<<=1;t--,x=pow(a,(s+1)>>1),f>>=1;while(t){f>>=1;if(pow((int)((ll)inv*x%mod*x%mod),f)!=1)x=(ll)x*pow(b,s)%mod;t--,s<<=1;}return min(x,mod-x); } struct poly {std::vector<int>A;poly(){A.resize(0);}poly(const int x){A.resize(1),A[0]=x;}inline int&operator[](const int x){return A[x];}inline int operator[](const int x)const{return A[x];}inline void clear(){A.clear();}inline unsigned int size()const{return A.size();}inline void resize(const unsigned int x){A.resize(x);}void RE(const int x){A.resize(x);for(rg int i=0;i<x;i++)A[i]=0; }void readin(const int MAX){A.resize(MAX);for(rg int i=0;i<MAX;i++)read(A[i]);}void putout()const{for(rg unsigned int i=0;i<A.size();i++)print(A[i]),putchar(' ');}inline poly operator +(const poly b)const{poly RES;RES.resize(max(size(),b.size()));for(rg unsigned int i=0;i<RES.size();i++)RES[i]=Md((i<size()?A[i]:0)+(i<b.size()?b[i]:0));return RES;}inline poly operator -(const poly b)const{poly RES;RES.resize(max(size(),b.size()));for(rg unsigned int i=0;i<RES.size();i++)RES[i]=Md((i<size()?A[i]:0)+mod-(i<b.size()?b[i]:0));return RES;}inline poly operator *(const int b)const{poly RES=*this;for(rg unsigned int i=0;i<RES.size();i++)RES[i]=(ll)RES[i]*b%mod;return RES;}inline poly operator /(const int b)const{poly RES=(*this)*pow(b,mod-2);return RES;}inline poly operator *(const poly b)const{const int RES=A.size()+b.size()-1;L=1;while(L<RES)L<<=1;poly c;c.A.resize(RES);memset(ha,0,L<<2);memset(hb,0,L<<2);for(rg unsigned int i=0;i<A.size();i++)ha[i]=A[i];for(rg unsigned int i=0;i<b.A.size();i++)hb[i]=b.A[i];DFT(ha),DFT(hb);for(rg int i=0;i<L;i++)ha[i]=(ll)ha[i]*hb[i]%mod;IDFT(ha);const int inv=pow(L,mod-2);for(rg int i=0;i<RES;i++)c.A[i]=(ll)ha[i]*inv%mod;return c;}inline poly inv()const{poly C;if(A.size()==1){C=*this;C[0]=pow(C[0],mod-2);return C;}C.resize((A.size()+1)>>1);for(rg unsigned int i=0;i<C.size();i++)C[i]=A[i];C=C.inv();L=1;while(L<(int)size()*2-1)L<<=1;for(rg unsigned int i=0;i<A.size();i++)ha[i]=A[i];for(rg unsigned int i=0;i<C.size();i++)hb[i]=C[i];memset(ha+A.size(),0,(L-A.size())<<2);memset(hb+C.size(),0,(L-C.size())<<2);DFT(ha),DFT(hb);for(rg int i=0;i<L;i++)ha[i]=(ll)(2-(ll)hb[i]*ha[i]%mod+mod)*hb[i]%mod;IDFT(ha);const int inv=pow(L,mod-2);C.resize(size());for(rg unsigned int i=0;i<size();i++)C[i]=(ll)ha[i]*inv%mod;return C;} /* inline poly inv()const{poly C;if(A.size()==1){C=*this;C[0]=pow(C[0],mod-2);return C;}C.resize((A.size()+1)>>1);for(rg unsigned int i=0;i<C.size();i++)C[i]=A[i];C=C.inv();poly D=(poly)2-*this*C;D.resize(size());D=D*C;D.resize(size());return D;}*///大常數版本 inline void Reverse(const int n){A.resize(n);for(rg int i=0,j=n-1;i<j;i++,j--)swap(A[i],A[j]);}inline poly operator /(const poly B)const{if(size()<B.size())return 0;poly a=*this,b=B;a.Reverse(size()),b.Reverse(B.size());b.resize(size()-B.size()+1);b=b.inv();b=b*a;b.Reverse(size()-B.size()+1);return b;}inline poly operator %(const poly B)const{poly C=(*this)-(*this)/B*B;C.resize(B.size()-1);return C;}inline poly diff()const{poly C;C.resize(size()-1);for(rg unsigned int i=1;i<size();i++)C[i-1]=(ll)A[i]*i%mod;return C;}inline poly inte()const{poly C;C.resize(size()+1);for(rg unsigned int i=0;i<size();i++)C[i+1]=(ll)A[i]*Inv[i+1]%mod;C[0]=0;return C;}inline poly ln()const{poly C=(diff()*inv()).inte();C.resize(size());return C;}inline poly exp()const{poly C;if(size()==1){C=*this;C[0]=1;return C;}C.resize((size()+1)>>1);for(rg unsigned int i=0;i<C.size();i++)C[i]=A[i];C=C.exp();C.resize(size());poly D=(poly)1-C.ln()+*this;D=D*C;D.resize(size());return D;}inline poly sqrt()const{poly C;if(size()==1){C=*this;C[0]=Quadratic_residue(C[0]);return C;}C.resize((size()+1)>>1);for(rg unsigned int i=0;i<C.size();i++)C[i]=A[i];C=C.sqrt();C.resize(size());C=(C+*this*C.inv())*(int)499122177;C.resize(size());return C;}inline poly operator >>(const unsigned int x)const{poly C;if(size()<x){C.resize(0);return C;}C.resize(size()-x);for(rg unsigned int i=0;i<C.size();i++)C[i]=A[i+x];return C;}inline poly operator <<(const unsigned int x)const{poly C;C.RE(size()+x);for(rg unsigned int i=0;i<size();i++)C[i+x]=A[i];return C;}inline poly Pow(const unsigned int x)const{for(rg unsigned int i=0;i<size();i++)if(A[i]){poly C=((((*this/A[i])>>i).ln()*x).exp()*pow(A[i],x))<<(min(i*x,size()));C.resize(size());return C;}return *this;}inline void cheng(const poly&B){for(rg unsigned int i=0;i<size();i++)A[i]=(ll)A[i]*B[i]%mod; }inline void jia(const poly&B){for(rg unsigned int i=0;i<size();i++)A[i]=Md(A[i]+B[i]); }inline void dft(){memset(ha,0,L<<2);for(rg unsigned int i=0;i<A.size();i++)ha[i]=A[i];DFT(ha);resize(L);for(rg int i=0;i<L;i++)A[i]=ha[i];}inline void idft(){memset(ha,0,L<<2);for(rg unsigned int i=0;i<A.size();i++)ha[i]=A[i];IDFT(ha);const int inv=pow(L,mod-2);for(rg int i=0;i<L;i++)A[i]=(ll)ha[i]*inv%mod;while(size()&&!A[size()-1])A.pop_back();} }; void fz(const int root,const int l,const int r,std::vector<int>&v,std::vector<poly>&A) {if(l==r){A[root].resize(2);A[root][0]=(mod-v[l])%mod;A[root][1]=1;return;}const int mid=(l+r)>>1;fz(root<<1,l,mid,v,A),fz(root<<1|1,mid+1,r,v,A);A[root]=A[root<<1]*A[root<<1|1]; } void calc(const int root,const int l,const int r,std::vector<int>&v,std::vector<poly>&A,std::vector<poly>&B) {if(l==r){v[l]=B[root][0];return;}const int mid=(l+r)>>1;B[root<<1]=B[root]%A[root<<1];B[root<<1|1]=B[root]%A[root<<1|1];calc(root<<1,l,mid,v,A,B),calc(root<<1|1,mid+1,r,v,A,B); } void multi_point_evaluation(const poly a,std::vector<int>&v) {std::vector<poly>A,B;A.resize(maxn),B.resize(maxn);fz(1,0,v.size()-1,v,A);B[1]=a%A[1];calc(1,0,v.size()-1,v,A,B); } void fz2(const int root,const int l,const int r,std::vector<int>&y,std::vector<poly>&A,std::vector<poly>&B) {if(l==r){B[root].resize(1),B[root][0]=y[l];return;}const int mid=(l+r)>>1;fz2(root<<1,l,mid,y,A,B),fz2(root<<1|1,mid+1,r,y,A,B);B[root]=B[root<<1]*A[root<<1|1]+B[root<<1|1]*A[root<<1]; } poly interpolation(std::vector<int>&x,std::vector<int>&y) {std::vector<poly>A,B;A.resize(maxn),B.resize(maxn);fz(1,0,x.size()-1,x,A);multi_point_evaluation(A[1].diff(),x);for(rg unsigned int i=0;i<x.size();i++)y[i]=(ll)y[i]*pow(x[i],mod-2)%mod;fz2(1,0,x.size()-1,y,A,B);return B[1]; } }///namespace of Poly Poly::poly a;std::vector<int>x,y; int n; int main() {Poly::init(maxn);///namespace of Polyread(n),x.resize(n),y.resize(n);for(rg int i=0;i<n;i++)read(x[i]),read(y[i]);(Poly::interpolation(x,y)).putout();return flush(),0; }總結
這個算法雖然有點常數,但是能很好的解決一些問題
總結
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