Taylor 展开
对于一个函数(f(x),)如果我们知道它在(x_0)处的各阶导数,那么:
[f(x)=sum_{i=0}^n frac{f^{(i)}(x_0)(x-x0)^i}{i!}
]
即 我们在(x_0)处逼近了(f(x).)
牛顿迭代
考虑求:
[G(F(x))equiv 0(mod x^n)
]
对于(n=1)特殊求出来
考虑已经解决了:
[G(F_0(x))equiv 0(mod x^{leftlceil frac{n}{2}
ight
ceil} )
]
考虑如何拓展到(x^n.)
在这里泰勒展开一下:
[G(F(x))=sum_{i=0}^infty frac{G^{(i)}(F_0(x))(F(x)-F_0(x))^i}{i!}
]
注意到当(ige 2)时 ((F(x)-F_0(x))^i)的最低非(0)次项的次数是严格大于(2leftlceilfrac{n}{2} ight ceil,)所以:
[G(F(x))equiv G(F_0(x))+(F(x)-F_0(x))G'(F_0(x))(mod x^n)
]
注意到由题设得:
[G(F(x))equiv 0(mod x^n)
]
所以:
[G(F_0(x))+(F(x)-F_0(x))G'(F_0(x))equiv 0(mod x^n)
]
整理可以得到:
[F(x)equiv F_0(x)-frac{G(F_0(x))}{G'(F_0(x))}(mod x^n)
]
例题:
1.多项式 exp
给定多项式(A(x),)求(e^{A(x)}(mod x^n).)
设(F(x)equiv e^{A(x)}(mod x^n),)两边取对数:
[ln F(x)equiv A(x)(mod x^n)
]
[ln F(x)-A(x)equiv 0(mod x^n)
]
将(A(x))看成常数,设:
[G(F(x))=ln F(x)-A(x)
]
则(G(F(x))equiv 0mod (x^n))
(G'(F(x))=frac{1}{F(x)})
[F(x)equiv F_0(x)-frac{ln F_0(x)-A(x)}{frac{1}{F_0(x)}}(mod x^n)
]
[F(x)equiv F_0(x)-F_0(x)(ln F_0(x)-A(x))(mod x^n)
]
[F(x)equiv F_0(x)(1-ln F_0(x)+A(x))(mod x^n)
]
递归求解(O(nlog n).)
代码:
#include<bits/stdc++.h>
using namespace std;
#define int long long
const int N=310000;
const int mod=998244353;
int rev[N],a[N],b[N],c[N],d[N],e[N],f[N],g[N];
int lnb[N],G[N],n,k[N];
inline int add(int x,int y){return (x+y)%mod;}
inline int mul(int x,int y){return 1ll*x*y%mod;}
inline int qpow(int a,int b){
int res=1;
while(b){
if(b&1)res=mul(res,a);
a=mul(a,a);b>>=1;
}
return res;
}
void NTT(int *A,int lim,int tp){
for(int i=0;i<lim;++i)if(i<rev[i])swap(A[i],A[rev[i]]);
for(int i=1;i<lim;i<<=1){
int gn=qpow(3,(mod-1)/(i<<1));
if(tp!=1)gn=qpow(gn,mod-2);
for(int j=0;j<lim;j+=(i<<1)){
int G=1;
for(int k=0;k<i;++k,G=mul(G,gn)){
int x=A[j+k],y=mul(G,A[i+j+k]);
A[j+k]=add(x,y);A[i+j+k]=add(x,mod-y);
}
}
}
if(tp==1)return;
int inv=qpow(lim,mod-2);
for(int i=0;i<lim;++i)A[i]=mul(A[i],inv);
}
void Inv(int d,int *a,int *b){
if(d==1){b[0]=qpow(a[0],mod-2);return;}
Inv((d+1)>>1,a,b);
int lim=1,len=0;
while(lim<(d<<1))lim<<=1,len++;
for(int i=1;i<lim;++i)rev[i]=(rev[i>>1]>>1)|((i&1)<<(len-1));
for(int i=0;i<d;++i)c[i]=a[i];
for(int i=d;i<lim;++i)c[i]=0;
NTT(c,lim,1);NTT(b,lim,1);
for(int i=0;i<lim;++i)b[i]=1ll*(2-1ll*b[i]*c[i]%mod+mod)%mod*b[i]%mod;
NTT(b,lim,-1);for(int i=d;i<lim;++i)b[i]=0;
}
inline void Dx(int *a,int *b,int L){for(int i=1;i<L;++i)b[i-1]=mul(i,a[i]);b[L-1]=0;}
inline void Int(int *a,int *b,int L){for(int i=1;i<L;++i)b[i]=mul(a[i-1],qpow(i,mod-2));b[0]=0;}
void Ln(int L,int *a,int *R){
Dx(a,e,L);Inv(L,a,f);
int lim=1,len=0;
while(lim<(L<<1))lim<<=1,len++;
for(int i=1;i<lim;++i)rev[i]=(rev[i>>1]>>1)|((i&1)<<(len-1));
NTT(e,lim,1);NTT(f,lim,1);
for(int i=0;i<lim;++i)e[i]=mul(e[i],f[i]);
NTT(e,lim,-1);Int(e,R,lim);
for(int i=0;i<lim;++i)e[i]=f[i]=0;
}
void Exp(int d,int *a,int *b){
if(d==1){b[0]=1;return;}
Exp((d+1)>>1,a,b);
Ln(d,b,lnb);
int lim=1,len=0;
while(lim<(d<<1))lim<<=1,len++;
for(int i=1;i<lim;++i)rev[i]=(rev[i>>1]>>1)|((i&1)<<(len-1));
for(int i=0;i<d;++i)lnb[i]=a[i]>=lnb[i]?a[i]-lnb[i]:a[i]-lnb[i]+mod;
for(int i=d;i<lim;++i)b[i]=lnb[i]=0;
lnb[0]++;lnb[0]%=mod;
NTT(lnb,lim,1);NTT(b,lim,1);
for(int i=0;i<lim;++i)b[i]=mul(b[i],lnb[i]);
NTT(b,lim,-1);
for(int i=d;i<lim;++i)b[i]=lnb[i]=0;
}
signed main(){
scanf("%lld",&n);
for(int i=0;i<n;++i)scanf("%lld",&a[i]);
int len=1;
while(len<=n)len<<=1;
Exp(len,a,k);
//Ln(len,a,k);
for(int i=0;i<n;++i)printf("%lld ",k[i]);
puts("");
return 0;
}
2.多项式开根
给定多项式(A(x),)求(B^2(x)equiv A(x)(mod x^n).)
[G(B(x))=B^2(x)-A(x)
]
求(G(B(x))equiv 0(mod x^n).)
令(F(x)=B(x).)套用牛顿迭代公式:
[F(x)equiv F_0(x)-frac{G(F_0(x))}{G'(F_0(x))}(mod x^n)
]
注意到,(A(x))是常数,(G(x))的导数(G'(x)=2B(x))
所以原式:
[F(x)equiv F_0(x)-frac{F_0^2(x)-A(x)}{2F_0(x)}(mod x^n)
]
[F(x)equiv frac{F_0^2(x)+A(x)}{2F_0(x)}
]
牛顿迭代即可。复杂度(O(nlog n).)
#include<bits/stdc++.h>
using namespace std;
#define int long long
const int N=310000;
const int mod=998244353;
int rev[N],a[N],b[N],c[N],F[N],G[N],ans[N];
int n,inv2,C[N];
inline int add(int x,int y){return (x+y)%mod;}
inline int mul(int x,int y){return 1ll*x*y%mod;}
inline int qpow(int a,int b){
int res=1;
while(b){
if(b&1)res=mul(res,a);
a=mul(a,a);b>>=1;
}
return res;
}
void NTT(int *A,int lim,int tp){
for(int i=0;i<lim;++i)if(i<rev[i])swap(A[i],A[rev[i]]);
for(int i=1;i<lim;i<<=1){
int gn=qpow(3,(mod-1)/(i<<1));
if(tp!=1)gn=qpow(gn,mod-2);
for(int j=0;j<lim;j+=(i<<1)){
int G=1;
for(int k=0;k<i;++k,G=mul(G,gn)){
int x=A[j+k],y=mul(G,A[i+j+k]);
A[j+k]=add(x,y);A[i+j+k]=add(x,mod-y);
}
}
}
if(tp==1)return;
int inv=qpow(lim,mod-2);
for(int i=0;i<lim;++i)A[i]=mul(A[i],inv);
}
void Inv(int d,int *a,int *b){
if(d==1){b[0]=qpow(a[0],mod-2);return;}
Inv((d+1)>>1,a,b);
int lim=1,len=0;
while(lim<(d<<1))lim<<=1,len++;
for(int i=1;i<lim;++i)rev[i]=(rev[i>>1]>>1)|((i&1)<<(len-1));
for(int i=0;i<d;++i)c[i]=a[i];
for(int i=d;i<lim;++i)c[i]=0;
NTT(c,lim,1);NTT(b,lim,1);
for(int i=0;i<lim;++i)b[i]=1ll*(2-1ll*b[i]*c[i]%mod+mod)%mod*b[i]%mod;
NTT(b,lim,-1);for(int i=d;i<lim;++i)b[i]=0;
}
void Sqrt(int d,int *a,int *b){
if(d==1){b[0]=1;return;}
Sqrt((d+1)>>1,a,b);
int lim=1,len=0;
while(lim<(d<<1))lim<<=1,len++;
for(int i=1;i<lim;++i)rev[i]=(rev[i>>1]>>1)|((i&1)<<(len-1));
Inv(d,b,G);
for(int i=0;i<d;++i)C[i]=a[i];
for(int i=d;i<lim;++i)C[i]=0;
NTT(C,lim,1);NTT(G,lim,1);
for(int i=0;i<lim;++i)C[i]=mul(G[i],C[i]);
NTT(C,lim,-1);//G=A/F_0
for(int i=0;i<d;++i)b[i]=1ll*(b[i]+C[i])%mod*inv2%mod;
for(int i=d;i<lim;++i)b[i]=G[i]=C[i]=0;
for(int i=0;i<d;++i)C[i]=G[i]=0;
}
signed main(){
scanf("%lld",&n);
for(int i=0;i<n;++i)scanf("%lld",&a[i]);
inv2=qpow(2,mod-2);
int len=1;
while(len<=n)len<<=1;
Sqrt(len,a,ans);
for(int i=0;i<n;++i)printf("%lld ",ans[i]);
puts("");
return 0;
}