多项式全家桶
一.多项式牛顿迭代法
已知多项式 \(G(x)\) ,求 \(F(x)\) ,满足:
\[G(F(x)) \equiv 0 \pmod {x^n}
\]
假设我们有一个 \(F_0(x)\) 满足:
\[G(F_0(x)) \equiv 0 \pmod{x^{\lceil \frac{n}{2} \rceil}}
\]
由定义可知:
\[F(x)-F_0(x)\equiv 0 \pmod{x^{\lceil \frac{n}{2} \rceil}}
\]
\[\Rightarrow \forall k \ge 2, \left( F(x)-F_0(x) \frac{}{} \right)^k \equiv 0 \pmod{x^n}
\]
由泰勒展开:
\[\sum_{i=0}^{\infty} \frac{G^{(i)}(F_0(x))}{i!}(F(x)-F_0(x))^i \equiv 0 \pmod{x^n}
\]
\[\Rightarrow G(F_0(x))+G'(F_0(x))(F(x)-F_0(x)) \equiv 0 \pmod {x^n}
\]
\[\Rightarrow F(x) \equiv F_0(x)-\frac{G(F_0(x))}{G'(F_0(x))} \pmod {x^n}
\]
二.多项式乘法逆
已知多项式 \(G(x)\) ,求 \(F(x)\) ,满足:
\[F(x)G(x) \equiv 1 \pmod {x^n}
\]
假设我们有一个 \(F_0(x)\) 满足:
\[F_0(x)G(x) \equiv 1 \pmod{x^{\lceil \frac{n}{2} \rceil}}
\]
由 (一) 可知:
\[\left( F(x)-F_0(x) \frac{}{} \right)^2 \equiv 0 \pmod{x^n}
\]
\[\Rightarrow F^2(x)-2F(x)F_0(x)+F_0^2(x) \equiv 0 \pmod{x^n}
\]
同乘 \(G(x)\):
\[\Rightarrow F(x)-2F_0(x)+F_0^2(x)G(x) \equiv 0 \pmod{x^n}
\]
\[\Rightarrow F(x) \equiv F_0(x)\left( 2-F_0(x)G(x) \frac{}{} \right)\pmod{x^n}
\]
注意 \(g\) 的长度也要控制为 \(len\) , 不然就退化成 \(\mathcal{O}(nlog^2n)\) 的了。
Poly Inv( Poly g ) {
Poly f = Poly( 1 , Inv( g[ 0 ] ) ) , A;
for( int len = 2 ; len < 2 * len( g ) ; len <<= 1 ) {
A = g; A.resize( len ); //!!!
Poly tmp = f * A; tmp.resize( len );
f = f * ( 2 - tmp ); f.resize( len );
}
return f;
}
虽然好写但不知道为什么常数那么大
三.多项式导数
\[F(x)=\sum_{i=0}^n f_ix^i
\]
\[\begin{aligned}
\Rightarrow F'(x)&= \sum_{i=1}^{n} if_i x^{i-1} \\
&= \sum_{i=0}^{n-1} (i+1)f_{i+1} x^i
\end{aligned}\]
Poly Der( Poly f ) {
for( int i = 0 ; i < len( f ) - 1 ; i ++ ) f[ i ] = Mul( i + 1 , f[ i + 1 ] );
f[ len( f ) - 1 ] = 0;
return f;
}
四.多项式积分
\[F(x)=\sum_{i=0}^n f_ix^i
\]
\[\begin{aligned}
\Rightarrow \int F(x)&= \sum_{i=0}^{n} \frac{f_i}{i+1} x^{i+1}\\
&= \sum_{i=1}^{n+1} \frac{f_{i-1}}{i} x^{i}\\
\end{aligned}\]
Poly Int( Poly f ) {
f.resize( len( f ) + 1 );
for( int i = len( f ) - 1 ; i >= 1 ; i -- ) f[ i ] = Mul( f[ i - 1 ] , iv[ i ] );
f[ 0 ] = 0;
return f;
}
五.多项式 ln
已知多项式 \(G(x)\) , 求 \(F(x)\) , 满足:
\[F(x) \equiv \ln G(x) \pmod{x^n}
\]
两边同时求导 (注意右式为复合函数):
\[F'(x) \equiv \frac{G'(x)}{G(x)} \pmod{x^n}
\]
算完后积分回来就好了。
因为保证 \(g(0)=1\) , 所以积分常数 \(C\) 为 \(0\)。
Poly Ln( Poly f ) {
return Int( Der( f ) * Inv( f ) );
}
六.多项式 exp
已知多项式 \(G(x)\) , 求 \(F(x)\) , 满足:
\[F(x) \equiv e^{G(x)} \pmod{x^n}
\]
两边取 ln:
\[\ln F(x) \equiv G(x) \pmod{x^n}
\]
\[\ln F(x)-G(x) \equiv 0 \pmod{x^n}
\]
令 \(H(F(x)) \equiv \ln F(x) - G(x) \equiv 0 \pmod{x^n}\) 。
两边求导:
\[H'(F(x)) \equiv \frac{1}{F(x)} \pmod{x^n}
\]
\[\]
用牛顿迭代展开:
\[F(x) \equiv F_0(x)-\frac{H
(F_0(x))}{H'(F_0(x))} \pmod {x^n}\]
\[\Rightarrow F(x) \equiv F_0(x)-\frac{\ln F_0(x)-G(x)}{F_0^{-1}(x)} \pmod {x^n}
\]
\[\Rightarrow F(x) \equiv F_0(x)\left( 1-\ln F_0(x)+G(x) \frac{}{} \right) \pmod {x^n}
\]
Poly Exp( Poly f ) {
Poly g = Poly( 1 , 1 );
for( mxlen = 2 ; mxlen < 4 * len( f ) ; mxlen <<= 1 ) {
Poly A = f; A.resize( mxlen );
g = g * ( ( 1 - Ln( g ) ) + A );
}
g.resize( len( f ) ); return g;
}
不知道为什么 mxlen
需要到 \(4\) 倍,希望有好心人指出。
七.多项式幂函数
已知多项式 \(G(x)\) , 求 \(F(x)\) , 满足:
\[F(x) \equiv G^k(x) \pmod{x^n}
\]
同时取 ln 得到:
\[\ln F(x) \equiv k \ln G(x) \pmod{x^n}
\]
然后 exp 回去即可。
常数一目了然,不言而喻
Poly pow( Poly f , int k ) {
f = Ln( f ); f = f * k; f = Exp( f );
return f;
}
八.多项式开根
已知多项式 \(G(x)\) , 求 \(F(x)\) , 满足:
\[F(x) \equiv \sqrt{G(x)} \pmod{x^n}
\]
法一
使用多项式幂函数计算 \(G^{\frac{1}{2}}(x)\) ,复杂度 \(\mathcal{O(n \log n)}\)
吸氧才能过
法二
\[F(x) \equiv \sqrt{G(x)} \pmod{x^n}
\]
\[\Rightarrow F^2(x) - G(x) \equiv 0 \pmod{x^n}
\]
令 \(H(F(x)) \equiv F^2(x)-G(x) \equiv 0\)
两边求导:
\[H'(F(x))=2F(x)
\]
用牛顿迭代展开:
\[F(x) \equiv F_0(x)-\frac{H
(F_0(x))}{H'(F_0(x))} \pmod {x^n}\]
\[\Rightarrow F(x) \equiv F_0(x)-\frac{F_0^2(x)-G(x)}{2F_0(x)} \pmod {x^n}
\]
\[\Rightarrow F(x) \equiv \frac{F_0^2(x)+G(x)}{2F_0(x)} \pmod {x^n}
\]
复杂度同样是 \(\mathcal{O(n \log n)}\)
当常数项不为 \(1\) 时,可以用二次剩余算出结果。
九.多项式除法
int Add( int x , int y ) { x += y; return x >= Mod ? x - Mod : x; }
int Sub( int x , int y ) { x -= y; return x < 0 ? x + Mod : x; }
int Mul( int x , int y ) { return 1ll * x * y % Mod; }
int Quick_pow( int x , int po ) { int Ans = 1; for( ; po ; po >>= 1 , x = Mul( x , x ) ) if( po & 1 ) Ans = Mul( Ans , x ); return Ans; }
int Inv( int x ) { return Quick_pow( x , Mod - 2 ); }
int iv[ MAXN + 5 ];
void Init() {
iv[ 1 ] = 1;
for( int i = 2 ; i <= MAXN ; i ++ ) iv[ i ] = Mul( Mod - Mod / i , iv[ Mod % i ] );
}
#define Poly vector< int >
#define len( x ) ( (int)x.size() )
Poly operator - ( int x , Poly f ) { for( int i = 0 ; i < len( f ) ; i ++ ) f[ i ] = Mod - f[ i ]; f[ 0 ] = Add( f[ 0 ] , x ); return f; }
Poly operator - ( Poly f , int x ) { f[ 0 ] = Sub( f[ 0 ] , x ); return f; }
Poly operator * ( Poly f , int x ) { for( int i = 0 ; i < len( f ) ; i ++ ) f[ i ] = Mul( f[ i ] , x ); return f; }
Poly operator + ( Poly f , Poly g ) {
int n = max( len( f ) , len( g ) ); f.resize( n ); g.resize( n );
for( int i = 0 ; i < n ; i ++ ) f[ i ] = Add( f[ i ] , g[ i ] );
return f;
}
Poly operator - ( Poly f , Poly g ) {
int n = max( len( f ) , len( g ) ); f.resize( n ); g.resize( n );
for( int i = 0 ; i < n ; i ++ ) f[ i ] = Sub( f[ i ] , g[ i ] );
return f;
}
const int G = 3 , IG = 332748118;
int lim , ilim , rev[ MAXN + 5 ];
void ntt( Poly &f , int op ) {
for( int i = 0 ; i < lim ; i ++ ) if( i < rev[ i ] ) swap( f[ i ] , f[ rev[ i ] ] );
for( int len = 2 , w ; len <= lim ; len <<= 1 ) {
w = Quick_pow( op == 1 ? G : IG , ( Mod - 1 ) / len );
for( int l = 0 ; l < lim ; l += len ) {
for( int i = l , wk = 1 ; i < l + len / 2 ; i ++ , wk = Mul( wk , w ) ) {
int t = Mul( wk , f[ i + len / 2 ] );
f[ i + len / 2 ] = Sub( f[ i ] , t ); f[ i ] = Add( f[ i ] , t );
}
}
}
if( op == -1 ) for( int i = 0 ; i < lim ; i ++ ) f[ i ] = Mul( f[ i ] , ilim );
}
int mxlen = 100000000;
Poly operator * ( Poly f , Poly g ) {
int n = len( f ) + len( g ) - 1; for( lim = 1 ; lim < n ; lim <<= 1 ); ilim = Inv( lim );
for( int i = 0 ; i < lim ; i ++ ) rev[ i ] = ( rev[ i >> 1 ] >> 1 ) | ( i & 1 ? lim >> 1 : 0 );
f.resize( lim ); g.resize( lim );
ntt( f , 1 ); ntt( g , 1 );
for( int i = 0 ; i < lim ; i ++ ) f[ i ] = Mul( f[ i ] , g[ i ] );
ntt( f , -1 ); f.resize( min( n , mxlen ) );
return f;
}
Poly Inv( Poly f ) {
Poly g = Poly( 1 , Inv( f[ 0 ] ) );
for( mxlen = 2 ; mxlen < 2 * len( f ) ; mxlen <<= 1 ) {
Poly A = f; A.resize( mxlen );
g = g * ( 2 - ( g * A ) );
}
g.resize( len( f ) ); return g;
}
Poly Der( Poly f ) {
for( int i = 0 ; i < len( f ) - 1 ; i ++ ) f[ i ] = Mul( i + 1 , f[ i + 1 ] );
f.resize( len( f ) - 1 );
return f;
}
Poly Int( Poly f ) {
f.resize( len( f ) + 1 );
for( int i = len( f ) - 1 ; i >= 1 ; i -- ) f[ i ] = Mul( f[ i - 1 ] , iv[ i ] );
f[ 0 ] = 0;
return f;
}
Poly Ln( Poly f ) {
Poly g = Int( Der( f ) * Inv( f ) );
g.resize( len( f ) ); return g;
}
Poly Exp( Poly f ) {
Poly g = Poly( 1 , 1 );
for( mxlen = 2 ; mxlen < 4 * len( f ) ; mxlen <<= 1 ) {
Poly A = f; A.resize( mxlen );
g = g * ( ( 1 - Ln( g ) ) + A );
}
g.resize( len( f ) ); return g;
}
Poly Pow( Poly f , int k ) {
f = Ln( f ); f = f * k; f = Exp( f );
return f;
}
Poly Sqrt( Poly f ) {
Poly g = Poly( 1 , 1 );
for( mxlen = 2 ; mxlen < 4 * len( f ) ; mxlen <<= 1 ) {
Poly A = f; A.resize( mxlen );
g = ( g + ( A * Inv( g ) ) ) * Inv( 2 );
}
g.resize( len( f ) ); return g;
}