FFT
坑
NTT
将\(FFT\)中的单位复数根改成原根即可。
卡常版NTT模版
struct Mul{ int Len; int wn[N], Lim; int rev[N]; inline void getReverse(int * a) { static int rev[N]; rev[0] = 0; for (register int i = 0; i < Len; i++) { rev[i] = (rev[i>>1] >> 1) | ((i&1) ? (Len >> 1) : 0); if (i < rev[i]) swap(a[i], a[rev[i]]); } } void NTT(int * a, int opt) { getReverse(a); for (register int i = 0; i < Len; i += 2) { int x = a[i], y = 1LL * wn[0] * a[i + 1] % mod; a[i] = Plus(x, y), a[i + 1] = Minus(x, y); } for (register int i = 0; i < Len; i += 4) { int x1 = a[i], y1 = 1LL * wn[0] * a[i + 2] % mod; int x2 = a[i + 1], y2 = 1LL * wn[Lim / 2] * a[i + 3] % mod; a[i] = Plus(x1, y1), a[i + 2] = Minus(x1, y1); a[i + 1] = Plus(x2, y2), a[i + 3] = Minus(x2, y2); } for (register int i = 8; i <= Len; i <<= 1) { int M = i >> 1; for (register int j = 0; j < Len; j += i) { for (register int k = 0; k < M; k += 4) { int x1 = a[j + k], y1 = 1LL * wn[Lim / M * k] * a[j + k + M] % mod; int x2 = a[j + k + 1], y2 = 1LL * wn[Lim / M * (k+1)] * a[j + k + M + 1] % mod; int x3 = a[j + k + 2], y3 = 1LL * wn[Lim / M * (k+2)] * a[j + k + M + 2] % mod; int x4 = a[j + k + 3], y4 = 1LL * wn[Lim / M * (k+3)] * a[j + k + M + 3] % mod; a[j + k] = Plus(x1, y1), a[j + k + M] = Minus(x1, y1); a[j + k + 1] = Plus(x2, y2), a[j + k + M + 1] = Minus(x2, y2); a[j + k + 2] = Plus(x3, y3), a[j + k + M + 2] = Minus(x3, y3); a[j + k + 3] = Plus(x4, y4), a[j + k + M + 3] = Minus(x4, y4); } } } if (opt == -1) { for (register int i = 0; i < Len; i++) a[i] = 1LL * a[i] * inv[Len] % mod; for(register int i = 1; i < Len; i++) if(i < Len-i) swap(a[i], a[Len-i]); } } inline void init() { Lim = Len; wn[0] = 1; wn[1] = power(7, (mod-1) / (Len << 1), mod); for (int i = 2; i < (Len << 1); i++) wn[i] = 1LL * wn[i-1] * wn[1] % mod; } inline int getLen(int l) { for (Len = 1; Len <= l; Len <<= 1); return Len; }}; 多项式求逆
也就是求
\[ A(x)B(x) = 1 \pmod{x^n}\]假设我们已知
\[ A(x)G(x) = 1 \pmod{x^{\lceil{n \over 2}\rceil}} \]两式相减得
\[ A(x)(B(x) - G(x)) = 0 \pmod{x^{\lceil{n \over 2}\rceil}}\]\[ B(x) - G(x) = 0 \pmod{x^{\lceil{n \over 2}\rceil}}\]
\[ B^2(x) + G^2(x) - 2B(x)G(x) = 0 \pmod{x^n}\]
两边同时乘上\(A(x)\)
\[ A(x)B^2(x) + A(x)G^2(x) = 2A(x)B(x)G(x) \]
因为\(A(x)B(x) = 1\), 所以可以得到
\[ B(x) + A(x)G^2(x) = 2G(x) \]
\[ B(x) = 2G(x) - A(x)G^2(x) \]
就这样解决了
时间复杂度为\(T(n) = T(\frac{n}{2}) +O(n \log n)\)
代码
inline void getInv(int * A, int * B, int len){ static int tmp1[N], tmp2[N]; B[0] = power(A[0], mod-2, mod); for (register int k = 2; k <= len; k <<= 1) { int Len = k << 1; cpy(tmp1, A, k, Len); cpy(tmp2, B, (k >> 1), Len); Calc.Len = Len; Calc.NTT(tmp1, 1); Calc.NTT(tmp2, 1); for (int i = 0; i < Len; i++) tmp1[i] = Minus(Plus(tmp2[i], tmp2[i]), 1LL * tmp1[i] * tmp2[i] % mod * tmp2[i] % mod); Calc.NTT(tmp1, -1); cpy(B, tmp1, k, Len); } return;} 多项式开根
也就是求
\[ B^2(x) = A(x) \pmod{x^n} \]
假设我们已知
\[ G^2(x) = A(x) \pmod{x^{\lceil{n \over 2}\rceil}} \]两式相减得
\[ B^2(x) - G^2(x) = 0 \pmod{x^{\lceil{n \over 2}\rceil}} \]然后平方一下
\[ B^4(x) + G^4(x) - 2B^2(x)G^2(x) = 0 \pmod{x^n} \]\[ B^4(x) + G^4(x) = 2B^2(x)G^2(x) \pmod{x^n} \]
配一下方
\[ B^4(x) + G^4(x) + 2B^2(x)G^2(x) = 4B^2(x)G^2(x) \pmod{x^n} \]\[ (B^2(x) + G^2(x))^2 = (2B(x)G(x))^2 \pmod{x^n} \]
\[ B^2(x) + G^2(x) = 2B(x)G(x) \pmod{x^n} \]
因为\(A(x)B(x) = 1\), 所以可以得到
\[ B(x) = {A(x) + G^2(x) \over {2G(x)}} \]
为了方便实现，我们常把它化成
\[ B(x) = {A(x) \over 2G(x)} + {G(x) \over 2} \]
然后就解决了
时间复杂度为\(T(n) = T(\frac{n}{2}) + O(n \log n) = O(n \log n)\)
多项式求导
模拟即可
代码
inline void getDeri(int * a, int len){ for (register int i = 0; i < len; i++) a[i] = 1LL * a[i+1] * (i+1) % mod;} 多项式积分
模拟即可
代码
inline void getInte(int * a, int len){ for (int i = len-1; i >= 1; i--) a[i] = 1LL * a[i-1] * inv[i] % mod; a[0] = 0;} 多项式求\(\ln\)
已知多项式\(A(x)\), 求\(B(x) = \ln (A(x))\)
根据链式法则， 我们可以得到
\[B'(x) = \frac{\mathbb{d}(\ln(A(x)))}{\mathbb{d}A(x)} \frac{\mathbb{d}A(x)}{\mathbb{d}x} = \frac{A'(x)}{A(x)}\]
所以
\[B(x) = \int \frac{A'(x)}{A(x)} \mathbb{d}x\]
代码
void getLn(int * A, int len){ static int tmp1[N], tmp2[N], tmp3[N]; int Len = len << 1; cpy(tmp1, A, len, Len); cpy(tmp2, A, len, Len); getDeri(tmp1, len); getInv(tmp2, tmp3, len); Calc.Len = Len; Calc.NTT(tmp1, 1); Calc.NTT(tmp3, 1); for (int i = 0; i < Len; i++) tmp1[i] = 1LL * tmp1[i] * tmp3[i] % mod; Calc.NTT(tmp1, -1); memset(tmp1 + len, 0, 4 * (Len - len)); getInte(tmp1, len); cpy(A, tmp1, len, Len);} 时间复杂度还是\(T(n) = T(\frac{n}{2}) + O(n \log n) = O(n \log n)\)
牛顿迭代
我们要求\(f(x) = 0\)的根
那么可以使用泰勒公式来近似\(f(x) = 0\)的根
我们设当前的近似值是\(x_{n}\)， 我们想要得到的近似值是\(x_{n+1}\)
截取泰勒公式的线性部分： \[f(x_{n}) + f'(x_{n})(x_{n+1}-x_n) = 0\]
解方程得： \[x_{n+1} = x_n - \frac{f(x_{n})}{f'(x_{n})}\]我们可以用这种方法来解多项式方程。
多项式\(\exp\)
我们要求的是 \(exp(A(x))\), 这相当于解一个方程： \(g(exp(A(x))) = \ln(exp(A(x))) - A(x) = 0\)
可以直接套用牛顿迭代法求解。