多项式插值

Keep away from polynomial. ---- Wild_Donkey

给 $(x_0, y_0), (x_1, y_1),...,(x_n, y_n)$ , 共 $n + 1$ 个点. 求一个 $n$ 次 $n + 1$ 项的多项式 $L$ , 使得多项式的图像过每一个点. 这个多项式 $L$ 便是拉格朗日多项式 (Lagrange polynomial).

拉格朗日基本多项式 (插值基函数)

$\ell_i$ 是一个多项式, 满足 $\ell(x_i) = 1$ , 且 $\ell(x_j) = 0\ (j \neq i)$ .

ℓ_{i} (x) = \prod_{j = 0, j \neq i}^{n} \frac{x - x_{j}}{x_{i} - x_{j}}

$\ell_i(x) = \prod_{j = 0, j \neq i}^{n} \frac{x - x_j}{x_i - x_j}$

这样当 $x = x_i$ 时, 所有 $\dfrac{x - x_j}{x_i - x_j}$ 的分子等于分母. 当 $x \neq x_i$ 时, 一定存在一个项分子为 $0$ , 所以乘起来还是 $0$ .

拉格朗日多项式

因为每个拉格朗日基本多项式 $\ell_i$ 在除了 $x_i$ 点以外的给定的横坐标上都是 $0$ , 所以我们把所有拉格朗日多项式加权求和就能得到符合要求的多项式 $L$ .

L (x) = \sum_{i = 0}^{n} y_{i} ℓ_{i} (x) L (x) = \sum_{i = 0}^{n} y_{i} \prod_{j = 0, j \neq i}^{n} \frac{x - x_{j}}{x_{i} - x_{j}}

$L(x) = \sum_{i = 0}^{n} y_i\ell_i(x) L(x) = \sum_{i = 0}^{n} y_i\prod_{j = 0, j \neq i}^{n} \frac{x - x_j}{x_i - x_j}$

重心拉格朗日插值法

为了方便计算, 预处理一个 $n + 1$ 次多项式 $\ell$ :

ℓ (x) = \prod_{i = 0}^{n} (x - x_{i})

$\ell(x) = \prod_{i = 0}^n (x - x_i)$

这个多项式可以进行 $O(n)$ 次耗时 $O(n)$ 的 $O(n)$ 项式和二项式乘法得到, 总耗时 $O(n^2)$ .

定义重心权 $w_i$ :

w_{i} = \frac{1}{\prod_{j = 0, j \neq i}^{n} (x_{i} - x_{j})}

$w_i = \frac 1{\displaystyle{\prod_{j = 0, j \neq i}^n (x_i - x_j)}}$

这样就可以得到:

\begin{aligned} ℓ_{i} (x) & = \frac{ℓ (x) w_{i}}{x - x_{i}} \\ L (x) & = \sum_{i = 0}^{n} \frac{ℓ (x) w_{i} y_{i}}{x - x_{i}} \\ L (x) & = ℓ (x) \sum_{i = 0}^{n} \frac{w_{i} y_{i}}{x - x_{i}} \end{aligned}

$\begin{aligned} \ell_i(x) &= \frac{\ell(x)w_i}{x - x_i}\\ L(x) &= \sum_{i = 0}^{n} \frac{\ell(x)w_iy_i}{x - x_i}\\ L(x) &= \ell(x) \sum_{i = 0}^{n} \frac{w_iy_i}{x - x_i}\\ \end{aligned}$

我们可以 $O(n)$ 地算出 $w_iy_i$ , 然后 $O(n)$ 地算出 $\dfrac {\ell(x)}{x - x_i}$ , 在除法过程中就可以直接统计 $\ell_i$ 堆 $L$ 的贡献. 所以在预处理之后, 计算 $L(x)$ 需要 $O(n^2)$ 的时间.

代码实现

模板题

代码还是很好理解的, 只需要把前面的式子用程序实现就可以了.

const unsigned long long Mod(998244353);
unsigned long long Pnt[2005][2], La[2005], L[2005], Ans(0), m, Now(1);
unsigned n;
unsigned A, B, C, D, t;
unsigned Cnt(0);
inline unsigned long long Inv(unsigned long long x) {
  unsigned long long Rt(1);
  unsigned y(998244351);
  while (y) { if (y & 1) Rt = Rt * x % Mod; x = x * x % Mod, y >>= 1; }
  return Rt;
}
signed main() {
  n = RD(), m = RD();
  for (unsigned i(0); i < n; ++i) Pnt[i][0] = RD(), Pnt[i][1] = RD();
  La[1] = 1, La[0] = Mod - Pnt[0][0];
  for (unsigned i(1); i < n; ++i) {
    for (unsigned j(i + 1); j; --j)
      La[j] = (La[j] * (Mod - Pnt[i][0]) + La[j - 1]) % Mod;
    La[0] = La[0] * (Mod - Pnt[i][0]) % Mod;
  }
  for (unsigned i(0); i < n; ++i) {
    unsigned long long Mul(1), Tmp(La[n]);
    for (unsigned j(0); j < n; ++j) if (j ^ i) Mul = Mul * (Mod + Pnt[i][0] - Pnt[j][0]) % Mod;
    Mul = Inv(Mul) * Pnt[i][1] % Mod;
    for (unsigned j(n - 1); ~j; --j) {
      L[j] = (L[j] + Mul * Tmp) % Mod;
      Tmp = (La[j] + Tmp * Pnt[i][0]) % Mod;
    }
  }
  for (unsigned i(0); i < n; ++i) {
    Ans = (Ans + L[i] * Now) % Mod;
    Now = Now * m % Mod;
  }
  printf("%llu\n", Ans);
  return Wild_Donkey;
}

加点

如果我们需要对已经插值的 $n$ 个点加入一个点, 如果重新插值, 需要 $O(n^2)$ 的时间.

考虑插入一个点 $(x_{n + 1}, y_{n + 1})$ 对 $L(x)$ 的影响, 它会使插入后 $\ell'(x) = \ell(x)(x - x_{n + 1})$ . 会使每个满足 $i \leq n$ 的 $w_i$ 都乘以 $\frac 1{x_i - x_{n + 1}}$ . 还会使 $L(x)$ 加上 $\frac{\ell'(x)w_{n + 1}y_{n + 1}}{x - x_{n + 1}}$ .

L^{'} (x) = ℓ^{'} (x) (\sum_{i = 0}^{n + 1} \frac{w_{i}^{'} y_{i}}{x - x_{i}})

$L'(x) = \ell'(x) (\sum_{i = 0}^{n + 1} \frac{w_i'y_i}{x - x_i})$

我们用 $O(n)$ 处理出 $w_i'\ (i \leq n)$ , 然后 $O(n)$ 直接算出 $w_{n + 1}'$ . 我们可以在 $O(n)$ 的时间内做乘法得到 $\ell'(x)$ . 维护了 $w$ 数组和多项式 $\ell(x)$ 就实现了对 $L(x)$ 的维护. 需要时 $O(n^2)$ 可以求出多项式 $L(x)$ .

如果问题只需要维护多项式 $L(x)$ 的一个点值 $y$ , 那么问题就更简单了, 我们可以直接维护 $\ell(x)$ 的点值, 这样就可以把多项式除法变成整数乘除, 实现直接维护 $y$ 并实现 $O(n)$ 插入.

第二型重心拉格朗日插值法

现在有另一组点, 每个点和我们要插值的点一一对应, 每一对点的横坐标相同, 而新给出的这些点纵坐标是 $1$ . 我们对新点进行插值, 得到多项式 $g(x)$ , 无论 $x$ 取什么值, $g(x) = 1$ . 如果我们代入拉格朗日插值公式, 可以得到:

g (x) = ℓ (x) \sum_{i = 0}^{n} \frac{w_{i}}{x - x_{i}}

$g(x) = \ell(x) \sum_{i = 0}^{n} \frac{w_i}{x - x_i}$

因为任何数除以 $1$ 还是它本身, 所以 $L(x) = \dfrac{L(x)}{g(x)}$ . 将 $L(x)$ , $g(x)$ 都用拉格朗日插值公式表示, 则得到:

\begin{aligned} L (x) & = \frac{L (x)}{g (x)} \\ = \frac{ℓ (x) \sum_{i = 0}^{n} \frac{w_{i} y_{i}}{x - x_{i}}}{ℓ (x) \sum_{i = 0}^{n} \frac{w_{i}}{x - x_{i}}} \\ = \frac{\sum_{i = 0}^{n} \frac{w_{i} y_{i}}{x - x_{i}}}{\sum_{i = 0}^{n} \frac{w_{i}}{x - x_{i}}} \end{aligned}

$\begin{aligned} L(x) &= \frac{L(x)}{g(x)}\\ &= \frac {\ell(x) \displaystyle{\sum_{i = 0}^{n} \frac{w_iy_i}{x - x_i}}}{\ell(x) \displaystyle{\sum_{i = 0}^{n} \frac{w_i}{x - x_i}}}\\ &= \frac {\displaystyle{\sum_{i = 0}^{n} \frac{w_iy_i}{x - x_i}}}{\displaystyle{\sum_{i = 0}^{n} \frac{w_i}{x - x_i}}}\\ \end{aligned}$

我们可以只维护 $w$ 数组, 实现 $O(n)$ 求值, $O(n)$ 插入.

差商/均差 (Divided Diffrences)

对于点 $(x_i, y_i), (x_{i + 1}, y_{i + 1}),...,(x_{i + j}, y_{i + j})$ , 用 $[y_]$

对于函数 $f(x)$ , 用 $f[x_i, x_{i + 1},...,x_{i + j}]$ 表示函数的点 $(x_i, f(x_i)), (x_{i + 1}, f(x_{i + 1})),..., (x_{i + j}, f(x_{i + j}))$ 的均差.

均差的预算规则是这样的:

单点的均差就是这个点的纵坐标: $f[x_i] = f(x_i)$

多点均差有递归定义: $f[x_i, x_{i + 1},...,x_{i + j}] = \dfrac{f[x_{i + 1},...,x_{i + j}] - f[x_i,...,x_{i + j - 1}]}{x_{i + j} - x_{i}}$

均差有前向和后向之分, 一般只会用到前向均差, 后向均差是把元素顺序倒过来写, 计算上也有一部分需要翻转.

我们可以递推地在 $O(n^2)$ 的时间内求出 $n$ 个点的序列的每个子串的均差.

均差的展开形式是这样的:

f [x_{l}, x_{l + 1}, . . ., x_{r}] = \sum_{i = l}^{r} \frac{f (x_{i})}{\prod_{j = l, j \neq i}^{r} x_{i} - x_{j}}

$f[x_l, x_{l + 1},...,x_{r}] = \sum_{i = l}^{r} \dfrac{f(x_i)}{\displaystyle{\prod_{j = l, j \neq i}^{r} x_i - x_j}}$

用归纳法证明, 单点的均差符合 $f[x_l] = \dfrac{f(x_l)}{1} = f(x_l)$ . 假设大小小于 $r - l + 1$ 的点集都满足这个式子, 验证大小为 $r - l + 1$ 的点集是否满足.

\begin{aligned} f [x_{l}, x_{l + 1}, . . ., x_{r}] & = \frac{f [x_{i + 1}, . . ., x_{i + j}] - f [x_{i}, . . ., x_{i + j - 1}]}{x_{i + j} - x_{i}} \\ f [x_{l}, x_{l + 1}, . . ., x_{r}] & = \frac{\sum_{i = l + 1}^{r} \frac{f (x_{i})}{\prod_{j = l + 1, j \neq i}^{r} x_{i} - x_{j}} - \sum_{i = l}^{r - 1} \frac{f (x_{i})}{\prod_{j = l, j \neq i}^{r - 1} x_{i} - x_{j}}}{x_{r} - x_{l}} \\ f [x_{l}, x_{l + 1}, . . ., x_{r}] & = \frac{\sum_{i = l + 1}^{r - 1} \frac{f (x_{i})}{\prod_{j = l + 1, j \neq i}^{r} x_{i} - x_{j}} - \sum_{i = l + 1}^{r - 1} \frac{f (x_{i})}{\prod_{j = l, j \neq i}^{r - 1} x_{i} - x_{j}} + \frac{f (x_{r})}{\prod_{j = l + 1}^{r - 1} x_{r} - x_{j}} - \frac{f (x_{l})}{\prod_{j = l + 1}^{r - 1} x_{l} - x_{j}}}{x_{r} - x_{l}} \\ f [x_{l}, x_{l + 1}, . . ., x_{r}] & = \frac{\sum_{i = l + 1}^{r - 1} \frac{f (x_{i})}{\prod_{j = l + 1, j \neq i}^{r} x_{i} - x_{j}} - \sum_{i = l + 1}^{r - 1} \frac{f (x_{i})}{\prod_{j = l, j \neq i}^{r - 1} x_{i} - x_{j}}}{x_{r} - x_{l}} + \frac{f (x_{r})}{\prod_{j = l}^{r - 1} x_{r} - x_{j}} + \frac{f (x_{l})}{\prod_{j = l + 1}^{r} x_{l} - x_{j}} \\ f [x_{l}, x_{l + 1}, . . ., x_{r}] & = \frac{\sum_{i = l + 1}^{r - 1} \frac{f (x_{i})}{\prod_{j = l + 1, j \neq i}^{r} x_{i} - x_{j}} - \frac{f (x_{i})}{\prod_{j = l, j \neq i}^{r - 1} x_{i} - x_{j}}}{x_{r} - x_{l}} + \frac{f (x_{r})}{\prod_{j = l}^{r - 1} x_{r} - x_{j}} + \frac{f (x_{l})}{\prod_{j = l + 1}^{r} x_{l} - x_{j}} \\ f [x_{l}, x_{l + 1}, . . ., x_{r}] & = \frac{\sum_{i = l + 1}^{r - 1} \frac{\frac{f (x_{i})}{x_{i} - x_{r}} - \frac{f (x_{i})}{x_{i} - x_{l}}}{\prod_{j = l + 1, j \neq i}^{r - 1} x_{i} - x_{j}}}{x_{r} - x_{l}} + \frac{f (x_{r})}{\prod_{j = l}^{r - 1} x_{r} - x_{j}} + \frac{f (x_{l})}{\prod_{j = l + 1}^{r} x_{l} - x_{j}} \\ f [x_{l}, x_{l + 1}, . . ., x_{r}] & = \frac{\sum_{i = l + 1}^{r - 1} \frac{\frac{f (x_{i}) ((x_{i} - x_{l}) - (x_{i} - x_{r}))}{(x_{i} - x_{r}) (x_{i} - x_{l})}}{\prod_{j = l + 1, j \neq i}^{r - 1} x_{i} - x_{j}}}{x_{r} - x_{l}} + \frac{f (x_{r})}{\prod_{j = l}^{r - 1} x_{r} - x_{j}} + \frac{f (x_{l})}{\prod_{j = l + 1}^{r} x_{l} - x_{j}} \\ f [x_{l}, x_{l + 1}, . . ., x_{r}] & = \frac{\sum_{i = l + 1}^{r - 1} \frac{\frac{f (x_{i}) (x_{r} - x_{l})}{(x_{i} - x_{r}) (x_{i} - x_{l})}}{\prod_{j = l + 1, j \neq i}^{r - 1} x_{i} - x_{j}}}{x_{r} - x_{l}} + \frac{f (x_{r})}{\prod_{j = l}^{r - 1} x_{r} - x_{j}} + \frac{f (x_{l})}{\prod_{j = l + 1}^{r} x_{l} - x_{j}} \\ f [x_{l}, x_{l + 1}, . . ., x_{r}] & = \sum_{i = l + 1}^{r - 1} \frac{\frac{f (x_{i})}{(x_{i} - x_{r}) (x_{i} - x_{l})}}{\prod_{j = l + 1, j \neq i}^{r - 1} x_{i} - x_{j}} + \frac{f (x_{r})}{\prod_{j = l}^{r - 1} x_{r} - x_{j}} + \frac{f (x_{l})}{\prod_{j = l + 1}^{r} x_{l} - x_{j}} \\ f [x_{l}, x_{l + 1}, . . ., x_{r}] & = \sum_{i = l + 1}^{r - 1} \frac{f (x_{i})}{\prod_{j = l, j \neq i}^{r} x_{i} - x_{j}} + \frac{f (x_{r})}{\prod_{j = l}^{r - 1} x_{r} - x_{j}} + \frac{f (x_{l})}{\prod_{j = l + 1}^{r} x_{l} - x_{j}} \\ f [x_{l}, x_{l + 1}, . . ., x_{r}] & = \sum_{i = l}^{r} \frac{f (x_{i})}{\prod_{j = l, j \neq i}^{r} x_{i} - x_{j}} \end{aligned}

$\begin{aligned} f[x_l, x_{l + 1},...,x_{r}] &= \dfrac{f[x_{i + 1},...,x_{i + j}] - f[x_i,...,x_{i + j - 1}]}{x_{i + j} - x_{i}}\\ f[x_l, x_{l + 1},...,x_{r}] &= \frac{\displaystyle{\sum_{i = l + 1}^{r} \dfrac{f(x_i)}{\displaystyle{\prod_{j = l + 1, j \neq i}^{r} x_i - x_j}} - \sum_{i = l}^{r - 1} \dfrac{f(x_i)}{\displaystyle{\prod_{j = l, j \neq i}^{r - 1} x_i - x_j}}}}{x_r - x_l}\\ f[x_l, x_{l + 1},...,x_{r}] &= \frac{\displaystyle{\sum_{i = l + 1}^{r - 1} \dfrac{f(x_i)}{\displaystyle{\prod_{j = l + 1, j \neq i}^{r} x_i - x_j}} - \sum_{i = l + 1}^{r - 1} \dfrac{f(x_i)}{\displaystyle{\prod_{j = l, j \neq i}^{r - 1} x_i - x_j}} + \dfrac{f(x_r)}{\displaystyle{\prod_{j = l + 1}^{r - 1} x_r - x_j}} - \dfrac{f(x_l)}{\displaystyle{\prod_{j = l + 1}^{r - 1} x_l - x_j}}}}{x_r - x_l}\\ f[x_l, x_{l + 1},...,x_{r}] &= \frac{\displaystyle{\sum_{i = l + 1}^{r - 1} \dfrac{f(x_i)}{\displaystyle{\prod_{j = l + 1, j \neq i}^{r} x_i - x_j}} - \sum_{i = l + 1}^{r - 1} \dfrac{f(x_i)}{\displaystyle{\prod_{j = l, j \neq i}^{r - 1} x_i - x_j}}}}{x_r - x_l} + \dfrac{f(x_r)}{\displaystyle{\prod_{j = l}^{r - 1} x_r - x_j}} + \dfrac{f(x_l)}{\displaystyle{\prod_{j = l + 1}^{r} x_l - x_j}}\\ f[x_l, x_{l + 1},...,x_{r}] &= \frac{\displaystyle{\sum_{i = l + 1}^{r - 1} \dfrac{f(x_i)}{\displaystyle{\prod_{j = l + 1, j \neq i}^{r} x_i - x_j}} - \dfrac{f(x_i)}{\displaystyle{\prod_{j = l, j \neq i}^{r - 1} x_i - x_j}}}}{x_r - x_l} + \dfrac{f(x_r)}{\displaystyle{\prod_{j = l}^{r - 1} x_r - x_j}} + \dfrac{f(x_l)}{\displaystyle{\prod_{j = l + 1}^{r} x_l - x_j}}\\ f[x_l, x_{l + 1},...,x_{r}] &= \frac{\displaystyle{\sum_{i = l + 1}^{r - 1} \dfrac{\dfrac{f(x_i)}{x_i - x_r} - \dfrac{f(x_i)}{x_i - x_l}}{\displaystyle{\prod_{j = l + 1, j \neq i}^{r - 1} x_i - x_j}}}}{x_r - x_l} + \dfrac{f(x_r)}{\displaystyle{\prod_{j = l}^{r - 1} x_r - x_j}} + \dfrac{f(x_l)}{\displaystyle{\prod_{j = l + 1}^{r} x_l - x_j}}\\ f[x_l, x_{l + 1},...,x_{r}] &= \frac{\displaystyle{\sum_{i = l + 1}^{r - 1} \dfrac{\dfrac{f(x_i)((x_i - x_l) - (x_i - x_r))}{(x_i - x_r)(x_i - x_l)}}{\displaystyle{\prod_{j = l + 1, j \neq i}^{r - 1} x_i - x_j}}}}{x_r - x_l} + \dfrac{f(x_r)}{\displaystyle{\prod_{j = l}^{r - 1} x_r - x_j}} + \dfrac{f(x_l)}{\displaystyle{\prod_{j = l + 1}^{r} x_l - x_j}}\\ f[x_l, x_{l + 1},...,x_{r}] &= \frac{\displaystyle{\sum_{i = l + 1}^{r - 1} \dfrac{\dfrac{f(x_i)(x_r - x_l)}{(x_i - x_r)(x_i - x_l)}}{\displaystyle{\prod_{j = l + 1, j \neq i}^{r - 1} x_i - x_j}}}}{x_r - x_l} + \dfrac{f(x_r)}{\displaystyle{\prod_{j = l}^{r - 1} x_r - x_j}} + \dfrac{f(x_l)}{\displaystyle{\prod_{j = l + 1}^{r} x_l - x_j}}\\ f[x_l, x_{l + 1},...,x_{r}] &= \sum_{i = l + 1}^{r - 1} \dfrac{\dfrac{f(x_i)}{(x_i - x_r)(x_i - x_l)}}{\displaystyle{\prod_{j = l + 1, j \neq i}^{r - 1} x_i - x_j}} + \dfrac{f(x_r)}{\displaystyle{\prod_{j = l}^{r - 1} x_r - x_j}} + \dfrac{f(x_l)}{\displaystyle{\prod_{j = l + 1}^{r} x_l - x_j}}\\ f[x_l, x_{l + 1},...,x_{r}] &= \sum_{i = l + 1}^{r - 1} \dfrac{f(x_i)}{\displaystyle{\prod_{j = l, j \neq i}^{r} x_i - x_j}} + \dfrac{f(x_r)}{\displaystyle{\prod_{j = l}^{r - 1} x_r - x_j}} + \dfrac{f(x_l)}{\displaystyle{\prod_{j = l + 1}^{r} x_l - x_j}}\\ f[x_l, x_{l + 1},...,x_{r}] &= \sum_{i = l}^{r} \dfrac{f(x_i)}{\displaystyle{\prod_{j = l, j \neq i}^{r} x_i - x_j}}\\ \end{aligned}$

展开式得证.

牛顿基本多项式

设多项式 $n_i(x)$ 满足在 $x_0, x_1,...,x_{j - 1}$ 处都取 $0$ 的多项式. 显然有:

n_{i} (x) = \prod_{j = 0}^{i - 1} (x - x_{j})

$n_i(x) = \prod_{j = 0}^{i - 1} (x - x_j)$

牛顿多项式

我们只要给每个牛顿基本多项式加权求和, 就能构造一个 $n$ 次多项式, 过全部 $n + 1$ 个点. 这个总和便是牛顿多项式 $N(x)$ .

假设给 $n_i(x)$ 加的权为 $a_i$ , 写出定义式:

\begin{aligned} N (x) & = \sum_{i = 0}^{n} a_{i} n_{i} (x) \\ = \sum_{i = 0}^{n} a_{i} \prod_{j = 0}^{i - 1} (x - x_{j}) \end{aligned}

$\begin{aligned} N(x) &= \sum_{i = 0}^{n} a_in_i(x)\\ &= \sum_{i = 0}^{n}a_i\prod_{j = 0}^{i - 1} (x - x_j) \end{aligned}$

因为 $n + 1$ 个横坐标不同的点可以唯一地确定一个 $n$ 次多项式, 因此相同点集满足 $L(x) = N(x)$ . 我们可以用拉格朗日插值公式来求 $a$ . 已经求出已经插了 $(x_0, y_0),...,(x_{n - 1}, y_{n - 1})$ 的拉格朗日多项式 $L(x)$ . 我们尝试给 $a_n$ 赋值, 使得 $L(x_n) + a_nn_n(x_n)$ 为 $y_n$ .

\begin{aligned} y_{n} & = L (x_{n}) + a_{n} n_{n} (x_{n}) \\ y_{n} & = \sum_{i = 0}^{n - 1} y_{i} \prod_{j = 0, j \neq i}^{n - 1} \frac{x_{n} - x_{j}}{x_{i} - x_{j}} + a_{n} \prod_{i = 0}^{n - 1} (x_{n} - x_{i}) \\ a_{n} \prod_{i = 0}^{n - 1} (x_{n} - x_{i}) & = y_{n} - \sum_{i = 0}^{n - 1} y_{i} \prod_{j = 0, j \neq i}^{n - 1} \frac{x_{n} - x_{j}}{x_{i} - x_{j}} \\ a_{n} & = \frac{y_{n}}{\prod_{i = 0}^{n - 1} (x_{n} - x_{i})} - \sum_{i = 0}^{n - 1} \frac{y_{i}}{x_{n} - x_{i}} \prod_{j = 0, j \neq i}^{n - 1} \frac{1}{x_{i} - x_{j}} \\ a_{n} & = \frac{y_{n}}{\prod_{i = 0}^{n - 1} (x_{n} - x_{i})} + \sum_{i = 0}^{n - 1} \frac{y_{i}}{x_{i} - x_{n}} \prod_{j = 0, j \neq i}^{n - 1} \frac{1}{x_{i} - x_{j}} \\ a_{n} & = \frac{y_{n}}{\prod_{i = 0}^{n - 1} (x_{n} - x_{i})} + \sum_{i = 0}^{n - 1} y_{i} \prod_{j = 0, j \neq i}^{n} \frac{1}{x_{i} - x_{j}} \\ a_{n} & = \frac{y_{n}}{\prod_{i = 0}^{n - 1} (x_{n} - x_{i})} + \sum_{i = 0}^{n - 1} \frac{y_{i}}{\prod_{j = 0, j \neq i}^{n} x_{i} - x_{j}} \\ a_{n} & = \sum_{i = 0}^{n} \frac{y_{i}}{\prod_{j = 0, j \neq i}^{n} x_{i} - x_{j}} \end{aligned}

$\begin{aligned} y_n &= L(x_n) + a_nn_n(x_n)\\ y_n &= \sum_{i = 0}^{n - 1} y_i\prod_{j = 0, j \neq i}^{n - 1} \frac{x_n - x_j}{x_i - x_j} + a_n\prod_{i = 0}^{n - 1} (x_n - x_i)\\ a_n\prod_{i = 0}^{n - 1} (x_n - x_i) &= y_n - \sum_{i = 0}^{n - 1} y_i\prod_{j = 0, j \neq i}^{n - 1} \frac{x_n - x_j}{x_i - x_j}\\ a_n &= \dfrac{y_n}{\displaystyle{\prod_{i = 0}^{n - 1} (x_n - x_i)}} - \sum_{i = 0}^{n - 1} \frac{y_i}{x_n - x_i} \prod_{j = 0, j \neq i}^{n - 1} \frac 1{x_i - x_j}\\ a_n &= \dfrac{y_n}{\displaystyle{\prod_{i = 0}^{n - 1} (x_n - x_i)}} + \sum_{i = 0}^{n - 1} \frac{y_i}{x_i - x_n} \prod_{j = 0, j \neq i}^{n - 1} \frac 1{x_i - x_j}\\ a_n &= \dfrac{y_n}{\displaystyle{\prod_{i = 0}^{n - 1} (x_n - x_i)}} + \sum_{i = 0}^{n - 1} y_i \prod_{j = 0, j \neq i}^{n} \frac 1{x_i - x_j}\\ a_n &= \dfrac{y_n}{\displaystyle{\prod_{i = 0}^{n - 1} (x_n - x_i)}} + \sum_{i = 0}^{n - 1} \frac {y_i}{\displaystyle {\prod_{j = 0, j \neq i}^{n} x_i - x_j}}\\ a_n &= \sum_{i = 0}^{n} \frac {y_i}{\displaystyle {\prod_{j = 0, j \neq i}^{n} x_i - x_j}}\\ \end{aligned}$

我们发现 $a_n = \displaystyle{\sum_{i = 0}^{n} \frac {y_i}{\displaystyle {\prod_{j = 0, j \neq i}^{n} x_i - x_j}}} = [y_0, y_1,...,y_n]$ .

所以牛顿多项式便是:

\begin{aligned} N (x) & = \sum_{i = 0}^{n} a_{i} n_{i} (x) \\ = \sum_{i = 0}^{n} a_{i} \prod_{j = 0}^{i - 1} (x - x_{j}) \\ = \sum_{i = 0}^{n} [y_{0}, y_{1}, . . ., y_{i}] \prod_{j = 0}^{i - 1} (x - x_{j}) \\ = \sum_{i = 0}^{n} (\sum_{j = 0}^{i} \frac{y_{j}}{\prod_{k = 0, k \neq j}^{i} x_{j} - x_{k}}) \prod_{j = 0}^{i - 1} (x - x_{j}) \end{aligned}

$\begin{aligned} N(x) &= \sum_{i = 0}^{n} a_in_i(x)\\ &= \sum_{i = 0}^{n}a_i\prod_{j = 0}^{i - 1} (x - x_j)\\ &= \sum_{i = 0}^{n}[y_0, y_1,...,y_i]\prod_{j = 0}^{i - 1} (x - x_j)\\ &= \sum_{i = 0}^{n} \Big(\displaystyle{\sum_{j = 0}^{i} \frac {y_j}{\displaystyle {\prod_{k = 0, k \neq j}^{i} x_j - x_k}}\Big)}\prod_{j = 0}^{i - 1} (x - x_j)\\ \end{aligned}$

例题

有一个 $n$ 次以内的多项式 $f(x)$ , 给 $n + 1$ 个点值, $f(0), f(1),..., f(n)$ . 求 $f(m), f(m + 1),..., f(m + n)$ .

这个题 $O(n^2)$ 无法通过, 而且由于询问也是 $n + 1$ 个点值, 所以插出系数来也无法用正确的复杂度询问. 发现横坐标很特殊, $x_i = i$ , 询问也很特殊, 横坐标也是连续的. 我们需要求出所以先把横坐标往插值公式里面带. 首先化简 $\displaystyle{\prod_{j = 0, j \neq i}^{n} (i - j)}$ .

\begin{aligned} \prod_{j = 0, j \neq i}^{n} (i - j) & = (\prod_{j = 0}^{i - 1} (i - j)) (\prod_{j = i + 1}^{n} (i - j)) \\ = (\prod_{j = 0}^{i - 1} (i - j)) (\prod_{j = i + 1}^{n} (j - i)) (- 1)^{n - i} \\ = i! (n - i)! (- 1)^{n - i} \end{aligned}

$\begin{aligned} \prod_{j = 0, j \neq i}^{n} (i - j) &= \Big(\prod_{j = 0}^{i - 1} (i - j)\Big)\Big(\prod_{j = i + 1}^n (i - j)\Big)\\ &= \Big(\prod_{j = 0}^{i - 1} (i - j)\Big)\Big(\prod_{j = i + 1}^n (j - i)\Big) (-1)^{n - i}\\ &= i!(n - i)!(-1)^{n - i} \end{aligned}$

如果默认 $x$ 为自然数且 $x > n$ , 还可以化简 $\displaystyle{\prod_{i = 0}^{n}} (x - i) = \dfrac {x!}{(x - n - 1)!}$ .

\begin{aligned} L (x) & = \sum_{i = 0}^{n} \frac{x! f (i)}{(x - n - 1)! i! (n - i)! (- 1)^{n - i} (x - i)} \\ = \frac{x!}{(x - n - 1)!} \sum_{i = 0}^{n} \frac{f (i)}{i! (n - i)! (- 1)^{n - i} (x - i)} \\ = \frac{x!}{(x - n - 1)!} \sum_{i = 0}^{n} \frac{f (i)}{i! (n - i)! (- 1)^{n - i}} \frac{1}{x - i} \end{aligned}

$\begin{aligned} L(x) &= \sum_{i = 0}^{n} \frac{x!f(i)}{(x - n - 1)!i!(n - i)!(-1)^{n - i}(x - i)}\\ &= \frac {x!}{(x - n - 1)!} \sum_{i = 0}^{n} \frac{f(i)}{i!(n - i)!(-1)^{n - i}(x - i)}\\ &= \frac {x!}{(x - n - 1)!} \sum_{i = 0}^{n}\frac {f(i)}{i!(n - i)!(-1)^{n - i}}\frac 1{x - i}\\ \end{aligned}$

我们可以 $O(n)$ 处理 $\dfrac{f(i)}{i!(n - i)!(-1)^{n - i}}$ , 和对于所有 $x$ 的 $\dfrac{x!}{(x - n - 1)!}$ . 注意 $x$ 较大, 无法直接计算 $x!$ , 所以我们可以先把 $x - n$ 到 $x$ 的数字相乘, 得到 $\dfrac{x!}{(x - n - 1)!}$ 然后枚举 $x$ 来递推.

但是我们要想计算一个点值, 仍然需要 $O(n)$ 枚举 $i$ .

定义序列 $A_i = \dfrac{f(i)}{i!(n - i)!(-1)^{n - i}}$ , $B_i = \dfrac 1{m + i - n}$ , $C_i = \displaystyle \sum_{j = 0}^{i} A_jB_{i - j}$ , 三个序列长度为 $2(n + 1)$ , $A_i = 0\ (i > n)$ .

L (x) = \frac{x!}{(x - n - 1)!} C_{x - m + n}

$L(x) = \frac{x!}{(x - n - 1)!}C_{x - m + n}$

我们用 $O(n\log n)$ 卷积算出 $C$ , 然后就可以 $O(1)$ 查询点值.

const unsigned long long Mod(998244353);
unsigned long long C1[160005], IFac[160005], MFac[160005], A[530000], B[530000];
unsigned long long Tmp(1), w;
unsigned m, n, N(1);
unsigned D, t;
unsigned Cnt(0), Ans(0);
inline void Mn(unsigned long long& x) { x -= ((x >= Mod) ? Mod : 0); }
inline unsigned long long W(unsigned x) {
  unsigned long long Now(3), Rt(1);
  while (x) { if (x & 1) Rt = Rt * Now % Mod;  Now = Now * Now % Mod, x >>= 1; }
  return Rt;
}
inline unsigned long long Inv(unsigned long long x) {
  unsigned long long Rt(1);
  unsigned y(998244351);
  while (y) { if (y & 1) Rt = Rt * x % Mod;y >>= 1, x = x * x % Mod; }
  return Rt;
}
inline void DIT(unsigned long long* f) {
  unsigned long long Now(1), Tmpw[20];
  unsigned Lg(0);
  Tmpw[0] = w;
  for (unsigned i(1); i < N; i <<= 1, ++Lg) Tmpw[Lg + 1] = Tmpw[Lg] * Tmpw[Lg] % Mod; --Lg;
  for (unsigned i(1); i < N; i <<= 1, --Lg) {
    for (unsigned j(0); j < N; ++j, Now = Now * Tmpw[Lg] % Mod) if (!(i & j)) {
      unsigned long long TmpA(f[j]), TmpB(f[j ^ i] * Now % Mod);
      f[j] = TmpA + TmpB, Mn(f[j]);
      f[j ^ i] = Mod + TmpA - TmpB, Mn(f[j ^ i]);
    }
  }
}
inline void DIF(unsigned long long* f) {
  unsigned long long Now(1);
  for (unsigned i(N >> 1); i; i >>= 1, w = w * w % Mod) {
    for (unsigned j(0); j < N; ++j, Now = Now * w % Mod) if (!(i & j)) {
      unsigned long long TmpA(f[j]), TmpB(f[j ^ i]);
      f[j] = TmpA + TmpB, Mn(f[j]);
      f[j ^ i] = (Mod + TmpA - TmpB) * Now % Mod;
    }
  }
}
signed main() {
  n = RD(), m = RD();
  while (N <= n) { N <<= 1; } N <<= 1;
  for (unsigned i(0); i <= n; ++i) A[i] = RD();
  for (unsigned i(1); i <= n; ++i) Tmp = Tmp * i % Mod;
  IFac[n] = Inv(Tmp);
  for (unsigned i(n - 1); ~i; --i) IFac[i] = IFac[i + 1] * (i + 1) % Mod;
  for (unsigned i(0); i <= n; ++i)
    A[i] = ((((n - i) & 1) ? (Mod - A[i]) : A[i]) * IFac[n - i] % Mod) * IFac[i] % Mod;
  C1[0] = 1;
  for (unsigned i(m - n); i <= m; ++i) C1[0] = C1[0] * i % Mod;
  for (unsigned i(1); i <= n; ++i) C1[i] = (C1[i - 1] * Inv(m - n + i - 1) % Mod) * (m + i) % Mod;
  for (unsigned i(n << 1); ~i; --i) B[i] = Inv(m + i - n);
  w = W(998244352 / N), DIF(A), w = W(998244352 / N), DIF(B);
  for (unsigned i(0); i < N; ++i) A[i] = A[i] * B[i] % Mod;
  w = Inv(W(998244352 / N)), DIT(A), w = Inv(N), N = (n << 1);
  for (unsigned i(n); i <= N; ++i) printf("%llu ", (C1[i - n] * A[i] % Mod) * w % Mod); putchar(0x0A);
  return Wild_Donkey;
}

posted @ 2022-02-20 21:34 Wild_Donkey 阅读(99) 评论(0) 编辑收藏举报

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