mirror of https://github.com/OI-wiki/OI-wiki
Tifa
GitHub
parent
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commit
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@ -98,7 +98,8 @@
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/math/poly/ln-exp /math/poly/elementary-func
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/math/poly/tri-func /math/poly/elementary-func
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/math/poly/inv-tri-func /math/poly/elementary-func
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/math/poly/lagrange /math/numerical/lagrange
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/math/poly/lagrange /math/numerical/interp
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/math/numerical/lagrange /math/numerical/interp
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/math/poly/cntt /math/poly/ntt
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/math/linear-algebra/gauss /math/numerical/gauss
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/math/integral /math/numerical/integral
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#include <bits/stdc++.h>
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using namespace std;
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constexpr uint32_t MOD = 998244353;
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struct mint {
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uint32_t v_;
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mint() : v_(0) {}
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mint(int64_t v) {
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int64_t x = v % (int64_t)MOD;
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v_ = (uint32_t)(x + (x < 0 ? MOD : 0));
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}
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friend mint inv(mint const &x) {
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int64_t a = x.v_, b = MOD;
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if ((a %= b) == 0) return 0;
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int64_t s = b, m0 = 0;
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for (int64_t q = 0, _ = 0, m1 = 1; a;) {
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_ = s - a * (q = s / a);
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s = a;
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a = _;
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_ = m0 - m1 * q;
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m0 = m1;
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m1 = _;
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}
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return m0;
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}
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mint &operator+=(mint const &r) {
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if ((v_ += r.v_) >= MOD) v_ -= MOD;
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return *this;
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}
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mint &operator-=(mint const &r) {
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if ((v_ -= r.v_) >= MOD) v_ += MOD;
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return *this;
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}
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mint &operator*=(mint const &r) {
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v_ = (uint32_t)((uint64_t)v_ * r.v_ % MOD);
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return *this;
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}
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mint &operator/=(mint const &r) { return *this = *this * inv(r); }
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friend mint operator+(mint l, mint const &r) { return l += r; }
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friend mint operator-(mint l, mint const &r) { return l -= r; }
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friend mint operator*(mint l, mint const &r) { return l *= r; }
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friend mint operator/(mint l, mint const &r) { return l /= r; }
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};
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template <class T>
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struct NewtonInterp {
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// {(x_0,y_0),...,(x_{n-1},y_{n-1})}
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vector<pair<T, T>> p;
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// dy[r][l] = [y_l,...,y_r]
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vector<vector<T>> dy;
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// (x-x_0)...(x-x_{n-1})
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vector<T> base;
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// [y_0]+...+[y_0,y_1,...,y_n](x-x_0)...(x-x_{n-1})
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vector<T> poly;
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void insert(T const &x, T const &y) {
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p.emplace_back(x, y);
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size_t n = p.size();
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if (n == 1) {
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base.push_back(1);
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} else {
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size_t m = base.size();
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base.push_back(0);
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for (size_t i = m; i; --i) base[i] = base[i - 1];
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base[0] = 0;
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for (size_t i = 0; i < m; ++i)
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base[i] = base[i] - p[n - 2].first * base[i + 1];
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}
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dy.emplace_back(p.size());
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dy[n - 1][n - 1] = y;
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if (n > 1) {
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for (size_t i = n - 2; ~i; --i)
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dy[n - 1][i] =
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(dy[n - 2][i] - dy[n - 1][i + 1]) / (p[i].first - p[n - 1].first);
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}
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poly.push_back(0);
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for (size_t i = 0; i < n; ++i) poly[i] = poly[i] + dy[n - 1][0] * base[i];
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}
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T eval(T const &x) {
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T ans{};
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for (auto it = poly.rbegin(); it != poly.rend(); ++it) ans = ans * x + *it;
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return ans;
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}
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};
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int main() {
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NewtonInterp<mint> ip;
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int n, k;
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cin >> n >> k;
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for (int i = 1, x, y; i <= n; ++i) {
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cin >> x >> y;
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ip.insert(x, y);
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}
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cout << ip.eval(k).v_;
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return 0;
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}
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@ -0,0 +1,10 @@
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id: oi-wiki:math/interp-1
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name: Interpolation
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solution:
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cpp: docs/math/code/numerical/interp/interp_1.cpp
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target:
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- 'test'
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testcases:
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answer: docs/math/examples/numerical/interp/interp_1.ans
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input: docs/math/examples/numerical/interp/interp_1.in
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version: 1000
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@ -0,0 +1,10 @@
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id: oi-wiki:math/interp-2
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name: Interpolation
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solution:
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cpp: docs/math/code/numerical/interp/interp_2.cpp
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target:
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- 'test'
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testcases:
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answer: docs/math/examples/numerical/interp/interp_2.ans
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input: docs/math/examples/numerical/interp/interp_2.in
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version: 1000
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@ -0,0 +1,10 @@
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id: oi-wiki:math/interp-3
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name: Interpolation
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solution:
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cpp: docs/math/code/numerical/interp/interp_3.cpp
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target:
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- 'test'
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testcases:
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answer: docs/math/examples/numerical/interp/interp_3.ans
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input: docs/math/examples/numerical/interp/interp_3.in
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version: 1000
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@ -1,10 +0,0 @@
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id: oi-wiki:math/lagrange-1
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name: Lagrange
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solution:
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cpp: docs/math/code/numerical/lagrange/lagrange_1.cpp
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target:
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- 'test'
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testcases:
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answer: docs/math/examples/numerical/lagrange/lagrange_1.ans
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input: docs/math/examples/numerical/lagrange/lagrange_1.in
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version: 1000
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@ -1,10 +0,0 @@
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id: oi-wiki:math/lagrange-2
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name: Lagrange
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solution:
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cpp: docs/math/code/numerical/lagrange/lagrange_2.cpp
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target:
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- 'test'
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testcases:
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answer: docs/math/examples/numerical/lagrange/lagrange_2.ans
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input: docs/math/examples/numerical/lagrange/lagrange_2.in
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version: 1000
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@ -0,0 +1 @@
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10201
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@ -0,0 +1,4 @@
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3 100
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1 4
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2 9
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3 16
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</g>
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<style xmlns="http://www.w3.org/1999/xhtml" id="ls082ltu.w37">rt.katakana-terminator-rt::before { content: attr(data-rt); }</style></svg>
|
||||
|
After Width: | Height: | Size: 13 KiB |
|
|
@ -0,0 +1,220 @@
|
|||
author: AtomAlpaca, billchenchina, Chrogeek, Early0v0, EndlessCheng, Enter-tainer, Henry-ZHR, hly1204, hsfzLZH1, Ir1d, Ghastlcon, kenlig, Marcythm, megakite, Peanut-Tang, qwqAutomaton, qz-cqy, StudyingFather, swift-zym, swiftqwq, Tiphereth-A, TrisolarisHD, x4Cx58x54, Xeonacid, xiaopangfeiyu, YanWQ-monad
|
||||
|
||||
## 引入
|
||||
|
||||
插值是一种通过已知的、离散的数据点推算一定范围内的新数据点的方法。插值法常用于函数拟合中。
|
||||
|
||||
例如对数据点：
|
||||
|
||||
| $x$ | $0$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ |
|
||||
| ------ | --- | -------- | -------- | -------- | --------- | --------- | --------- |
|
||||
| $f(x)$ | $0$ | $0.8415$ | $0.9093$ | $0.1411$ | $-0.7568$ | $-0.9589$ | $-0.2794$ |
|
||||
|
||||
|
||||
|
||||
其中 $f(x)$ 未知，插值法可以通过按一定形式拟合 $f(x)$ 的方式估算未知的数据点。
|
||||
|
||||
例如，我们可以用分段线性函数拟合 $f(x)$：
|
||||
|
||||
|
||||
|
||||
这种插值方式叫做 [线性插值](https://en.wikipedia.org/wiki/Linear_interpolation)。
|
||||
|
||||
我们也可以用多项式拟合 $f(x)$：
|
||||
|
||||
|
||||
|
||||
这种插值方式叫做 [多项式插值](https://en.wikipedia.org/wiki/Polynomial_interpolation)。
|
||||
|
||||
多项式插值的一般形式如下：
|
||||
|
||||
???+ note "多项式插值"
|
||||
对已知的 $n+1$ 的点 $(x_0,y_0),(x_1,y_1),\dots,(x_n,y_n)$，求 $n$ 阶多项式 $f(x)$ 满足
|
||||
|
||||
$$
|
||||
f(x_i)=y_i,\qquad\forall i=0,1,\dots,n
|
||||
$$
|
||||
|
||||
下面介绍多项式插值中的两种方式：Lagrange 插值法与 Newton 插值法。不难证明这两种方法得到的结果是相等的。
|
||||
|
||||
## Lagrange 插值法
|
||||
|
||||
由于要求构造一个函数 $f(x)$ 过点 $P_1(x_1, y_1), P_2(x_2,y_2),\cdots,P_n(x_n,y_n)$. 首先设第 $i$ 个点在 $x$ 轴上的投影为 $P_i^{\prime}(x_i,0)$.
|
||||
|
||||
考虑构造 $n$ 个函数 $f_1(x), f_2(x), \cdots, f_n(x)$，使得对于第 $i$ 个函数 $f_i(x)$，其图像过 $\begin{cases}P_j^{\prime}(x_j,0),(j\neq i)\\P_i(x_i,y_i)\end{cases}$，则可知题目所求的函数 $f(x)=\sum\limits_{i=1}^nf_i(x)$.
|
||||
|
||||
那么可以设 $f_i(x)=a\cdot\prod_{j\neq i}(x-x_j)$，将点 $P_i(x_i,y_i)$ 代入可以知道 $a=\dfrac{y_i}{\prod_{j\neq i} (x_i-x_j)}$，所以
|
||||
|
||||
$$
|
||||
f_i(x)=y_i\cdot\dfrac{\prod_{j\neq i} (x-x_j)}{\prod_{j\neq i} (x_i-x_j)}=y_i\cdot\prod_{j\neq i}\dfrac{x-x_j}{x_i-x_j}
|
||||
$$
|
||||
|
||||
那么我们就可以得出 Lagrange 插值的形式为：
|
||||
|
||||
$$
|
||||
f(x)=\sum_{i=1}^ny_i\cdot\prod_{j\neq i}\dfrac{x-x_j}{x_i-x_j}
|
||||
$$
|
||||
|
||||
朴素实现的时间复杂度为 $O(n^2)$，可以优化到 $O(n\log^2 n)$，参见 [多项式快速插值](../poly/multipoint-eval-interpolation.md#多项式的快速插值)。
|
||||
|
||||
???+ note "[Luogu P4781【模板】拉格朗日插值](https://www.luogu.com.cn/problem/P4781)"
|
||||
给出 $n$ 个点对 $(x_i,y_i)$ 和 $k$，且 $\forall i,j$ 有 $i\neq j \iff x_i\neq x_j$ 且 $f(x_i)\equiv y_i\pmod{998244353}$ 和 $\deg(f(x))<n$（定义 $\deg(0)=-\infty$），求 $f(k)\bmod{998244353}$.
|
||||
|
||||
??? note "题解"
|
||||
本题中只用求出 $f(k)$ 的值，所以在计算上式的过程中直接将 $k$ 代入即可。
|
||||
|
||||
$$
|
||||
f(k)=\sum_{i=1}^{n}y_i\prod_{j\neq i }\frac{k-x_j}{x_i-x_j}
|
||||
$$
|
||||
|
||||
本题中，还需要求解逆元。如果先分别计算出分子和分母，再将分子乘进分母的逆元，累加进最后的答案，时间复杂度的瓶颈就不会在求逆元上，时间复杂度为 $O(n^2)$.
|
||||
|
||||
因为在固定模 $998244353$ 意义下运算，计算乘法逆元的时间复杂度我们在这里暂且认为是常数时间。
|
||||
|
||||
??? note "代码实现"
|
||||
```cpp
|
||||
--8<-- "docs/math/code/numerical/interp/interp_1.cpp"
|
||||
```
|
||||
|
||||
### 横坐标是连续整数的 Lagrange 插值
|
||||
|
||||
如果已知点的横坐标是连续整数，我们可以做到 $O(n)$ 插值。
|
||||
|
||||
设要求 $n$ 次多项式为 $f(x)$，我们已知 $f(1),\cdots,f(n+1)$（$1\le i\le n+1$），考虑代入上面的插值公式：
|
||||
|
||||
$$
|
||||
\begin{aligned}
|
||||
f(x)&=\sum\limits_{i=1}^{n+1}y_i\prod\limits_{j\ne i}\frac{x-x_j}{x_i-x_j}\\
|
||||
&=\sum\limits_{i=1}^{n+1}y_i\prod\limits_{j\ne i}\frac{x-j}{i-j}
|
||||
\end{aligned}
|
||||
$$
|
||||
|
||||
后面的累乘可以分子分母分别考虑，不难得到分子为：
|
||||
|
||||
$$
|
||||
\dfrac{\prod\limits_{j=1}^{n+1}(x-j)}{x-i}
|
||||
$$
|
||||
|
||||
分母的 $i-j$ 累乘可以拆成两段阶乘来算：
|
||||
|
||||
$$
|
||||
(-1)^{n+1-i}\cdot(i-1)!\cdot(n+1-i)!
|
||||
$$
|
||||
|
||||
于是横坐标为 $1,\cdots,n+1$ 的插值公式：
|
||||
|
||||
$$
|
||||
f(x)=\sum\limits_{i=1}^{n+1}(-1)^{n+1-i}y_i\cdot\frac{\prod\limits_{j=1}^{n+1}(x-j)}{(i-1)!(n+1-i)!(x-i)}
|
||||
$$
|
||||
|
||||
预处理 $(x-i)$ 前后缀积、阶乘阶乘逆，然后代入这个式子，复杂度为 $O(n)$.
|
||||
|
||||
???+ note " 例题 [CF622F The Sum of the k-th Powers](https://codeforces.com/contest/622/problem/F)"
|
||||
给出 $n,k$，求 $\sum\limits_{i=1}^ni^k$ 对 $10^9+7$ 取模的值。
|
||||
|
||||
??? note "题解"
|
||||
本题中，答案是一个 $k+1$ 次多项式，因此我们可以线性筛出 $1^i,\cdots,(k+2)^i$ 的值然后进行 $O(n)$ 插值。
|
||||
|
||||
也可以通过组合数学相关知识由差分法的公式推得下式：
|
||||
|
||||
$$
|
||||
f(x)=\sum_{i=1}^{n+1}\binom{x-1}{i-1}\sum_{j=1}^{i}(-1)^{i+j}\binom{i-1}{j-1}y_{j}=\sum\limits_{i=1}^{n+1}y_i\cdot\frac{\prod\limits_{j=1}^{n+1}(x-j)}{(x-i)\cdot(-1)^{n+1-i}\cdot(i-1)!\cdot(n+1-i)!}
|
||||
$$
|
||||
|
||||
??? note "代码实现"
|
||||
```cpp
|
||||
--8<-- "docs/math/code/numerical/interp/interp_2.cpp"
|
||||
```
|
||||
|
||||
## Newton 插值法
|
||||
|
||||
Newton 插值法是基于高阶差分来插值的方法，优点是支持 $O(n)$ 插入新数据点。
|
||||
|
||||
为了实现 $O(n)$ 插入新数据点，我们令：
|
||||
|
||||
$$
|
||||
f(x)=\sum_{j=0}^n a_jn_j(x)
|
||||
$$
|
||||
|
||||
其中 $n_j(x):=\prod_{i=0}^{j-1}(x-x_i)$ 称为 **Newton 基**（Newton basis）。
|
||||
|
||||
若解出 $a_j$，则可得到 $f(x)$ 的插值多项式。我们按如下方式定义 **前向差商**（forward divided differences）：
|
||||
|
||||
$$
|
||||
\begin{aligned}
|
||||
\lbrack y_k\rbrack & := y_k, & k=0,\dots,n, \\
|
||||
[y_k,\dots,y_{k+j}] & := \dfrac{[y_{k+1},\dots,y_{k+j}]-[y_k,\dots,y_{k+j-1}]}{x_{k+j}-x_k}, & k=0,\dots,n-j,~j=1,\dots,n.
|
||||
\end{aligned}
|
||||
$$
|
||||
|
||||
则：
|
||||
|
||||
$$
|
||||
\begin{aligned}
|
||||
f(x)&=[y_0]+[y_0,y_1](x-x_0)+\dots+[y_0,\dots,y_n](x-x_0)\dots(x-x_{n-1})\\
|
||||
&=\sum_{j=0}^n [y_0,\dots,y_j]n_j(x)
|
||||
\end{aligned}
|
||||
$$
|
||||
|
||||
此即 Newton 插值的形式。朴素实现的时间复杂度为 $O(n^2)$.
|
||||
|
||||
若样本点是等距的（即 $x_i=x_0+ih$，$i=1,\dots,n$），令 $x=x_0+sh$，Newton 插值的公式可化为：
|
||||
|
||||
$$
|
||||
f(x)=\sum_{j=0}^n \binom{s}{j}j!h^j[y_0,\dots,y_j]
|
||||
$$
|
||||
|
||||
上式称为 **Newton 前向差分公式**（Newton forward divided difference formula）。
|
||||
|
||||
???+ note
|
||||
若样本点是等距的，我们还可以推出：
|
||||
|
||||
$$
|
||||
[y_k,\dots,y_{k+j}]=\frac{1}{j!h^j}\Delta^{(j)}y_k
|
||||
$$
|
||||
|
||||
其中 $\Delta^{(j)}y_k$ 为 **前向差分**（forward differences），定义如下：
|
||||
|
||||
$$
|
||||
\begin{aligned}
|
||||
\Delta^{(0)}y_k & := y_k, & k=0,\dots,n, \\
|
||||
\Delta^{(j)}y_k & := \Delta^{(j-1)} y_{k+1}-\Delta^{(j-1)} y_k, & k=0,\dots,n-j,~j=1,\dots,n.
|
||||
\end{aligned}
|
||||
$$
|
||||
|
||||
??? note " 代码实现（[Luogu P4781【模板】拉格朗日插值](https://www.luogu.com.cn/problem/P4781)）"
|
||||
```cpp
|
||||
--8<-- "docs/math/code/numerical/interp/interp_3.cpp"
|
||||
```
|
||||
|
||||
### 横坐标是连续整数的 Newton 插值
|
||||
|
||||
例如：求某三次多项式 $f(x)=\sum_{i=0}^{3} a_ix^i$ 的多项式系数，已知 $f(1)$ 至 $f(6)$ 的值分别为 $1, 5, 14, 30, 55, 91$。
|
||||
|
||||
$$
|
||||
\begin{array}{cccccccccccc}
|
||||
1 & & 5 & & 14 & & 30 & & 55 & & 91 & \\
|
||||
& 4 & & 9 & & 16 & & 25 & & 36 & \\
|
||||
& & 5 & & 7 & & 9 & & 11 & \\
|
||||
& & & 2 & & 2 & & 2 & \\
|
||||
\end{array}
|
||||
$$
|
||||
|
||||
第一行为 $f(x)$ 的连续的前 $n$ 项；之后的每一行为之前一行中对应的相邻两项之差。观察到，如果这样操作的次数足够多（前提是 $f(x)$ 为多项式），最终总会返回一个定值。
|
||||
|
||||
计算出第 $i-1$ 阶差分的首项为 $\sum_{j=1}^{i}(-1)^{i+j}\binom{i-1}{j-1}f(j)$，第 $i-1$ 阶差分的首项对 $f(k)$ 的贡献为 $\binom{k-1}{i-1}$ 次。
|
||||
|
||||
$$
|
||||
f(k)=\sum_{i=1}^n\binom{k-1}{i-1}\sum_{j=1}^{i}(-1)^{i+j}\binom{i-1}{j-1}f(j)
|
||||
$$
|
||||
|
||||
时间复杂度为 $O(n^2)$.
|
||||
|
||||
## C++ 中的实现
|
||||
|
||||
自 C++ 20 起，标准库添加了 [`std::midpoint`](https://en.cppreference.com/w/cpp/numeric/midpoint) 和 [`std::lerp`](https://en.cppreference.com/w/cpp/numeric/lerp) 函数，分别用于求中点和线性插值。
|
||||
|
||||
## 参考资料
|
||||
|
||||
1. [Interpolation - Wikipedia](https://en.wikipedia.org/wiki/Interpolation)
|
||||
2. [Newton polynomial - Wikipedia](https://en.wikipedia.org/wiki/Newton_polynomial)
|
||||
|
|
@ -1,169 +0,0 @@
|
|||
author: Ir1d, Marcythm, x4Cx58x54, YanWQ-monad, AtomAlpaca, billchenchina, Chrogeek, Early0v0, EndlessCheng, Enter-tainer, Ghastlcon, Henry-ZHR, hly1204, hsfzLZH1, kenlig, Peanut-Tang, qwqAutomaton, qz-cqy, rui\_er, StudyingFather, swift-zym, Tiphereth-A, TrisolarisHD, Xeonacid
|
||||
|
||||
???+ note " 例题 [Luogu P4781【模板】拉格朗日插值](https://www.luogu.com.cn/problem/P4781)"
|
||||
给出 $n$ 个点对 $(x_i,y_i)$ 和 $k$，且 $\forall i,j$ 有 $i\neq j \iff x_i\neq x_j$ 且 $f(x_i)\equiv y_i\pmod{998244353}$ 和 $\deg(f(x))<n$（定义 $\deg(0)=-\infty$），求 $f(k)\bmod{998244353}$。
|
||||
|
||||
### 方法 1：差分法
|
||||
|
||||
差分法适用于 $x_i=i$ 的情况。
|
||||
|
||||
如，用差分法求某三次多项式 $f(x)=\sum_{i=0}^{3} a_ix^i$ 的多项式形式，已知 $f(1)$ 至 $f(6)$ 的值分别为 $1, 5, 14, 30, 55, 91$。
|
||||
|
||||
$$
|
||||
\begin{array}{cccccccccccc}
|
||||
1 & & 5 & & 14 & & 30 & & 55 & & 91 & \\
|
||||
& 4 & & 9 & & 16 & & 25 & & 36 & \\
|
||||
& & 5 & & 7 & & 9 & & 11 & \\
|
||||
& & & 2 & & 2 & & 2 & \\
|
||||
\end{array}
|
||||
$$
|
||||
|
||||
第一行为 $f(x)$ 的连续的前 $n$ 项；之后的每一行为之前一行中对应的相邻两项之差。观察到，如果这样操作的次数足够多（前提是 $f(x)$ 为多项式），最终总会返回一个定值。
|
||||
|
||||
计算出第 $i-1$ 阶差分的首项为 $\sum_{j=1}^{i}(-1)^{i+j}\binom{i-1}{j-1}f(j)$，第 $i-1$ 阶差分的首项对 $f(k)$ 的贡献为 $\binom{k-1}{i-1}$ 次。
|
||||
|
||||
$$
|
||||
f(k)=\sum_{i=1}^n\binom{k-1}{i-1}\sum_{j=1}^{i}(-1)^{i+j}\binom{i-1}{j-1}f(j)
|
||||
$$
|
||||
|
||||
时间复杂度为 $O(n^2)$。这种方法对给出的点的限制性较强。
|
||||
|
||||
### 方法 2：待定系数法
|
||||
|
||||
设 $f(x)=\sum_{i=0}^{n-1} a_ix^i$ 将每个 $x_i$ 代入 $f(x)$，有 $f(x_i)=y_i$，这样就可以得到一个由 $n$ 条 $n$ 元一次方程所组成的方程组，然后使用 **高斯消元** 解该方程组求出每一项 $a_i$，即确定了 $f(x)$ 的表达式。
|
||||
|
||||
如果您不知道什么是高斯消元，请看 [高斯消元](./gauss.md)。
|
||||
|
||||
时间复杂度 $O(n^3)$，对给出点的坐标无要求。
|
||||
|
||||
### 方法 3：拉格朗日插值法
|
||||
|
||||
在 [多项式部分简介](../poly/intro.md) 里我们已经定义了多项式除法。
|
||||
|
||||
那么我们会有：
|
||||
|
||||
$$
|
||||
f(x)\equiv f(a)\pmod{(x-a)}
|
||||
$$
|
||||
|
||||
因为 $f(x)-f(a)=(a_0-a_0)+a_1(x^1-a^1)+a_1(x^2-a^2)+\cdots +a_n(x^n-a^n)$，显然有 $(x-a)$ 这个因式。
|
||||
|
||||
这样我们就可以列一个关于 $f(x)$ 的多项式线性同余方程组：
|
||||
|
||||
$$
|
||||
\begin{cases}
|
||||
f(x)\equiv y_1\pmod{(x-x_1)}\\
|
||||
f(x)\equiv y_2\pmod{(x-x_2)}\\
|
||||
\vdots\\
|
||||
f(x)\equiv y_n\pmod{(x-x_n)}
|
||||
\end{cases}
|
||||
$$
|
||||
|
||||
令
|
||||
|
||||
$$
|
||||
\begin{aligned}
|
||||
M(x)&=\prod_{i=1}^n{(x-x_i)},\\
|
||||
m_i(x)&=\dfrac M{x-x_i}
|
||||
\end{aligned}
|
||||
$$
|
||||
|
||||
则 $m_i(x)$ 在模 $(x-x_i)$ 意义下的乘法逆元为
|
||||
|
||||
$$
|
||||
m_i(x_i)^{-1}=\prod_{j\ne i}{(x_i-x_j)^{-1}}
|
||||
$$
|
||||
|
||||
故
|
||||
|
||||
$$
|
||||
\begin{aligned}
|
||||
f(x)&\equiv\sum_{i=1}^n{y_i\left(m_i(x)\right)\left(m_i(x_i)^{-1}\right)}&\pmod{M(x)}\\
|
||||
&\equiv\sum_{i=1}^n{y_i\prod_{j\ne i}{\dfrac {x-x_j}{x_i-x_j}}}&\pmod{M(x)}
|
||||
\end{aligned}
|
||||
$$
|
||||
|
||||
又因为 $\deg\left(f(x)\right)<n$ 所以在模 $M(x)$ 意义下 $f(x)$ 就是唯一的，即：
|
||||
|
||||
$$
|
||||
f(x)=\sum_{i=1}^n{y_i\prod_{j\ne i}{\dfrac {x-x_j}{x_i-x_j}}}
|
||||
$$
|
||||
|
||||
这就是拉格朗日插值的表达式。
|
||||
|
||||
??? note "通常意义下拉格朗日插值的一种推导"
|
||||
由于要求构造一个函数 $f(x)$ 过点 $P_1(x_1, y_1), P_2(x_2,y_2),\cdots,P_n(x_n,y_n)$。首先设第 $i$ 个点在 $x$ 轴上的投影为 $P_i^{\prime}(x_i,0)$。
|
||||
|
||||
考虑构造 $n$ 个函数 $f_1(x), f_2(x), \cdots, f_n(x)$，使得对于第 $i$ 个函数 $f_i(x)$，其图像过 $\begin{cases}P_j^{\prime}(x_j,0),(j\neq i)\\P_i(x_i,y_i)\end{cases}$，则可知题目所求的函数 $f(x)=\sum\limits_{i=1}^nf_i(x)$。
|
||||
|
||||
那么可以设 $f_i(x)=a\cdot\prod_{j\neq i}(x-x_j)$，将点 $P_i(x_i,y_i)$ 代入可以知道 $a=\dfrac{y_i}{\prod_{j\neq i} (x_i-x_j)}$，所以
|
||||
|
||||
$f_i(x)=y_i\cdot\dfrac{\prod_{j\neq i} (x-x_j)}{\prod_{j\neq i} (x_i-x_j)}=y_i\cdot\prod_{j\neq i}\dfrac{x-x_j}{x_i-x_j}$。
|
||||
|
||||
那么我们就可以从另一个角度推导出通常意义下（而非模意义下）拉格朗日插值的式子为：
|
||||
|
||||
$f(x)=\sum_{i=1}^ny_i\cdot\prod_{j\neq i}\dfrac{x-x_j}{x_i-x_j}$。
|
||||
|
||||
??? note "代码实现"
|
||||
因为在固定模 $998244353$ 意义下运算，计算乘法逆元的时间复杂度我们在这里暂且认为是常数时间。
|
||||
|
||||
```cpp
|
||||
--8<-- "docs/math/code/numerical/lagrange/lagrange_1.cpp"
|
||||
```
|
||||
|
||||
本题中只用求出 $f(k)$ 的值，所以在计算上式的过程中直接将 $k$ 代入即可。
|
||||
|
||||
$$
|
||||
f(k)=\sum_{i=1}^{n}y_i\prod_{j\neq i }\frac{k-x_j}{x_i-x_j}
|
||||
$$
|
||||
|
||||
本题中，还需要求解逆元。如果先分别计算出分子和分母，再将分子乘进分母的逆元，累加进最后的答案，时间复杂度的瓶颈就不会在求逆元上，时间复杂度为 $O(n^2)$。
|
||||
|
||||
### 横坐标是连续整数的拉格朗日插值
|
||||
|
||||
如果已知点的横坐标是连续整数，我们可以做到 $O(n)$ 插值。
|
||||
|
||||
设要求 $n$ 次多项式为 $f(x)$，我们已知 $f(1),\cdots,f(n+1)$（$1\le i\le n+1$），考虑代入上面的插值公式：
|
||||
|
||||
$$
|
||||
\begin{aligned}
|
||||
f(x)&=\sum\limits_{i=1}^{n+1}y_i\prod\limits_{j\ne i}\frac{x-x_j}{x_i-x_j}\\
|
||||
&=\sum\limits_{i=1}^{n+1}y_i\prod\limits_{j\ne i}\frac{x-j}{i-j}
|
||||
\end{aligned}
|
||||
$$
|
||||
|
||||
后面的累乘可以分子分母分别考虑，不难得到分子为：
|
||||
|
||||
$$
|
||||
\dfrac{\prod\limits_{j=1}^{n+1}(x-j)}{x-i}
|
||||
$$
|
||||
|
||||
分母的 $i-j$ 累乘可以拆成两段阶乘来算：
|
||||
|
||||
$$
|
||||
(-1)^{n+1-i}\cdot(i-1)!\cdot(n+1-i)!
|
||||
$$
|
||||
|
||||
于是横坐标为 $1,\cdots,n+1$ 的插值公式：
|
||||
|
||||
$$
|
||||
f(x)=\sum\limits_{i=1}^{n+1}y_i\cdot\frac{\prod\limits_{j=1}^{n+1}(x-j)}{(x-i)\cdot(-1)^{n+1-i}\cdot(i-1)!\cdot(n+1-i)!}
|
||||
$$
|
||||
|
||||
预处理 $(x-i)$ 前后缀积、阶乘阶乘逆，然后代入这个式子，复杂度为 $O(n)$。
|
||||
|
||||
???+ note " 例题 [CF622F The Sum of the k-th Powers](https://codeforces.com/contest/622/problem/F)"
|
||||
给出 $n,k$，求 $\sum\limits_{i=1}^ni^k$ 对 $10^9+7$ 取模的值。
|
||||
|
||||
本题中，答案是一个 $k+1$ 次多项式，因此我们可以线性筛出 $1^i,\cdots,(k+2)^i$ 的值然后进行 $O(n)$ 插值。
|
||||
|
||||
也可以通过组合数学相关知识由差分法的公式推得下式：
|
||||
|
||||
$$
|
||||
f(x)=\sum_{i=1}^{n+1}\binom{x-1}{i-1}\sum_{j=1}^{i}(-1)^{i+j}\binom{i-1}{j-1}y_{j}=\sum\limits_{i=1}^{n+1}y_i\cdot\frac{\prod\limits_{j=1}^{n+1}(x-j)}{(x-i)\cdot(-1)^{n+1-i}\cdot(i-1)!\cdot(n+1-i)!}
|
||||
$$
|
||||
|
||||
??? note "代码实现"
|
||||
```cpp
|
||||
--8<-- "docs/math/code/numerical/lagrange/lagrange_2.cpp"
|
||||
```
|
||||
|
|
@ -89,7 +89,7 @@ $$
|
|||
|
||||
### Lagrange 插值公式法
|
||||
|
||||
考虑 Lagrange 插值公式
|
||||
考虑 [Lagrange 插值公式](../numerical/interp.md#lagrange-插值法)
|
||||
|
||||
$$
|
||||
\begin{aligned}
|
||||
|
|
|
|||
|
|
@ -322,7 +322,7 @@ nav:
|
|||
- 非公平组合游戏: math/game-theory/partizan-game.md
|
||||
- 反常游戏: math/game-theory/misere-game.md
|
||||
- 数值算法:
|
||||
- 拉格朗日插值: math/numerical/lagrange.md
|
||||
- 插值: math/numerical/interp.md
|
||||
- 数值积分: math/numerical/integral.md
|
||||
- 高斯消元: math/numerical/gauss.md
|
||||
- 牛顿迭代法: math/numerical/newton.md
|
||||
|
|
|
|||
Loading…
Reference in New Issue