【计算方法】拉格朗日插值法
概念
设f(x)在N+1个点(x0,y0)…(xn,yn)处的值已知，其中值xk在区间【a，b】上，xk互不相同，满足a≤x0＜x1<…
插值法的思路在于通过构造一个N次多项式P(x)，使其通过N+1个点，然后求x处的函数值y。
拉格朗日逼近
例题
【问题描述】考虑[0.0,1.2]内的函数y=f(x)=cos(x)。利用多个（2,3,4等）节点构造拉格朗日插值多项式。
【输入形式】在屏幕上依次输入在区间[0.0,1.2]内的一个值x*，构造插值多项式后求其P(x*)值，和多个节点的x坐标。
【输出形式】输出插值多项式系数矩阵，拉格朗日系数多项式矩阵和P(x*)值（保留6位有效数字）。
【样例1输入】
0.3
0 0.6 1.2
【样例1输出】
-0.400435
-0.0508461
1
1.38889 -2.5 1
-2.77778 3.33333 -0
1.38889 -0.833333 0
0.948707
【样例1说明】输入：x为0.3，3个节点的x坐标分别为x0=0，x1=0.6和x2=1.2。
输出：插值多项式系数矩阵，则插值多项式P2(x)表示为-0.400435x**2-0.0508461x+1；
拉格朗日系数多项式矩阵，则P2(x)表示为：y0(1.38889x2-2.5x+1)+y1*(-2.77778x2+3.33333x-0)+y2*(1.38889x**2-0.83333x+0)；
当x*为0.3时，P2(0.3)值为0.948707。
【评分标准】根据输入得到的输出准确
ACcode:
c++ code(后面有python代码）
#include <bits/stdc++.h>
#define pr printf
#define sc scanf
#define for0(i, n) for (i = 0; i < n; i++)
#define for1n(i, n) for (i = 1; i <= n; i++)
#define forab(i, a, b) for (i = a; i <= b; i++)
#define forba(i, a, b) for (i = b; i >= a; i--)
#define pb push_back
#define eb emplace_back
#define fi first
#define se second
#define int long long
#define pii pair<int, int>
#define pdd pair<double, double>
#define vd vector<double>
#define vdd vector<vector<double>>
#define vi vector<int>
#define vii vector<vector<int>>
#define vt3 vector<tuple<int, int, int>>
#define mem(ara, n) memset(ara, n, sizeof(ara))
#define memb(ara) memset(ara, false, sizeof(ara))
#define all(x) (x).begin(), (x).end()
#define sq(x) ((x) * (x))
#define sz(x) x.size()
const int N = 2e5 + 100;
const int mod = 1e9 + 7;
namespace often
{
inline void input(int &res)
{
char c = getchar();
res = 0;
int f = 1;
while (!isdigit(c))
{
f ^= c == '-';
c = getchar();
}
while (isdigit(c))
{
res = (res << 3) + (res << 1) + (c ^ 48);
c = getchar();
}
res = f ? res : -res;
}
inline int qpow(int a, int b)
{
int ans = 1, base = a;
while (b)
{
if (b & 1)
ans = (ans * base % mod + mod) % mod;
base = (base * base % mod + mod) % mod;
b >>= 1;
}
return ans;
}
int fact(int n)
{
int res = 1;
for (int i = 1; i <= n; i++)
res = res * 1ll * i % mod;
return res;
}
int C(int n, int k)
{
return fact(n) * 1ll * qpow(fact(k), mod - 2) % mod * 1ll * qpow(fact(n - k), mod - 2) % mod;
}
int exgcd(int a, int b, int &x, int &y)
{
if (b == 0)
{
x = 1, y = 0;
return a;
}
int res = exgcd(b, a % b, x, y);
int t = y;
y = x - (a / b) * y;
x = t;
return res;
}
int invmod(int a, int mod)
{
int x, y;
exgcd(a, mod, x, y);
x %= mod;
if (x < 0)
x += mod;
return x;
}
}
using namespace often;
using namespace std;
vd operator*(vd const &y, vdd const &l)
{
int m = sz(y);
vd c(m);
for (int ii = 0; ii < m; ii++)
for (int jj = 0; jj < m; jj++)
c[ii] += y[jj] * l[jj][ii];
return c;
}
signed main()
{
int __ = 1;
//input(__);
auto polymul = [&](vd &v, double const &d, double const &err) {
int m = sz(v);
vd _ = v;
_.emplace_back(0);
for (int g = 0; g < m; g++)
_[g + 1] += v[g] * d, _[g] /= err;
_[m] /= err;
v = _;
};
auto polyval = [&](vd const &c, double const &_x) -> double {
double res = 0.0;
int m = sz(c);
for (int ii = 0; ii < m; ii++)
res += c[ii] * pow(_x, (double)(m - ii - 1));
return res;
};
while (__--)
{
double _x;
cin >> _x;
vd x, y;
double X, Y;
while (cin >> X)
Y = cos(X), x.emplace_back(X), y.emplace_back(Y);
int n = sz(x);
int i, j, k;
vdd l(n, vd(n));
for0(k, n)
{
vd v(1, 1);
for0(j, n)
{
if (j != k)
{
double d = -x[j];
double err = x[k] - x[j];
polymul(v, d, err);
}
}
l[k] = v;
}
vd c = y*l;
for0(i, n)
pr("%g\n", c[i]);
for0(i, n)
{
for0(j, n)
pr("%g ", l[i][j]);
puts("");
}
pr("%g\n", polyval(c, _x));
}
return 0;
}
python --code
'''
Author: csc
Date: 2021-04-24 10:52:25
LastEditTime: 2021-04-24 14:06:57
LastEditors: Please set LastEditors
Description: In User Settings Edit
FilePath: \code_py\lagrange.py
'''
import numpy as np
def lagrange(_x, n, x, y):
l = np.zeros([n, n], dtype=np.double)
for k in range(n):
v = 1
for j in range(n):
if j != k:
v = np.polymul(v, np.poly1d([1, -x[j]]))/(x[k]-x[j])
l[k, :] = v
c = y@l
for i in range(n):
print('%g' % c[i])
for i in range(n):
for j in range(n):
print('%g' % l[i][j], end=' ')
print()
_y = np.polyval(c, _x)
print('%g' % _y, end=' ')
def main():
_x = float(input())
x = np.array(list(map(float, input().split())))
n = len(x)
y = np.cos(x)
lagrange(_x, n, x, y)
if __name__ == '__main__':
main()