牛顿插值法（伪代码 c/c++ python实现）

插值法利用函数f(x)在某区间中若干点的函数值，作出适当的特定函数，在这些点上取已知值，在区间的其他点上用这特定函数的值作为函数f(x)的近似值。

牛顿插值法通过构造差商表来得到计算公式，进而得到计算结果。

伪代码

newton_interpolation (input a[][], input n, input x)//牛顿插值法，使用差商表，a[0][]存储的是已知的x的值，a[1][]存储的是已知的y的值，求x所对应的y值i <- 1while i <= 2 do:j <- i - 1k <- 0while j < n do:a[i][j] <- (a[i-1][j] - a[i-1][j-1])/(a[0][j] - a[0][k]);j <- j + 1endk <- k + 1i <- i + 1endi <- 2while i <= n do:sum <- 1j <- i - 2while j >= 0 do:sum <- (x - a[0][j]) * sumendsum <- sum * a[i][i-1]y = sum + yendreturn y

c/c++

#include <iostream>using namespace std;double newton_interpolation(double a[10][10], int n, double x);int main(){double a[10][10],x;int n;memset(a,0,sizeof(a));cout << "输入已有数据的数量"<<endl;cin >> n;cout << "输入"<<n<<"个x的值"<<endl;for (int i = 0;i<n;i++)cin >> a[0][i];cout << "输入"<<n<<"个y的值"<<endl;for (int i = 0;i<n;i++)cin >> a[1][i];cout << "输入所要求的x的值:"<<endl;cin >> x;cout << "y = "<<newton_interpolation(a,n,x)<<endl;cout << "所得到的差商表为:"<<endl;for (int i = 0;i <= n;i++){for (int j = 0; j < n;j++)cout << a[i][j]<<"\t\t";cout << endl;}}double newton_interpolation(double a[10][10], int n, double x){int i,j,k;double y = a[1][0],sum;for (i = 2;i <= n;i++)for (j = i-1,k=0;j < n; j++,k++)a[i][j] = (a[i-1][j] - a[i-1][j-1])/(a[0][j]-a[0][k]);for (i = 2; i <= n; i++){sum = 1;for (j = i-2; j >= 0; j--)sum = (x - a[0][j]) * sum;cout << a[i][i-1] << endl;sum = sum * a[i][i-1];y = sum + y;}return y;}

python

def newton_interpolation1(a, n, x):    i = 2    y = a[1][0]    while i <= n:        j = i - 1        k = 0        while j < n:            a[i][j] = (a[i-1][j] - a[i-1][j-1]) / (a[0][j] - a[0][k])            j = j + 1            k = k + 1        i = i + 1    i = 2    while i <= n:        sum = 1        j = i - 2        while j >= 0:            sum = (x - a[0][j]) * sum            j = j - 1        sum = sum * a[i][i - 1]        y = sum + y        i = i + 1    return ydef newton_interpolation(a, n, x):    y = a[1][0]    for i in range(2, n+1):        k = 0        for j in range(i-1, n):            a[i][j] = (a[i-1][j] - a[i-1][j-1]) / (a[0][j] - a[0][k])            k = k + 1    for i in range(2, n+1):        sum = 1        for j in range(i-2, -1, -1):            sum = (x - a[0][j]) * sum        sum = sum * a[i][i-1]        y = sum + y    return ya = [[4.0, 9.0, 6.25], [2.0, 3.0, 2.5],[0, 0, 0], [0, 0, 0]]print newton_interpolation(a, 3, 7)

值得注意的是，使用python的时候，输入的数组中的数字人为要变成小数，输入整数的话，会出现错误。

我写的python分别用while循环和for循环都写了一遍，其实就是提高python的编程能力的。