拉格朗日插值算法及牛顿插值算法的C++实现

拉格朗日插值法

声明：本算法源自此博客：http://blog.csdn.net/xiaowei_cqu/article/details/8584966

本算法实现了批数据处理的插值实现，为上面网址博客提及的算法改良版（注：上面所述网址博客算法只能实现单个数据的插值，无法实现批数据插值）。

算法代码：

#include <iostream>  #include <string>  #include <vector>  using namespace std;void Lagrange(int N, int N1, vector<double>&X, vector<double>&Y, vector<double>&xp, vector<double> &result);int main(){char a = 'n';do{cout << "请输入所需插值点N的值：" << endl;int N;int N1;cin >> N;vector<double>X(N, 0);vector<double>Y(N, 0);cout << "请输入"<<N<<"个插值点对应的值X[i]：" << endl;for (int a = 0; a<N; a++)   {cin >> X[a];if (a >= N)break;   }cout << "请输入" << N << "个插值点对应的函数值Y[i]：" << endl;for (int a = 0; a<N; a++)   {cin >> Y[a];if (a >= N)break;   }cout << "请输入所求xp次数N1的值：" << endl;cin >> N1;vector<double>xp(N1, 0);vector<double> result(N1, 0);cout << "请输入" << N1 << "个所求向量xp[i]的值：" << endl;for (int a = 0; a<N1; a++)  {cin >> xp[a];if (a >= N1)break;  }Lagrange(N, N1, X, Y, xp, result);for (int a = 0; a < N1; a++)  {cout << result[a] << endl;  }cout << "是否要继续？(y/n)：";cin >> a;  } while (a == 'y');return 0;}void Lagrange(int N, int N1,vector<double>&X, vector<double>&Y, vector<double>&xp, vector<double> &result){double temp1=0;for (int h = 0; h < N1; h++)   {for (int i = 0; i < N; i++)    {double temp = Y[i];for (int j = 0; j < N; j++)                    {if (i != j) {temp = temp*(xp[h] - X[j]);temp = temp/(X[i] - X[j]); }     }temp1 += temp;    }result[h] = temp1;temp1 = 0 ;   }}

牛顿插值法：

其公式：

算法流程：

算法代码：

#include<iostream>  #include<string>  #include<vector>  using namespace std;double ChaShang(int n, vector<double>&X, vector<double>&Y);void Newton(int n1, vector<double>& xp, vector<double>&X, vector<double>&Y, vector<double> &result);int main(){char a = 'n';do{int n,n1;cout << "请输入插值点的个数" << endl;cin >> n;vector<double>X(n, 0);vector<double>Y(n, 0);cout << "请输入插值点X[i]的值" << endl;for (int i = 0; i<n; i++){cin >> X[i] ;}cout << "请输入插值点Y[i]的值" << endl;for (int i = 0; i<n; i++){cin >> Y[i];}cout << "请输入所求点的个数" << endl;cin >> n1;vector<double> xp(n1, 0);vector<double> result(n1, 0);cout << "请输入所求插值点xp[i]的值：" << endl;for (int i = 0; i<n1; i++){cin >> xp[i];}Newton(n1,xp, X, Y,result);cout << "输出所求插值点的函数值：" << endl;for (int h = 0; h < n1; h++){cout<< result[h]<<endl;}cout << "是否要继续？(y/n)：";cin >> a;} while (a == 'y');return 0;}double ChaShang(int n, vector<double>&X, vector<double>&Y){double f = 0;double temp = 0;for (int i = 0; i<n + 1; i++){temp = Y[i];for (int j = 0; j<n + 1; j++)if (i != j) temp /= (X[i] - X[j]);f += temp;}return f;}void Newton(int n1, vector<double>& xp, vector<double>&X, vector<double> &Y, vector<double> &result){double temp1 = 0;for (int h = 0; h < n1; h++){for (int i = 0; i < X.size(); i++){double temp = 1;double f = ChaShang(i, X, Y);for (int j = 0; j < i; j++){temp = temp*(xp[h] - X[j]);}temp1 += f*temp;}result[h] = temp1;temp1 = 0;}}

实验结果：

给定数据值：

所求插值点及结果：

总结：本算法实现的结果与引用博客的算法结果完全一致。