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1 //题目要求:用牛顿差商公式进行插值 2 #include <iostream> 3 #include <cmath> 4 #include <cstring> 5 using namespace std; 6 #define numofx 20 //定义数据量的大小 7 struct data{ 8 double x; 9 double fx; //定义进行运算的点(x,fx) 10 }pointdata[numofx]; 11 12 //求差商f(x0,x1,x2...xn)的公式 13 long double chashang(int n){ //计算f[x0,x1,x2...xn] 14 if(n<=0){ 15 return 0; 16 } 17 18 long double result=0; 19 for(int i=0;i<=n;i++){ 20 //计算这0~n的计算值的和 21 long double resproduct=1; 22 for(int j=0;j<=n;j++){ 23 if(j!=i){ 24 resproduct*=(pointdata[i].x-pointdata[j].x); 25 } 26 } 27 long double temp=pointdata[i].fx/resproduct; 28 result+=temp; 29 } 30 return result; 31 } 32 33 int main(){ 34 int numofpoint; 35 cout<<"将要输入的插值点的个数为:"; 36 cin>>numofpoint; 37 cout<<"请依次输入x,fx的值:\n"; 38 for(int i=0;i<numofpoint;i++){ 39 cin>>pointdata[i].x>>pointdata[i].fx; 40 } 41 42 //打印输出牛顿差商公式 ,未简约 43 cout<<"由以上插值点计算出的牛顿插值多项式为:\n"; 44 cout<<"f(x)="<<pointdata[0].fx; 45 for(int i=1;i<numofpoint;i++){ 46 long double temp4=chashang(i); 47 if(temp4>=0){ 48 cout<<"+"<<temp4<<"*"; 49 } 50 else{ 51 cout<<temp4<<"*"; 52 } 53 54 for(int j=0;j<i;j++){ 55 cout<<"(x-"<<pointdata[j].x<<")"; 56 } 57 } 58 59 //计算某个x的牛顿插值多项式近似值 60 double x; 61 cout<<"\n\n请输入所求变量x的值:"; 62 cin>>x; 63 cout<<"牛顿插值多项式近似结果为:"; 64 65 long double res=pointdata[0].fx; //牛顿差商公式的f(x0)项 66 for(int i=0;i<numofpoint-1;i++){ //牛顿插值多项式项f[x0,x1...xn](x-x0)(x-x1)...(x-xn) n>=1 67 68 long double temp=1; 69 for(int j=0;j<=i;j++){ 70 temp=temp*(x-pointdata[j].x); 71 } 72 73 res=res+chashang(i+1)*temp; 74 } 75 cout<<res<<endl; 76 }
拉格朗日插值的程序:
1 #include <iostream> 2 #include <fstream> 3 using namespace std; 4 5 class lagrange{ 6 7 private: 8 int i,j,n; 9 double mult,sum,z; 10 double *x,*y; 11 12 public: 13 void interpolation(){ 14 ifstream fin("a.txt"); 15 fin>>n; 16 x=new double[n]; 17 y=new double[n]; 18 for(i=0;i<n;i++){ 19 fin>>x[i]>>y[i]; 20 } 21 fin.close(); 22 cout<<"\n输入需要插值的点:"; 23 cin>>z; 24 sum=0.0; 25 26 for(i=0;i<n;i++){ 27 mult=1.0; 28 for(j=0;j<n;j++){ 29 if(j!=i){ 30 mult*=(z-x[j])/(x[i]-x[j]); 31 } 32 } 33 sum+=mult*y[i]; 34 } 35 cout<<"\n插补的值="<<sum<<endl; 36 37 } 38 39 ~lagrange(){ 40 delete[] x,y; 41 } 42 43 }; 44 45 int main(){ 46 lagrange interp; 47 interp.interpolation(); 48 }
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原文地址:http://www.cnblogs.com/liugl7/p/4888692.html