计算方法_拉格朗日插值_C++实现_方法2

#include<cstring>
#include<string.h>
#include<stdio.h>
#include<iostream>
#include<stdlib.h>
#include<math.h>
using namespace std;

int main( )
{
int n;
char c;
void sub44();
cout<<"拉格朗日插值法 " ;
sub44();

return 0;
}

void sub44()
{

int n,i,j;
double xm,ym=0;
double x[200],y[200], l[200];
cout <<endl<<"请输入已知点的个数n-1=";
cin >> n;
cout << endl;
cout <<"请输入待求点的横坐标xm="<<endl;
cin >> xm;
cout << endl;

cout <<endl<<"请分别输入已知点的横坐标"<<endl;
for(i=0;i<=n;i++)
{
cout << "x[" << i << "]=";
cin >> x[i];
cout << endl;
}

cout <<endl<<"请分别输入已知点的纵坐标"<<endl;
for(i=0;i<=n;i++)
{
cout << "y[" << i << "]=";
cin >> y[i];
cout << endl;
}

for(i=0;i<=n;i++)
{
double b=1,c=1;
for(j=0;j<=n;j++)
{
if(i!=j){b=b*(xm-x[j]);}
}
for(j=0;j<=n;j++)
{
if(i!=j){c=c*(x[i]-x[j]);}
}
l[i]=b/c;
}

for(i=0;i<=n;i++)
{
ym=ym+y[i]*l[i];
}

cout <<"你输入待求点xm的值,即ym=?"<< endl<<ym<<endl;
}