C语言实现拉格朗日插值法

        由于本学期有一门计算方法的课需要C语言编程，在这里记录一下。

        由于这是第一章且后面的课大约也有相似的属性，比如对n个点的使用和函数的调用，这里就进行一个小封装，方便以后吃老本。

        欢迎需要参考的同学复制粘贴，互相交流.

//POINTS.h
#pragma once
#include
using namespace std;
#ifndef _POINTS_
#define _POINTS_
class POINTS
{
public:
	typedef struct MYPOINT
	{
		double x;
		double y;
	}mypoint;
	int len;
	int size;
	virtual void run(double val);
	virtual void loaddata(double* vallist);
};
#endif // !_POINTS_

//POINTS.cpp
#include "POINTS.h"
#include
using namespace std;
void POINTS::run(double val)
{
	cout << "计算函数" << endl;
}
void POINTS::loaddata(double* vallist)
{
	cout << "数据处理函数" << endl;
}

//LAGRANGE.h
#pragma once
#include "POINTS.h"
#ifndef _LAGRANGE_
#define _LAGRANGE_
class LAGRANGE :
    public POINTS
{
public:
    LAGRANGE(int n);
    vector datalist;
    virtual void run(double val);
    virtual void loaddata(double* vallist);
};
#endif // !LAGRANGE

//LAGRANGE.cpp
#include "LAGRANGE.h"
#include
using namespace std;
LAGRANGE::LAGRANGE(int n)
	:datalist(n)
{
	this->size = n;
	this->len = 0;
}
void LAGRANGE::run(double val)
{
	double sum = 0.0;
	vector templist;
	for (size_t i = 0; i < this->size; i++)
	{
		double tempval = 1.0;
		for (size_t j = 0; j < this->size; j++)
		{
			if (j != i)
				tempval *= ((val - this->datalist.at(j).x) /
					(this->datalist.at(i).x - this->datalist.at(j).x));
			else
				continue;
		}
		templist.push_back(tempval);
	}
	for (size_t i = 0; i < this->size; i++)
	{
		sum += this->datalist.at(i).y * templist.at(i);
	}
	cout << "拉格朗日" << this->size << "项插值计算为：" << sum << endl;
}

void LAGRANGE::loaddata(double* vallist)
{
	cout << "正在处理数据..." << endl;
	for (size_t i = 0; i < this->size * 2; i++)
	{
		if (((i + 1) % 2) == 0)
		{
			this->datalist.at(this->len).y = vallist[i];
			this->len++;
		}
		else
		{
			this->datalist.at(this->len).x = vallist[i];
		}
		
	}
}

//main.cpp
#include
#include"POINTS.h"
#include"LAGRANGE.h"
using namespace std;
int main()
{
	cout << "例1" << endl;
	//实例化工具类
	LAGRANGE lgr(2);
	//构建运算参数表
	double val[4] = { 100.0,10.0,121.0,11.0 };
	//调用处理数据函数
	lgr.loaddata(val);
	//传入目标x运算,例1
	lgr.run(115);
	//传入目标x运算,例2
	cout << "例2" << endl;
	LAGRANGE lgr1(3);
	double val1[6] = { 100.0,10.0,121.0,11.0,144.0,12.0 };
	lgr1.loaddata(val1);
	lgr1.run(115);
	//课上的题目求ln3.16
	cout << "求ln3.16" << endl;
	LAGRANGE lgr2(2);
	double val2[4] = { 3.1,1.1314,3.2,1.1632 };
	lgr2.loaddata(val2);
	lgr2.run(3.16);
	system("pause");
	return 0;
}