newton.c

计算方法 NEWTON插值的C程序

////////////////////////////////////////////////////////
//               (x_i,y_i)的牛顿插值多项式            //
////////////////////////////////////////////////////////

#include <stdio.h>
#define MAX_N 20         //定义(x_i,y_i)的最大维数

typedef struct tagPOINT  //定义点的结构
{
	double x;
	double y;
} POINT;

int main()
{
	int n, i, j;
	POINT points[MAX_N+1];
	double diff[MAX_N+1];
	double x, tmp, newton;
	char c;
	printf("\nInput n value: "); //输入被插值点的数目
	scanf("%d",&n);
	if (n > MAX_N)
	{
		printf("The input is larger than MAX_N, please redefine the MAX_N. \n");
		return 1;
	}
	if (n <= 0)
	{
		printf("Please input a number between 1 and %d. \n", MAX_N );
		return 1;
	}
	
	//输入被插值点(x_i,y_i)
	printf("Now input the (x_i,y_i),i=0,...,%d: \n", n);
	for(i = 0; i <= n; i++)
		scanf("%lf %lf", &points[i].x, &points[i].y);
	for(i = 0; i <= n; i++)  diff[i] = points[i].y;
	for(i = 0; i < n; i++ )
	{
		for (j = n; j > i; j--)
			//计算f(x_0,..,x_n)的插商
			diff[j] = (diff[j] - diff[j-1])/(points[j].x - points[j-1-i].x);
	}
	while(1)
	{
		printf("Do you want to continue the calculation? 'y' or 'n'? : ");
		getchar();      //先把缓冲区的字符读出，防止出错
		c = getchar();  //然后再接受用户的输入
		switch(c) 
		{
		case 'y': 
			{
				printf("\nNow input the x value: ");   //输入计算牛顿插值多项式的x值
				scanf("%lf", &x);	
				tmp = 1;
				newton = diff[0];
				for(i = 0; i < n; i++)
				{
					tmp = tmp * (x - points[i].x);
					newton = newton + tmp * diff[i+1];
				}
				printf("newton(%f) = %f\n", x, newton);    //输出结果
				break;
			}
		case 'n':
			{
				printf("Thank you for using, goodbye.\n");
				return 0;
			}
		default:			
			{
				printf("\nPlesae press y or n to choose! \n");
				break;
			}
		}
	}
}
/*
if(c == 'n')
{
printf("Thank you for using, goodbye.\n");
return 0;
}
else if(c == 'y')
{

  printf("Now input the x value: ");   //输入计算牛顿插值多项式的x值
  scanf("%lf", &x);
  for(i = 0; i <= n; i++)  diff[i] = points[i].y;
  for(i = 0; i < n; i++ )
  {
  for (j = n; j > i; j--)
  //计算f(x_0,..,x_n)的插商
  diff[j] = (diff[j] - diff[j-1])/(points[j].x - points[j-1-i].x);
  }
  tmp = 1;
  newton = diff[0];
  for(i = 0; i < n; i++)
  {
  tmp = tmp * (x - points[i].x);
  newton = newton + tmp * diff[i+1];
  }
  printf("newton(%f) = %f\n", x, newton);    //输出结果
  return 0;
  }
  else 
  {
  printf("Plesae press y or n to choose! \n");
  continue;
	}*/