niudun.c

实现 了 数值分析 课程 中得 几个 算法

/* 这份源代码文件已被未注册的SourceFormatX格式化过 */
/* 如果您想不再添加此类信息，请您注册这个共享软件  */
/* 更多相关信息请访问网站: http://cn.textrush.com  */


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

/**
 *The result :
 * Input n value: 2
 *Now input the (x_i,y_i),i=0,...,2:
 *11 0.190809 12 0.207912 13 0.224951
 *Now input the x value: 11.5
 *newton(11.500000)=0.199369
 *请按任意键继续. . .
 */

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

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

int main()
{
  int n;
  int i, j;
  POINT points[MAX_N + 1];
  double diff[MAX_N + 1];
  double x, tmp, newton = 0;
  printf("\n Input n value: "); //输入被插值点的数目
  scanf("%d", &n);
  if (n > MAX_N)
  {
    printf("The input n 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);
  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--)
    {
      diff[j] = (diff[j] - diff[j - 1]) / (points[j].x - points[j - 1-i].x);
    }
    //计算f(x_0,...,x_n)的差商
  }
  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

  system("pause");
}