Hermit插值法C语言
时间: 2023-06-20 12:03:21 浏览: 52
Hermit插值法是一种用于数据拟合的方法,可以用于生成曲线或曲面,具有高精度和快速计算的优点。下面是一个基于C语言的Hermit插值法示例:
```c
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#define MAX 100
struct point {
double x;
double y;
};
double hermit_interpolation(double x, struct point *p, int n)
{
double h[MAX], b[MAX], c[MAX], d[MAX], f[MAX];
int i, j, k;
for (i = 0; i < n; i++) {
h[i] = p[i+1].x - p[i].x;
b[i] = (p[i+1].y - p[i].y) / h[i];
}
c[0] = 0.0;
c[n-1] = 0.0;
for (i = 1; i < n-1; i++) {
c[i] = 2.0 * (h[i-1] + h[i]) - h[i-1] * c[i-1] / 3.0;
d[i] = (b[i] - b[i-1]) / h[i-1] - (b[i+1] - b[i]) / h[i];
c[i] = (d[i] - h[i-1] * c[i-1] / 3.0) / c[i];
}
f[n-1] = b[n-1];
for (i = n-2; i >= 0; i--)
f[i] = b[i] - c[i] * f[i+1] - (2.0 * h[i] * f[i+1] + h[i] * f[i]) / 3.0;
for (i = 0; i < n; i++) {
if (x >= p[i].x && x <= p[i+1].x) {
k = i;
break;
}
}
double A, B, C, D;
A = f[k] * pow(p[k+1].x - x, 3) / (6.0 * h[k]);
B = f[k+1] * pow(x - p[k].x, 3) / (6.0 * h[k]);
C = (p[k+1].y - f[k] * pow(h[k], 2) / 6.0) * (x - p[k].x) / h[k];
D = (p[k].y - f[k+1] * pow(h[k], 2) / 6.0) * (p[k+1].x - x) / h[k];
return A + B + C + D;
}
int main()
{
int n, i;
double x, y;
struct point p[MAX];
printf("Enter the number of data points: ");
scanf("%d", &n);
printf("Enter the data points:\n");
for (i = 0; i < n; i++) {
printf("x[%d] = ", i);
scanf("%lf", &p[i].x);
printf("y[%d] = ", i);
scanf("%lf", &p[i].y);
}
printf("Enter the value of x for which y is to be found: ");
scanf("%lf", &x);
y = hermit_interpolation(x, p, n-1);
printf("The interpolated value of y at x = %lf is %lf\n", x, y);
return 0;
}
```
在该示例中,我们首先定义了一个结构体来存储数据点的坐标。然后我们定义了一个函数 `hermit_interpolation` 来进行Hermit插值。该函数接受一个待插值的x值,数据点集合以及数据点的数量作为输入参数。该函数的主要任务是计算插值多项式,并使用它来计算给定x值的插值y值。在计算插值多项式时,我们使用了Hermit插值法的公式,该公式需要计算一些常数和系数,并使用它们来计算插值多项式。最后,我们在主函数中读取数据点,输入待插值的x值,并使用 `hermit_interpolation` 函数计算插值y值。