利用lagrange插值法计算函数值

最新推荐文章于 2020-04-03 12:05:57 发布

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最新推荐文章于 2020-04-03 12:05:57 发布

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文章标签： F#

博客给出了一组x和y的值，要求计算f(13)的值，并提供了用C语言编写的求解代码。代码通过用户输入n的个数、已知的x和y值以及要求的x值，最终计算并输出对应的y值。

x: 10 15 20
y: 1 1.1761 1.3010
求f(13);

答案：
#include <stdio.h>
#include <stdlib.h>

void main( void )
{
intn;
float*x = NULL;
float*y = NULL;
floatxFound;
floatyGet = 0.0;
float yGetTemp = 1.0;

printf("请输入你要输入的n的个数:");
scanf("%d", &n);

printf("请输入已知x的值(如:10 29 23 ……):");
x = (float *)malloc(sizeof(float) * n);
for (int client = 0; client < n; client++)
scanf("%f", (x + client));

printf("\n");

printf("请输入已知Y的值(如:23 23 23 ……):");
y = (float *)malloc(sizeof(float) * n);
for (client = 0; client < n; client++)
scanf("%f", (y + client));

printf("\n请输入要求的x的值:");
scanf("%f", &xFound);

int temp;
for (client = 0; client < n; client++)
{
for (temp = 0; temp < n; temp++)
{
if (temp == client)
continue;
else
yGetTemp *= ( (xFound - *(x + temp)) / ( *(x + client) - *(x + temp)));
}

yGet += (yGetTemp * (*(y + client)));
yGetTemp = 1.0;
}
printf("\nThe result is:%f\n", yGet);
}