我一直坚持这一段时间了。我正在用C编写一个算法,用拉格朗日插值法提取多项式的系数。C中的算法使用拉格朗日插值计算多项式的系数
我的代码部分的工作,例如,如果我们做的第一个例子这里http://en.wikipedia.org/wiki/Lagrange_polynomial#Example_1然后代码可以打印出第2个系数(0和4.834848)
同样有例如3这篇文章中,将打印2个系数6和-11。
我需要能够准确地获得任何一组点的所有系数。请告知代码所需的更改。
在此先感谢!
已更新为最新的代码,7:57 PM,GMT于8月5日。现在9个系数正在工作,看起来很丑。将调查明天n度的迭代过程!
#include<ncursesw/ncurses.h>
#include<math.h>
#include <stdio.h>
#include <string.h>
#include <errno.h>
#include <stdlib.h>
#define MAX 200
float coeff[MAX], coefftwo[MAX], coeffthree[MAX], coefffour[MAX];
int count;
void main()
{
int n,i,j ;
char ch;
float x[MAX],y[MAX],fp2, coeff1, coeff2;
printf("\n\nn = ");
scanf("%i", &count);
for(i=0; i < count; i++)
{
printf("\n\n The value of x%i= ", i);
scanf("%f",&x[i]);
printf("\n The value of f(x%i)= ", i);
scanf("%f",&y[i]);
}
for(i=0;i<count;i++)
{
coeff1 = 1.0;
coeff2 = 0.0;
coeff3 = 0.0;
coeff4 = 0.0;
coeff5 = 0.0;
coeff6 = 0.0;
coeff7 = 0.0;
coeff8 = 0.0;
coeff9 = 0.0;
for(j=0; j<count; j++)
{
if(i!=j) {
coeff1 = coeff1 * (array[i]-array[j]);
coeff2 -= array[j];
for (int k=j; k < count; k++) {
if ((j!=k) && (k!=i)) {
coeff3 += array[j] * array[k];
for(int l=k; l < count; l++) {
if ((l!=j) && (l!=k) && (l!=i)) {
coeff4 -= array[j] * array[k] * array[l];
for (int m = l; m < count; m++) {
if ((m!=l) && (m!=k) && (m!=j) && (m!=i)) { coeff5 += array[j] * array[k] * array[l] * array[m];
for (int n = m; n < count; n++) {
if ((n!=m) && (n!=l) && (n!=k) && (n!=j) && (n!=i)) {
coeff6 -= array[j] * array[k] * array[l] * array[m] * array[n];
for (int o = n; o < count; o++) {
if ((o!=n) && (o!=m) && (o!=l) && (o!=k) && (o!=j) && (o!=i)) {
coeff7 += array[j] * array[k] * array[l] * array[m] * array[n] * array[o];
for (int p = o; p < count; p++) {
if ((p!=o) && (p!=n) && (p!=m) && (p!=l) && (p!=k) && (p!=j) && (p!=i)) {
coeff8 -= array[j] * array[k] * array[l] *array[m] *array[n] * array[o] * array[p];
for (int q = p; q < count; q++) {
if ((q!=p) && (q!=o) && (q!=n) && (q!=m) && (q!=l) && (q!=k) && (q!=j) && (q!=i)) {
coeff9 += array[j] * array[k] * array[l] * array[m] * array[n] * array[o] * array[p] * array[q];
}
}
}
}
}
}
}
}
}
}
}
}
}
}
}
}
coeff[i] = y[i]/coeff1;
coefftwo[i] = y[i] * coeff2/coeff1;
coeffthree[i] = y[i] * coeff3/coeff1;
coefffour[i] = y[i] * coeff4/coeff1;
coefffive[i] = y[i] * coeff5/coeff1;
coeffsix[i] = y[i] * coeff6/coeff1;
coeffseven[i] = y[i] * coeff7/coeff1;
coeffeight[i] = y[i] * coeff8/coeff1;
coeffnine[i] = y[i] * coeff9/coeff1;
}
float coefficientone = 0.0;
float coefficienttwo = 0.0;
float coefficientthree = 0.0;
float coefficientfour = 0.0;
float coefficientfive = 0.0;
float coefficientsix = 0.0;
float coefficientseven = 0.0;
float coefficienteight = 0.0;
float coefficientnine = 0.0;
for (int i = 0; i< count; i++){
coefficientone = coefficientone + coeff[i];
coefficienttwo = coefficienttwo + coefftwo[i];
coefficientthree = coefficientthree + coeffthree[i];
coefficientfour = coefficientfour + coefffour[i];
coefficientfive = coefficientfive + coefffive[i];
coefficientsix = coefficientsix + coeffsix[i];
coefficientseven = coefficientseven + coeffseven[i];
coefficienteight = coefficienteight + coeffeight[i];
coefficientnine = coefficientnine + coeffnine[i];
}
printf("coefficient 1 = %f\n", coefficientone);
printf("coefficient 2 = %f\n", coefficienttwo);
printf("coefficient 3 = %f\n", coefficientthree);
printf("coefficient 4 = %f\n", coefficientfour);
printf("coefficient 5 = %f\n", coefficientfive);
printf("coefficient 6 = %f\n", coefficientsix);
printf("coefficient 7 = %f\n", coefficientseven);
printf("coefficient 8 = %f\n", coefficienteight);
printf("coefficient 9 = %f\n", coefficientnine);
}
你能给它的一个例子*失败* – Beta
当然 - ? 用以下输入,计数= 4,阵列[] =在维基百科文章的“X”值将来自实施例1,Y [] = “f(x)”值;我们得到以下输出,系数1 = 0.000000,系数2 = 4.834848,系数3 = 4.834848,系数4 = 10.576050。前两个系数是正确的,第三个和第四个系数都不正确(请参阅示例中的正确值) – JaboJG
'count'以什么值开头?你的使用表明它以'1'开始,在这种情况下你的数组索引可能是关闭的。也许你的循环打算达到'
usr2564301