Newton多项式插值法

Newton插值法公式：

$P_n(x)=f_0+{\sum }_{k=1}^n{a}_k(x-x_0)(x-x_1)...(x-x_{k-1}),a_k=f_{0,1,2,...,k},k=n$

例：

已知$(x_0,f(x_0)),(x_1,f(x_1)),...,(x_n,f(x_n))$共n+1点的坐标值，如何计算插值函数，关键在于如何建立差商表

#include
int main()
{
	int i,j,n;
	double f[40][40],x[40];
	scanf("%d",&n);
	for(i=0;i<=n;i++)
		scanf("%lf%lf",&x[i],&f[i][0]);
	printf("\n Divided Difference Table: \n");
	printf("=========================\n");
	for(j=1;j<=n;j++)
	{
		for(i=0;i<=n-j;i++)
		{
			f[i][j]=(f[i+1][j-1]-f[i][j-1])/(x[i+j]-x[i]);
		}
	}
	printf("i x(i) f(i) f(i,i+1) f(i,i+1,i+2),...... \n");
	for(i=0;i<=n;i++)
	{
		printf("%d %10.5lf",i,x[i]);
		for(j=0;j<=n-i;j++)
		{
			printf("%10.5lf",f[i][j]);
		}
		printf("\n");
	}
	return 0;
}
/*
5
0.0  -6.0
0.1  -5.89483
0.3  -5.65014
0.6  -5.17788
1.0  -4.28172
1.1  -3.99583

*/