三次样条插值函数(C++版)

1. double spl2(double x[],                                                                /*x坐标序列*/
2.                         double y[],                                                                /*y坐标序列*/
3.                         int n,                                                                        /*输入数据个数*/
4.                         double ddy1, double ddyn,                                /*第一点和最末点二阶导数*/
5.                         double t[],                                                                /*插值点的x坐标序列*/
6.                         int m,                                                                        /*插值点个数*/
7.                         double z[]                                                                /*差值点的y坐标序列*/
8. )        
9. {
10.         int i,j;
11.         double h0,h1,alpha,beta,*s,*dy;
12.         s= new double[n];
13.         dy = new double[n];
15.         dy[0]=-0.5;
16.         h0=x[1]-x[0];
17.         s[0]=3.0*(y[1]-y[0])/(2.0*h0)-ddy1*h0/4.0;
19.         for (j=1;j<=n-2;j++)
20.         {
21.                 h1=x[j+1]-x[j];
22.                 alpha=h0/(h0+h1);
23.                 beta=(1.0-alpha)*(y[j]-y[j-1])/h0;
24.                 beta=3.0*(beta+alpha*(y[j+1]-y[j])/h1);
25.                 dy[j]=-alpha/(2.0+(1.0-alpha)*dy[j-1]);
26.                 s[j]=(beta-(1.0-alpha)*s[j-1]);
27.                 s[j]=s[j]/(2.0+(1.0-alpha)*dy[j-1]);
28.                 h0=h1;
29.         }
30.         dy[n-1]=(3.0*(y[n-1]-y[n-2])/h1+ddyn*h1/2.0-s[n-2])/(2.0+dy[n-2]);
31.         for (j=n-2;j>=0;j--)        dy[j]=dy[j]*dy[j+1]+s[j];
32.         for (j=0;j<=n-2;j++)        s[j]=x[j+1]-x[j];
34.         for (j=0;j<=m-1;j++)
35.         {
36.                 if (t[j]>=x[n-1]) i=n-2;
37.                 else
38.                 {
39.                         i=0;
40.                         while (t[j]>x[i+1]) i=i+1;
41.                 }
42.                 h1=(x[i+1]-t[j])/s[i];
43.                 h0=h1*h1;
44.                 z[j]=(3.0*h0-2.0*h0*h1)*y[i];
45.                 z[j]=z[j]+s[i]*(h0-h0*h1)*dy[i];
47.                 h1=(t[j]-x[i])/s[i];
48.                 h0=h1*h1;
49.                 z[j]=z[j]+(3.0*h0-2.0*h0*h1)*y[i+1];
50.                 z[j]=z[j]-s[i]*(h0-h0*h1)*dy[i+1];
51.         }
52.         delete[] s;
53.         delete[] dy;
54.         return 0.0;
55. }