/*********************下面程序是C语言程序（标准C）******************/
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/* 计算给定M0,Mn值的三次样条插值多项式 */
/*给定离散点(1.1,0.4),(1.2,0.8),(1.4,1.65),(1.5,1.8),M0=Mn=0,*/
/*用M关系式构造三次样条插值多项式S(x),计算S(1.25)。 */
/*************************************************************/
#include stdio.h
#define Max_N 20
main()
{int i,k,n;
double h[Max_N+1],b[Max_N+1],c[Max_N+1],d[Max_N+1],M[Max_N+1];
double u[Max_N+1],v[Max_N+1],yy[Max_N+1],x[Max_N+1],y[Max_N+1];
double xx,p,q,S;
printf("\nPlease input n value:"); /*输入插值点数n*/
do
{ scanf("%d",n);
if(nMax_N)
printf("\nplease re-input n value:");
}
while(nMax_N||n=0);
printf("Input x[i],i=0,...%d:\n",n-1);
for(i=0;in;i++) scanf("%lf",x[i]);
printf("Input y[i],i=0,...%d:\n",n-1);
for(i=0;in;i++) scanf("%lf",y[i]);
printf("\nInput the M0,Mn value:");
scanf("%lf%lf",M[0],M[n]);
printf("\nInput the x value:"); /*输入计算三次样条插值函数的x值*/
scanf("%lf",xx);
if((xxx[n-1])||(xxx[0]))
{printf("Please input a number between %f and %f.\n",x[0],x[n-1]);
return;
}
/*计算M关系式中各参数的值*/
h[0]=x[1]-x[0];
for(i=1;in;i++)
{h[i]=x[i+1]-x[i];
b[i]=h[i]/(h[i]+h[i-1]);
c[i]=1-b[i];
d[i]=6*((y[i+1]-y[i])/h[i]-(y[i]-y[i-1])/h[i-1])/(h[i]+h[i-1]);
}
/*用追赶法计算Mi,i=1,...,n-1*/
d[1]-=c[1]*M[0];
d[n-1]-=b[n-1]*M[n];
b[n-1]=0; c[1]=0; v[0]=0;
for(i=1;in;i++)
{u[i]=2-c[i]*v[i-1];
v[i]=b[i]/u[i];
yy[i]=(d[i]-c[i]*y[i-1])/u[i];
}
for(i=1;in;i++)
{M[n-i]=yy[n-i]-v[n-i]*M[n-i+1];
}
/*计算三次样条插值函数在x处的值*/
k=0;
while(xx=x[k]) k++;
k=k-1;
p=x[k+1]-xx;
q=xx-x[k];
S=(p*p*p*M[k]+q*q*q*M[k+1])/(6*h[k])+(p*y[k]+q*y[k+1])/h[k]-h[k]*(p*M[k]+q*M[k+1])/6;
printf("S(%f)=%f\n",xx,S); /*输出*/
getch();
}
/*----------------------------------- End of file ------------------------------------*/
/*程序输入输出：
Please input n value:4
Input x[i],i=0,...3:
1.1 1.2 1.4 1.5
Input y[i],i=0,...3:
0.4 0.8 1.65 1.8
Input the M0,Mn value: 0 0
Input the x value:1.25
S(1.250000)=1.033171
*/
spline函数可以实现三次样条插值
x = 0:10;
y = sin(x);
xx = 0:.25:10;
yy = spline(x,y,xx);
plot(x,y,'o',xx,yy)
另外fnplt csapi这两个函数也是三次样条插值函数，具体你可以help一下！
void SPL(int n, double *x, double *y, int ni, double *xi, double *yi)； 是你所要。
已知 n 个点 x,y; x 必须已按顺序排好。要插值 ni 点，横坐标 xi[], 输出 yi[]。
程序里用double 型，保证计算精度。
SPL调用现成的程序。
现成的程序很多。端点处理方法不同，结果会有不同。想同matlab比较，你需 尝试 调用 spline（）函数 时，令 end1 为 1, 设 slope1 的值，令 end2 为 1 设 slope2 的值。
#include
#include
int spline (int n, int end1, int end2,
double slope1, double slope2,
double x[], double y[],
double b[], double c[], double d[],
int *iflag)
{
int nm1, ib, i, ascend;
double t;
nm1 = n - 1;
*iflag = 0;
if (n 2)
{ /* no possible interpolation */
*iflag = 1;
goto LeaveSpline;
}
ascend = 1;
for (i = 1; i n; ++i) if (x[i] = x[i-1]) ascend = 0;
if (!ascend)
{
*iflag = 2;
goto LeaveSpline;
}
if (n = 3)
{
d[0] = x[1] - x[0];
c[1] = (y[1] - y[0]) / d[0];
for (i = 1; i nm1; ++i)
{
d[i] = x[i+1] - x[i];
b[i] = 2.0 * (d[i-1] + d[i]);
c[i+1] = (y[i+1] - y[i]) / d[i];
c[i] = c[i+1] - c[i];
}
/* ---- Default End conditions */
b[0] = -d[0];
b[nm1] = -d[n-2];
c[0] = 0.0;
c[nm1] = 0.0;
if (n != 3)
{
c[0] = c[2] / (x[3] - x[1]) - c[1] / (x[2] - x[0]);
c[nm1] = c[n-2] / (x[nm1] - x[n-3]) - c[n-3] / (x[n-2] - x[n-4]);
c[0] = c[0] * d[0] * d[0] / (x[3] - x[0]);
c[nm1] = -c[nm1] * d[n-2] * d[n-2] / (x[nm1] - x[n-4]);
}
/* Alternative end conditions -- known slopes */
if (end1 == 1)
{
b[0] = 2.0 * (x[1] - x[0]);
c[0] = (y[1] - y[0]) / (x[1] - x[0]) - slope1;
}
if (end2 == 1)
{
b[nm1] = 2.0 * (x[nm1] - x[n-2]);
c[nm1] = slope2 - (y[nm1] - y[n-2]) / (x[nm1] - x[n-2]);
}
/* Forward elimination */
for (i = 1; i n; ++i)
{
t = d[i-1] / b[i-1];
b[i] = b[i] - t * d[i-1];
c[i] = c[i] - t * c[i-1];
}
/* Back substitution */
c[nm1] = c[nm1] / b[nm1];
for (ib = 0; ib nm1; ++ib)
{
i = n - ib - 2;
c[i] = (c[i] - d[i] * c[i+1]) / b[i];
}
b[nm1] = (y[nm1] - y[n-2]) / d[n-2] + d[n-2] * (c[n-2] + 2.0 * c[nm1]);
for (i = 0; i nm1; ++i)
{
b[i] = (y[i+1] - y[i]) / d[i] - d[i] * (c[i+1] + 2.0 * c[i]);
d[i] = (c[i+1] - c[i]) / d[i];
c[i] = 3.0 * c[i];
}
c[nm1] = 3.0 * c[nm1];
d[nm1] = d[n-2];
}
else
{
b[0] = (y[1] - y[0]) / (x[1] - x[0]);
c[0] = 0.0;
d[0] = 0.0;
b[1] = b[0];
c[1] = 0.0;
d[1] = 0.0;
}
LeaveSpline:
return 0;
}
double seval (int n, double u,
double x[], double y[],
double b[], double c[], double d[],
int *last)
{
int i, j, k;
double w;
i = *last;
if (i = n-1) i = 0;
if (i 0) i = 0;
if ((x[i] u) || (x[i+1] u))
{
i = 0;
j = n;
do
{
k = (i + j) / 2;
if (u x[k]) j = k;
if (u = x[k]) i = k;
}
while (j i+1);
}
*last = i;
w = u - x[i];
w = y[i] + w * (b[i] + w * (c[i] + w * d[i]));
return (w);
}
void SPL(int n, double *x, double *y, int ni, double *xi, double *yi)
{
double *b, *c, *d;
int iflag,last,i;
b = (double *) malloc(sizeof(double) * n);
c = (double *)malloc(sizeof(double) * n);
d = (double *)malloc(sizeof(double) * n);
if (!d) { printf("no enough memory for b,c,d\n");}
else {
spline (n,0,0,0,0,x,y,b,c,d,iflag);
if (iflag==0) printf("I got coef b,c,d now\n"); else printf("x not in order or other error\n");
for (i=0;ini;i++) yi[i] = seval(ni,xi[i],x,y,b,c,d,last);
free(b);free(c);free(d);
};
}
main(){
double x[6]={0.,1.,2.,3.,4.,5};
double y[6]={0.,0.5,2.0,1.6,0.5,0.0};
double u[8]={0.5,1,1.5,2,2.5,3,3.5,4};
double s[8];
int i;
SPL(6, x,y, 8, u, s);
for (i=0;i8;i++) printf("%lf %lf \n",u[i],s[i]);
return 0;
}
