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#include stdio.h
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#include math.h
double fun(double x,double x0,double x1,double x2,double y0,double y1,double y2)
{
double yx=0;
yx=y0*(x-x1)*(x-x2)/((x0-x1)*(x0-x2))+
y1*(x-x0)*(x-x2)/((x1-x0)*(x1-x2))+
y2*(x-x0)*(x-x1)/((x2-x0)*(x2-x1));//3点插值公式
return yx;
}
int main(int argc, char *argv[])
{
double x,x0,x1,x2,y0,y1,y2;
printf("输入待求值x：\n");
scanf("%lf",x);
x0=x-0.1;x1=x+0.1;x2=x+0.15;//需要输入3个插值点，即对应的x值和函数y值，这里简单计算的可以手动输入
y0=sin(x0);y1=sin(x1);y2=sin(x2);
printf("sin(%lf)=%lf-------fun(%lf)=%lf\n",x,sin(x),x,fun(x,x0,x1,x2,y0,y1,y2));
return 0;
}
我记得大三学的计算方法课上有，课后作业实现了的。不过在实验室那个电脑上，如果你有条件的话先参考《数值分析》书上吧。
至于c语言和c++的区别，这个程序应该没什么区别，反正都拿数组做。
#includeiostream
#includeiomanip
using
namespace
std;
const
int
MAX
=
50;
float
x[MAX],
y[MAX],
h[MAX];
float
c[MAX],
a[MAX],
fxym[MAX];
float
f(int
x1,
int
x2,
int
x3){
float
a
=
(y[x3]
-
y[x2])
/
(x[x3]
-
x[x2]);
float
b
=
(y[x2]
-
y[x1])
/
(x[x2]
-
x[x1]);
return
(a
-
b)/(x[x3]
-
x[x1]);
}
//求差分
void
cal_m(int
n){
//用追赶法求解出弯矩向量M……
float
B[MAX];
B[0]
=
c[0]
/
2;
for(int
i
=
1;
i
n;
i++)
B[i]
=
c[i]
/
(2
-
a[i]*B[i-1]);
fxym[0]
=
fxym[0]
/
2;
for(i
=
1;
i
=
n;
i++)
fxym[i]
=
(fxym[i]
-
a[i]*fxym[i-1])
/
(2
-
a[i]*B[i-1]);
for(i
=
n-1;
i
=
0;
i--)
fxym[i]
=
fxym[i]
-
B[i]*fxym[i+1];
}
void
printout(int
n);
int
main(){
int
n,i;
char
ch;
do{
cout"Please
put
in
the
number
of
the
dots:";
cinn;
for(i
=
0;
i
=
n;
i++){
cout"Please
put
in
X"i':';
cinx[i];
//coutendl;
cout"Please
put
in
Y"i':';
ciny[i];
//coutendl;
}
for(i
=
0;
i
n;
i++)
//求
步长
h[i]
=
x[i+1]
-
x[i];
cout"Please
输入边界条件\n
1:
已知两端的一阶导数\n
2:两端的二阶导数已知\n
默认:自然边界条件\n";
int
t;
float
f0,
f1;
cint;
switch(t){
case
1:cout"Please
put
in
Y0\'
Y"n"\'\n";
cinf0f1;
c[0]
=
1;
a[n]
=
1;
fxym[0]
=
6*((y[1]
-
y[0])
/
(x[1]
-
x[0])
-
f0)
/
h[0];
fxym[n]
=
6*(f1
-
(y[n]
-
y[n-1])
/
(x[n]
-
x[n-1]))
/
h[n-1];
break;
case
2:cout"Please
put
in
Y0\"
Y"n"\"\n";
cinf0f1;
c[0]
=
a[n]
=
0;
fxym[0]
=
2*f0;
fxym[n]
=
2*f1;
break;
default:cout"不可用\n";//待定
};//switch
for(i
=
1;
i
n;
i++)
fxym[i]
=
6
*
f(i-1,
i,
i+1);
for(i
=
1;
i
n;
i++){
a[i]
=
h[i-1]
/
(h[i]
+
h[i-1]);
c[i]
=
1
-
a[i];
}
a[n]
=
h[n-1]
/
(h[n-1]
+
h[n]);
cal_m(n);
cout"\n输出三次样条插值函数：\n";
printout(n);
cout"Do
you
to
have
anther
try
?
y/n
:";
cinch;
}while(ch
==
'y'
||
ch
==
'Y');
return
0;
}
void
printout(int
n){
coutsetprecision(6);
for(int
i
=
0;
i
n;
i++){
couti+1":
["x[i]"
,
"x[i+1]"]\n""\t";
/*
coutfxym[i]/(6*h[i])"
*
("x[i+1]"
-
x)^3
+
""
*
(x
-
"x[i]")^3
+
"
(y[i]
-
fxym[i]*h[i]*h[i]/6)/h[i]"
*
("x[i+1]"
-
x)
+
"
(y[i+1]
-
fxym[i+1]*h[i]*h[i]/6)/h[i]"(x
-
"x[i]")\n";
coutendl;*/
float
t
=
fxym[i]/(6*h[i]);
if(t
0)coutt"*("x[i+1]"
-
x)^3";
else
cout-t"*(x
-
"x[i+1]")^3";
t
=
fxym[i+1]/(6*h[i]);
if(t
0)cout"
+
"t"*(x
-
"x[i]")^3";
else
cout"
-
"-t"*(x
-
"x[i]")^3";
cout"\n\t";
t
=
(y[i]
-
fxym[i]*h[i]*h[i]/6)/h[i];
if(t
0)cout"+
"t"*("x[i+1]"
-
x)";
else
cout"-
"-t"*("x[i+1]"
-
x)";
t
=
(y[i+1]
-
fxym[i+1]*h[i]*h[i]/6)/h[i];
if(t
0)cout"
+
"t"*(x
-
"x[i]")";
else
cout"
-
"-t"*(x
-
"x[i]")";
coutendlendl;
}
coutendl;
}
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。