三次参数样条插值方法程序设计

1． 次参数样条插值方法程序设计要求：1) 三学习三次参数样条曲线计算原理；2) 对给定的n个已知数据点Pi(xi,yi) i=1,…,n，确定已知端点条件为抛物端。3) 确定计算参数所满足的线性方程组及边界条件4) 设计求解线性组的方法5) 设计绘制生成曲线的方法6) 编写上述过程的程序7) 程序调试、测试，并利用给定坐标绘出图形编写设计报告。

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fnpltcsapi这两个函数也是三次样条插值函数，具体你可以help一下！

现在电脑上没有matlab,一会给你程序，呵呵！

#include<iostream>

#include<iomanip>

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:"

cin>>n

for(i

=

0

i

<=

n

i++){

cout<<"Please

put

in

X"<<i<<':'

cin>>x[i]

//cout<<endl

cout<<"Please

put

in

Y"<<i<<':'

cin>>y[i]

//cout<<endl

}

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

cin>>t

switch(t){

case

1:cout<<"Please

put

in

Y0\'

Y"<<n<<"\'\n"

cin>>f0>>f1

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"

cin>>f0>>f1

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

:"

cin>>ch

}while(ch

==

'y'

||

ch

==

'Y')

return

0

}

void

printout(int

n){

cout<<setprecision(6)

for(int

i

=

0

i

<

n

i++){

cout<<i+1<<":

["<<x[i]<<"

,

"<<x[i+1]<<"]\n"<<"\t"

/*

cout<<fxym[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"

cout<<endl*/

float

t

=

fxym[i]/(6*h[i])

if(t

>

0)cout<<t<<"*("<<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]<<")"

cout<<endl<<endl

}

cout<<endl

}

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