三次样条函数c语言,c语言三次样条插值函数

求助:求三次样条插值函数的C++程序

#includeiostream

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#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;

}

三次样条插值用c语言具体怎么做

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。

谁来帮帮我啊:运用三次样条插值函数法对一条已知的折线进行圆滑(编程实现)。

matlab算么?

function output = spline(x,y,xx)

%SPLINE Cubic spline data interpolation.

% PP = SPLINE(X,Y) provides the piecewise polynomial form of the

% cubic spline interpolant to the data values Y at the data sites X,

% for use with the evaluator PPVAL and the spline utility UNMKPP.

% X must be a vector.

% If Y is a vector, then Y(j) is taken as the value to be matched at X(j),

% hence Y must be of the same length as X -- see below for an exception

% to this.

% If Y is a matrix or ND array, then Y(:,...,:,j) is taken as the value to

% be matched at X(j), hence the last dimension of Y must equal length(X) --

% see below for an exception to this.

%

% YY = SPLINE(X,Y,XX) is the same as YY = PPVAL(SPLINE(X,Y),XX), thus

% providing, in YY, the values of the interpolant at XX. For information

% regarding the size of YY see PPVAL.

%

% Ordinarily, the not-a-knot end conditions are used. However, if Y contains

% two more values than X has entries, then the first and last value in Y are

% used as the endslopes for the cubic spline. If Y is a vector, this

% means:

% f(X) = Y(2:end-1), Df(min(X))=Y(1), Df(max(X))=Y(end).

% If Y is a matrix or N-D array with SIZE(Y,N) equal to LENGTH(X)+2, then

% f(X(j)) matches the value Y(:,...,:,j+1) for j=1:LENGTH(X), then

% Df(min(X)) matches Y(:,:,...:,1) and Df(max(X)) matches Y(:,:,...:,end).

%

% Example:

% This generates a sine-like spline curve and samples it over a finer mesh:

% x = 0:10; y = sin(x);

% xx = 0:.25:10;

% yy = spline(x,y,xx);

% plot(x,y,'o',xx,yy)

%

% Example:

% This illustrates the use of clamped or complete spline interpolation where

% end slopes are prescribed. In this example, zero slopes at the ends of an

% interpolant to the values of a certain distribution are enforced:

% x = -4:4; y = [0 .15 1.12 2.36 2.36 1.46 .49 .06 0];

% cs = spline(x,[0 y 0]);

% xx = linspace(-4,4,101);

% plot(x,y,'o',xx,ppval(cs,xx),'-');

%

% Class support for inputs x, y, xx:

% float: double, single

%

% See also INTERP1, PPVAL, UNMKPP, MKPP, SPLINES (The Spline Toolbox).

% Carl de Boor 7-2-86

% Copyright 1984-2004 The MathWorks, Inc.

% $Revision: 5.18.4.3 $ $Date: 2004/03/02 21:48:06 $

output=[];

% Check that data are acceptable and, if not, try to adjust them appropriately

[x,y,sizey,endslopes] = chckxy(x,y);

n = length(x); yd = prod(sizey);

% Generate the cubic spline interpolant in ppform

dd = ones(yd,1); dx = diff(x); divdif = diff(y,[],2)./dx(dd,:);

if n==2

if isempty(endslopes) % the interpolant is a straight line

pp=mkpp(x,[divdif y(:,1)],sizey);

else % the interpolant is the cubic Hermite polynomial

pp = pwch(x,y,endslopes,dx,divdif); pp.dim = sizey;

end

elseif n==3isempty(endslopes) % the interpolant is a parabola

y(:,2:3)=divdif;

y(:,3)=diff(divdif')'/(x(3)-x(1));

y(:,2)=y(:,2)-y(:,3)*dx(1);

pp = mkpp(x([1,3]),y(:,[3 2 1]),sizey);

else % set up the sparse, tridiagonal, linear system b = ?*c for the slopes

b=zeros(yd,n);

b(:,2:n-1)=3*(dx(dd,2:n-1).*divdif(:,1:n-2)+dx(dd,1:n-2).*divdif(:,2:n-1));

if isempty(endslopes)

x31=x(3)-x(1);xn=x(n)-x(n-2);

b(:,1)=((dx(1)+2*x31)*dx(2)*divdif(:,1)+dx(1)^2*divdif(:,2))/x31;

b(:,n)=...

(dx(n-1)^2*divdif(:,n-2)+(2*xn+dx(n-1))*dx(n-2)*divdif(:,n-1))/xn;

else

x31 = 0; xn = 0; b(:,[1 n]) = dx(dd,[2 n-2]).*endslopes;

end

dxt = dx(:);

c = spdiags([ [x31;dxt(1:n-2);0] ...

[dxt(2);2*[dxt(2:n-1)+dxt(1:n-2)];dxt(n-2)] ...

[0;dxt(2:n-1);xn] ],[-1 0 1],n,n);

% sparse linear equation solution for the slopes

mmdflag = spparms('autommd');

spparms('autommd',0);

s=b/c;

spparms('autommd',mmdflag);

% construct piecewise cubic Hermite interpolant

% to values and computed slopes

pp = pwch(x,y,s,dx,divdif); pp.dim = sizey;

end

if nargin==2, output = pp; else output = ppval(pp,xx); end

三次样条插值C语言

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。

(5)三次样条和B样条

首先什么是样条? 是 区间上的 个不同的点,当满足如下两个条件的时候, 就称为一个 次的样条函数

(1)

(2)

也就是说,在每个小区间上是 次多项式,并且整体是 阶连续的。注意这里对样条的定义并没有要求在每个点 上函数值相等,如果真的满足了 ,那么这就称为样条插值函数。

接下来有一个结论:

结论:对于 区间上由 个点所构建的所有的 次样条函数张成的函数空间 ,其维数 。

这也就是说,对于任意一个样条 ,它写成基函数的形式应该是

   

而实际上,根据基函数的选择不同,对应的样条当然也不同,其中比较有名的是三次样条和B样条。

我们考虑 这一情况,并且是插值样条,也就说 ,并且由于三次样条要求二阶连续,那么对于所有内部的节点 ,应该要求这些点处一阶导数和二阶导数应该左右相等。计算一下不难发现,这样还缺少两个条件,这里需要边界条件,根据边界条件的不同,插值样条也不同。如自然样条要求边界点的二阶导数为0。

构造三次样条插值函数的方法如下,可以从每个区间的二阶导数出发做一个线性插值,然后根据内部条件还有边界条件构造方程组,最后解一个三对角的行列式。

B样条的理论挺复杂的,在CAGD等领域是重点研究的方向。这里挖个坑,不写了。

总结:插值先告一段落了。之前写的都是一元函数的插值,实际上多元插值在研究领域中更重要,包括多元样条,有限元等。从下一篇开始,写数值积分。

三次样条插值 C++程序

#includeiostream.h

#includeiomanip.h

#includemath.h

void main()

{

float a[37],b[37];

cout" ""度数"" ""sin(x)值"" ""一阶导值"" ""二阶导值"endl;

for(int i=0;i37;i++)

coutsetw(11)setprecision(3)

10*i

setw(11)setprecision(3)

sin(i*31.4/180)

setw(11)setprecision(3)

cos(i*31.4/180)

setw(11)setprecision(3)

-sin(i*31.4/180)

endl;

cout"一个周期内的积分值:0"endl;

}


本文题目:三次样条函数c语言,c语言三次样条插值函数
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