#includestdio.h
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#includestdlib.h
#includeiostream.h
typedef struct data
{
float x;
float y;
}Data;//變量x和函數(shù)值y的結(jié)構(gòu)
Data d[20];//最多二十組數(shù)據(jù)
float f(int s,int t)//牛頓插值法，用以返回插商
{
if(t==s+1)
return (d[t].y-d[s].y)/(d[t].x-d[s].x);
else
return (f(s+1,t)-f(s,t-1))/(d[t].x-d[s].x);
}
float Newton(float x,int count)
{
int n;
while(1)
{
cout"請輸入n值(即n次插值):";//獲得插值次數(shù)
cinn;
if(n=count-1)// 插值次數(shù)不得大于count－1次
break;
else
system("cls");
}
//初始化t，y，yt。
float t=1.0;
float y=d[0].y;
float yt=0.0;
//計算y值
for(int j=1;j=n;j++)
{
t=(x-d[j-1].x)*t;
yt=f(0,j)*t;
//coutf(0,j)endl;
y=y+yt;
}
return y;
}
float lagrange(float x,int count)
{
float y=0.0;
for(int k=0;kcount;k++)//這兒默認為count－1次插值
{
float p=1.0;//初始化p
for(int j=0;jcount;j++)
{//計算p的值
if(k==j)continue;//判斷是否為同一個數(shù)
p=p*(x-d[j].x)/(d[k].x-d[j].x);
}
y=y+p*d[k].y;//求和
}
return y;//返回y的值
}
void main()
{
float x,y;
int count;
while(1)
{
cout"請輸入x[i],y[i]的組數(shù)，不得超過20組:";//要求用戶輸入數(shù)據(jù)組數(shù)
cincount;
if(count=20)
break;//檢查輸入的是否合法
system("cls");
}
//獲得各組數(shù)據(jù)
for(int i=0;icount;i++)
{
cout"請輸入第"i+1"組x的值:";
cind[i].x;
cout"請輸入第"i+1"組y的值:";
cind[i].y;
system("cls");
}
cout"請輸入x的值:";//獲得變量x的值
cinx;
while(1)
{
int choice=3;
cout"請您選擇使用哪種插值法計算："endl;
cout" (0):退出"endl;
cout" (1):Lagrange"endl;
cout" (2):Newton"endl;
cout"輸入你的選擇：";
cinchoice;//取得用戶的選擇項
if(choice==2)
{
cout"你選擇了牛頓插值計算方法，其結(jié)果為：";
y=Newton(x,count);break;//調(diào)用相應(yīng)的處理函數(shù)
}
if(choice==1)
{
cout"你選擇了拉格朗日插值計算方法，其結(jié)果為：";
y=lagrange(x,count);break;//調(diào)用相應(yīng)的處理函數(shù)
}
if(choice==0)
break;
system("cls");
cout"輸入錯誤!!!!"endl;
}
coutx" , "yendl;//輸出最終結(jié)果
}
void
SPL(int
n,
double
*x,
double
*y,
int
ni,
double
*xi,
double
*yi)；
是你所要。
已知
n
個點
x,y;
x
必須已按順序排好。要插值
ni
點，橫坐標
xi[],
輸出
yi[]。
程序里用double
型，保證計算精度。
SPL調(diào)用現(xiàn)成的程序。
現(xiàn)成的程序很多。端點處理方法不同，結(jié)果會有不同。想同matlab比較，你需
嘗試
調(diào)用
spline（）函數(shù)
時，令
end1
為
1,
設(shè)
slope1
的值，令
end2
為
1
設(shè)
slope2
的值。
#include
stdio.h
#include
math.h
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;
}
程序代碼如下。
希望能幫助到你！
牛頓插值法
#includestdio.h
#includemath.h
#define
n
4
void
difference(float
*x,float
*y,int
n)
{
float
*f;
int
k,i;
f=(float
*)malloc(n*sizeof(float));
for(k=1;k=n;k
)
{
f[0]=y[k];
for(i=0;ik;i
)
f[i
1]=(f[i]-y[i])/(x[k]-x[i]);
y[k]=f[k];
}
return;
}
main()
{
int
i;
float
varx=0.895,b;
float
x[n
1]={0.4,0.55,0.65,0.8,0.9};
float
y[n
1]={0.41075,0.57815,0.69675,0.88811,1.02652};
difference(x,(float
*
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