10个重要的算法C语言实现源代码

包括 拉格朗日，牛顿插值，高斯，龙贝格，牛顿迭代，牛顿-科特斯，雅克比，秦九昭，幂法，高斯塞德尔 。都是经典的数学算法，希望能开托您的思路。转自kunli.info 1.拉格朗日插值多

包括拉格朗日，牛顿插值，高斯，龙贝格，牛顿迭代，牛顿-科特斯，雅克比，秦九昭，幂法，高斯塞德尔 。都是经典的数学算法，希望能开托您的思路。转自kunli.info

1.拉格朗日插值多项式 ，用于离散数据的拟合

C/C++ code

#include
<
stdio.h
>

#include
<
conio.h
>

#include
<
alloc.h
>


float
lagrange(
float

*
x,
float

*
y,
float
xx,
int
n)
/*
拉格朗日插值算法
*/

{
int
i,j;

float

*
a,yy
=
0.0
;
/*
a作为临时变量，记录拉格朗日插值多项式
*/

a
=
(
float

*
)malloc(n
*
sizeof
(
float
));

for
(i
=
0
;i
<=
n
–
1
;i
++
)
{ a[i]
=
y[i];

for
(j
=
0
;j
<=
n
–
1
;j
++
)

if
(j
!=
i) a[i]
*=
(xx
–
x[j])
/
(x[i]
–
x[j]);
yy
+=
a[i];
}
free(a);

return
yy;
}
main()
{
int
i,n;

float
x[
20
],y[
20
],xx,yy;
printf(
“
Input n:
“
);
scanf(
“
%d
“
,
&
n);

if
(n
>=
20
) {printf(
“
Error!The value of n must in (0,20).
“
); getch();
return

1
;}

if
(n
<=
0
) {printf(
“
Error! The value of n must in (0,20).
“
); getch();
return

1
;}

for
(i
=
0
;i
<=
n
–
1
;i
++
)
{ printf(
“
x[%d]:
“
,i);
scanf(
“
%f
“
,
&
x[i]);
}
printf(
“
/n
“
);

for
(i
=
0
;i
<=
n
–
1
;i
++
)
{ printf(
“
y[%d]:
“
,i);scanf(
“
%f
“
,
&
y[i]);}
printf(
“
/n
“
);
printf(
“
Input xx:
“
);
scanf(
“
%f
“
,
&
xx);
yy
=
lagrange(x,y,xx,n);
printf(
“
x=%f,y=%f/n
“
,xx,yy);
getch();
}

2.牛顿插值多项式，用于离散数据的拟合

C/C++ code

#include
<
stdio.h
>

#include
<
conio.h
>

#include
<
alloc.h
>


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
;i
<
k;i
++
)
f[i
+
1
]
=
(f[i]
–
y[i])
/
(x[k]
–
x[i]);
y[k]
=
f[k];
}

return
;
}
main()
{
int
i,n;

float
x[
20
],y[
20
],xx,yy;
printf(
“
Input n:
“
);
scanf(
“
%d
“
,
&
n);

if
(n
>=
20
) {printf(
“
Error! The value of n must in (0,20).
“
); getch();
return

1
;}

if
(n
<=
0
) {printf(
“
Error! The value of n must in (0,20).
“
);getch();
return

1
;}

for
(i
=
0
;i
<=
n
–
1
;i
++
)
{ printf(
“
x[%d]:
“
,i);
scanf(
“
%f
“
,
&
x[i]);
}
printf(
“
/n
“
);

for
(i
=
0
;i
<=
n
–
1
;i
++
)
{ printf(
“
y[%d]:
“
,i);scanf(
“
%f
“
,
&
y[i]);}
printf(
“
/n
“
);
difference(x,(
float

*
)y,n);
printf(
“
Input xx:
“
);
scanf(
“
%f
“
,
&
xx);
yy
=
y[
20
];

for
(i
=
n
–
1
;i
>=
0
;i
—
) yy
=
yy
*
(xx
–
x[i])
+
y[i];
printf(
“
NewtonInter(%f)=%f
“
,xx,yy);
getch();
}

3.高斯列主元消去法,求解其次线性方程组

C/C++ code

#include
<
stdio.h
>

#include
<
math.h
>


#define
N 20


int
main()
{
int
n,i,j,k;

int
mi,tmp,mx;

float
a[N][N],b[N],x[N];
printf(
“
/nInput n:
“
);
scanf(
“
%d
“
,
&
n);

if
(n
>
N)
{ printf(
“
The input n should in(0,N)!/n
“
);
getch();

return

1
;
}

if
(n
<=
0
)
{ printf(
“
The input n should in(0,N)!/n
“
);
getch();

return

1
;
}
printf(
“
Now input a(i,j),i,j=0…%d:/n
“
,n
–
1
);

for
(i
=
0
;i
<
n;i
++
)
{
for
(j
=
0
;j
<
n;j
++
)
scanf(
“
%f
“
,
&
a[i][j]);}
printf(
“
Now input b(i),i,j=0…%d:/n
“
,n
–
1
);

for
(i
=
0
;i
<
n;i
++
)
scanf(
“
%f
“
,
&
b[i]);

for
(i
=
0
;i
<
n
–
2
;i
++
)
{
for
(j
=
i
+
1
,mi
=
i,mx
=
fabs(a[i][j]);j
<
n
–
1
;j
++
)

if
(fabs(a[j][i])
>
mx)
{ mi
=
j;
mx
=
fabs(a[j][i]);
}

if
(i
<
mi)
{ tmp
=
b[i];b[i]
=
b[mi];b[mi]
=
tmp;

for
(j
=
i;j
<
n;j
++
)
{ tmp
=
a[i][j];
a[i][j]
=
a[mi][j];
a[mi][j]
=
tmp;
}
}

for
(j
=
i
+
1
;j
<
n;j
++
)
{ tmp
=-
a[j][i]
/
a[i][i];
b[j]
+=
b[i]
*
tmp;

for
(k
=
i;k
<
n;k
++
)
a[j][k]
+=
a[i][k]
*
tmp;
}
}
x[n
–
1
]
=
b[n
–
1
]
/
a[n
–
1
][n
–
1
];

for
(i
=
n
–
2
;i
>=
0
;i
—
)
{ x[i]
=
b[i];

for
(j
=
i
+
1
;j
<
n;j
++
)
x[i]
-=
a[i][j]
*
x[j];
x[i]
/=
a[i][i];
}

for
(i
=
0
;i
<
n;i
++
)
printf(
“
Answer:/n x[%d]=%f/n
“
,i,x[i]);
getch();

return

0
;
}

#include
<
math.h
>

#include
<
stdio.h
>


#define
NUMBER 20


#define
Esc 0x1b


#define
Enter 0x0d

float
A[NUMBER][NUMBER
+
1
] ,ark;

int
flag,n;
exchange(
int
r,
int
k);

float
max(
int
k);
message();

main()
{

float
x[NUMBER];

int
r,k,i,j;

char
celect;
clrscr();

printf(
“
/n/nUse Gauss.
“
);
printf(
“
/n/n1.Jie please press Enter.
“
);
printf(
“
/n/n2.Exit press Esc.
“
);
celect
=
getch();

if
(celect
==
Esc)
exit(
0
);
printf(
“
/n/n input n=
“
);
scanf(
“
%d
“
,
&
n);
printf(
“
/n/nInput matrix A and B:
“
);

for
(i
=
1
;i
<=
n;i
++
)
{
printf(
“
/n/nInput a%d1–a%d%d and b%d:
“
,i,i,n,i);

for
(j
=
1
;j
<=
n
+
1
;j
++
) scanf(
“
%f
“
,
&
A[i][j]);
}

for
(k
=
1
;k
<=
n
–
1
;k
++
)
{
ark
=
max(k);

if
(ark
==
0
)
{
printf(
“
/n/nIt’s wrong!
“
);message();
}

else

if
(flag
!=
k)
exchange(flag,k);

for
(i
=
k
+
1
;i
<=
n;i
++
)

for
(j
=
k
+
1
;j
<=
n
+
1
;j
++
)
A[i][j]
=
A[i][j]
–
A[k][j]
*
A[i][k]
/
A[k][k];
}
x[n]
=
A[n][n
+
1
]
/
A[n][n];

for
( k
=
n
–
1
;k
>=
1
;k
—
)
{

float
me
=
0
;

for
(j
=
k
+
1
;j
<=
n;j
++
)
{
me
=
me
+
A[k][j]
*
x[j];
}
x[k]
=
(A[k][n
+
1
]
–
me)
/
A[k][k];
}

for
(i
=
1
;i
<=
n;i
++
)
{
printf(
“
/n/nx%d=%f
“
,i,x[i]);
}
message();
}

exchange(
int
r,
int
k)
{

int
i;

for
(i
=
1
;i
<=
n
+
1
;i
++
)
A[
0
][i]
=
A[r][i];

for
(i
=
1
;i
<=
n
+
1
;i
++
)
A[r][i]
=
A[k][i];

for
(i
=
1
;i
<=
n
+
1
;i
++
)
A[k][i]
=
A[
0
][i];
}

float
max(
int
k)
{

int
i;

float
temp
=
0
;

for
(i
=
k;i
<=
n;i
++
)

if
(fabs(A[i][k])
>
temp)
{
temp
=
fabs(A[i][k]);
flag
=
i;
}

return
temp;
}

message()
{
printf(
“
/n/n Go on Enter ,Exit press Esc!
“
);

switch
(getch())
{

case
Enter: main();

case
Esc: exit(
0
);

default
:{printf(
“
/n/nInput error!
“
);message();}
}
}

4.龙贝格求积公式，求解定积分

C/C++ code

#include
<
stdio.h
>

#include
<
math.h
>


#define
f(x) (sin(x)/x)


#define
N 20


#define
MAX 20


#define
a 2


#define
b 4


#define
e 0.00001


float
LBG(
float
p,
float
q,
int
n)
{
int
i;

float
sum
=
0
,h
=
(q
–
p)
/
n;

for
(i
=
1
;i
<
n;i
++
)
sum
+=
f(p
+
i
*
h);
sum
+=
(f(p)
+
f(q))
/
2
;

return
(h
*
sum);
}

void
main()
{
int
i;

int
n
=
N,m
=
0
;

float
T[MAX
+
1
][
2
];
T[
0
][
1
]
=
LBG(a,b,n);
n
*=
2
;

for
(m
=
1
;m
<
MAX;m
++
)
{
for
(i
=
0
;i
<
m;i
++
)
T[i][
0
]
=
T[i][
1
];
T[
0
][
1
]
=
LBG(a,b,n);
n
*=
2
;

for
(i
=
1
;i
<=
m;i
++
)
T[i][
1
]
=
T[i
–
1
][
1
]
+
(T[i
–
1
][
1
]
–
T[i
–
1
][
0
])
/
(pow(
2
,
2
*
m)
–
1
);

if
((T[m
–
1
][
1
]
<
T[m][
1
]
+
e)
&&
(T[m
–
1
][
1
]
>
T[m][
1
]
–
e))
{ printf(
“
Answer=%f/n
“
,T[m][
1
]); getch();

return
;
}
}
}

C/C++ code

5
.牛顿迭代公式，求方程的近似解

C/C++ code

#include
<
stdio.h
>

#include
<
math.h
>

#include
<
conio.h
>


#define
N 100


#define
PS 1e-5


#define
TA 1e-5


float
Newton(
float
(
*
f)(
float
),
float
(
*
f1)(
float
),
float
x0 )
{
float
x1,d
=
0
;

int
k
=
0
;

do

{ x1
=
x0
–
f(x0)
/
f1(x0);

if
((k
++>
N)
||
(fabs(f1(x1))
<
PS))
{ printf(
“
/nFailed!
“
);
getch();
exit();
}
d
=
(fabs(x1)
<
1
?
x1
–
x0:(x1
–
x0)
/
x1);
x0
=
x1;
printf(
“
x(%d)=%f/n
“
,k,x0);
}

while
((fabs(d))
>
PS
&&
fabs(f(x1))
>
TA) ;

return
x1;
}

float
f(
float
x)
{
return
x
*
x
*
x
+
x
*
x
–
3
*
x
–
3
; }

float
f1(
float
x)
{
return

3.0
*
x
*
x
+
2
*
x
–
3
; }

void
main()
{
float
f(
float
);

float
f1(
float
);

float
x0,y0;
printf(
“
Input x0:
“
);
scanf(
“
%f
“
,
&
x0);
printf(
“
x(0)=%f/n
“
,x0);
y0
=
Newton(f,f1,x0);
printf(
“
/nThe root is x=%f/n
“
,y0);
getch();
}

6. 牛顿-科特斯求积公式，求定积分

C/C++ code

#include
<
stdio.h
>

#include
<
math.h
>


int
NC(a,h,n,r,f)

float
(
*
a)[];

float
h;

int
n,f;

float

*
r;
{
int
nn,i;

float
ds;

if
(n
>
1000
||
n
<
2
)
{
if
(f)
printf(
“
/n Faild! Check if 1<n<1000!/n
“
,n);

return
(
–
1
);
}

if
(n
==
2
)
{
*
r
=
0.5
*
((
*
a)[
0
]
+
(
*
a)[
1
])
*
(h);

return
(
0
);
}

if
(n
–
4
==
0
)
{
*
r
=
0
;

*
r
=*
r
+
0.375
*
(h)
*
((
*
a)[n
–
4
]
+
3
*
(
*
a)[n
–
3
]
+
3
*
(
*
a)[n
–
2
]
+
(
*
a)[n
–
1
]);

return
(
0
);
}

if
(n
/
2
–
(n
–
1
)
/
2
<=
0
)
nn
=
n;

else

nn
=
n
–
3
;
ds
=
(
*
a)[
0
]
–
(
*
a)[nn
–
1
];

for
(i
=
2
;i
<=
nn;i
=
i
+
2
)
ds
=
ds
+
4
*
(
*
a)[i
–
1
]
+
2
*
(
*
a)[i];

*
r
=
ds
*
(h)
/
3
;

if
(n
>
nn)

*
r
=*
r
+
0.375
*
(h)
*
((
*
a)[n
–
4
]
+
3
*
(
*
a)[n
–
3
]
+
3
*
(
*
a)[n
–
2
]
+
(
*
a)[n
–
1
]);

return
(
0
);
}
main()
{

float
h,r;

int
n,ntf,f;

int
i;

float
a[
16
];
printf(
“
Input the x[i](16):/n
“
);

for
(i
=
0
;i
<=
15
;i
++
)
scanf(
“
%d
“
,
&
a[i]);
h
=
0.2
;
f
=
0
;
ntf
=
NC(a,h,n,
&
r,f);

if
(ntf
==
0
)
printf(
“
/nR=%f/n
“
,r);

else

printf(
“
/n Wrong!Return code=%d/n
“
,ntf);
getch();
}

7.雅克比迭代，求解方程近似解

C/C++ code

#include
<
stdio.h
>

#include
<
math.h
>


#define
N 20


#define
MAX 100


#define
e 0.00001


int
main()
{
int
n;

int
i,j,k;

float
t;

float
a[N][N],b[N][N],c[N],g[N],x[N],h[N];
printf(
“
/nInput dim of n:
“
); scanf(
“
%d
“
,
&
n);

if
(n
>
N)
{ printf(
“
Faild! Check if 0<n<N!/n
“
); getch();
return

1
; }

if
(n
<=
0
)
{printf(
“
Faild! Check if 0<n<N!/n
“
); getch();
return

1
;}
printf(
“
Input a[i,j],i,j=0…%d:/n
“
,n
–
1
);

for
(i
=
0
;i
<
n;i
++
)

for
(j
=
0
;j
<
n;j
++
)
scanf(
“
%f
“
,
&
a[i][j]);
printf(
“
Input c[i],i=0…%d:/n
“
,n
–
1
);

for
(i
=
0
;i
<
n;i
++
)
scanf(
“
%f
“
,
&
c[i]);

for
(i
=
0
;i
<
n;i
++
)

for
(j
=
0
;j
<
n;j
++
)
{ b[i][j]
=-
a[i][j]
/
a[i][i]; g[i]
=
c[i]
/
a[i][i]; }

for
(i
=
0
;i
<
MAX;i
++
)
{
for
(j
=
0
;j
<
n;j
++
)
h[j]
=
g[j];
{
for
(k
=
0
;k
<
n;k
++
)
{
if
(j
==
k)
continue
; h[j]
+=
b[j][k]
*
x[k]; }
}
t
=
0
;

for
(j
=
0
;j
<
n;j
++
)

if
(t
<
fabs(h[j]
–
x[j])) t
=
fabs(h[j]
–
x[j]);

for
(j
=
0
;j
<
n;j
++
)
x[j]
=
h[j];

if
(t
<
e)
{ printf(
“
x_i=/n
“
);

for
(i
=
0
;i
<
n;i
++
)
printf(
“
x[%d]=%f/n
“
,i,x[i]);
getch();

return

0
;
}
printf(
“
after %d repeat , return/n
“
,MAX);
getch();

return

1
;
}
getch();
}

8.秦九昭算法

C/C++ code

#include
<
math.h
>


float
qin(
float
a[],
int
n,
float
x)
{
float
r
=
0
;

int
i;

for
(i
=
n;i
>=
0
;i
—
)
r
=
r
*
x
+
a[i];

return
r;
}
main()
{
float
a[
50
],x,r
=
0
;

int
n,i;

do

{ printf(
“
Input frequency:
“
);
scanf(
“
%d
“
,
&
n);
}

while
(n
<
1
);
printf(
“
Input value:
“
);

for
(i
=
0
;i
<=
n;i
++
)
scanf(
“
%f
“
,
&
a[i]);
printf(
“
Input frequency:
“
);
scanf(
“
%f
“
,
&
x);
r
=
qin(a,n,x);
printf(
“
Answer:%f
“
,r);
getch();
}

9.幂法

C/C++ code

#include
<
stdio.h
>

#include
<
math.h
>


#define
N 100


#define
e 0.00001


#define
n 3


float
x[n]
=
{
0
,
0
,
1
};

float
a[n][n]
=
{{
2
,
3
,
2
},{
10
,
3
,
4
},{
3
,
6
,
1
}};

float
y[n];
main()
{
int
i,j,k;

float
xm,oxm;
oxm
=
0
;

for
(k
=
0
;k
<
N;k
++
)
{
for
(j
=
0
;j
<
n;j
++
)
{ y[j]
=
0
;

for
(i
=
0
;i
<
n;i
++
)
y[j]
+=
a[j][i]
*
x[i];
}
xm
=
0
;

for
(j
=
0
;j
<
n;j
++
)

if
(fabs(y[j])
>
xm) xm
=
fabs(y[j]);

for
(j
=
0
;j
<
n;j
++
)
y[j]
/=
xm;

for
(j
=
0
;j
<
n;j
++
)
x[j]
=
y[j];

if
(fabs(xm
–
oxm)
<
e)
{ printf(
“
max:%f/n/n
“
,xm);
printf(
“
v[i]:/n
“
);

for
(k
=
0
;k
<
n;k
++
) printf(
“
%f/n
“
,y[k]);

break
;
}
oxm
=
xm;
}
getch();
}

10.高斯塞德尔

C/C++ code

#include
<
math.h
>

#include
<
stdio.h
>


#define
N 20


#define
M 99


float
a[N][N];

float
b[N];

int
main()
{
int
i,j,k,n;

float
sum,no,d,s,x[N];
printf(
“
/nInput dim of n:
“
);
scanf(
“
%d
“
,
&
n);

if
(n
>
N)
{ printf(
“
Faild! Check if 0<n<N!/n
“
); getch();

return

1
;
}

if
(n
<=
0
)
{ printf(
“
Faild! Check if 0<n<N!/n
“
);getch();
return

1
;}
printf(
“
Input a[i,j],i,j=0…%d:/n
“
,n
–
1
);

for
(i
=
0
;i
<
n;i
++
)

for
(j
=
0
;j
<
n;j
++
)
scanf(
“
%f
“
,
&
a[i][j]);
printf(
“
Input b[i],i=0…%d:/n
“
,n
–
1
);

for
(i
=
0
;i
<
n;i
++
) scanf(
“
%f
“
,
&
b[i]);

for
(i
=
0
;i
<
n;i
++
) x[i]
=
0
;
k
=
0
;
printf(
“
/nk=%dx=
“
,k);

for
(i
=
0
;i
<
n;i
++
) printf(
“
%12.8f
“
,x[i]);

do

{ k
++
;

if
(k
>
M){printf(
“
/nError!/n”);getch();}



break
;
}
no
=
0.0
;

for
(i
=
0
;i
<
n;i
++
)
{ s
=
x[i];
sum
=
0.0
;

for
(j
=
0
;j
<
n;j
++
)

if
(j
!=
i) sum
=
sum
+
a[i][j]
*
x[j];
x[i]
=
(b[i]
–
sum)
/
a[i][i];
d
=
fabs(x[i]
–
s);

if
(no
<
d) no
=
d;
}
printf(
“
/nk=%2dx=
“
,k);

for
(i
=
0
;i
<
n;i
++
) printf(
“
%f
“
,x[i]);
}

while
(no
>=
0.1e-6
);

if
(no
<
0.1e-6
)
{ printf(
“
/n/n answer=/n
“
);
printf(
“
/nk=%d
“
,k);

for
(i
=
0
;i
<
n;i
++
)
printf(
“
/n x[%d]=%12.8f
“
,i,x[i]);
}
getch();
}