10个重要的算法C语言实现源代码(留着以后看)

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#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(); }

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#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(); }

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#include
<
stdio.h
>
#include
<
math.h
>

#define
N 20

int
main() {
int
n,i,j,k;
int
mi,tmp,mx;
float
a

,b
,x
; 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

=
A
[n
+
1
]
/
A

;
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();} } }

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#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
; } } }

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5
.牛顿迭代公式，求方程的近似解

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#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(); }

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#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

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