雅可比迭代法和高斯赛德尔迭代法

#include
<
iostream
>

#include
"
math.h
"

using

namespace
std;

#define
n 3

double
a

=
{{
5
,
2
,
1
},{
-
1
,
4
,
2
},{
2
,
-
3
,
10
}}, x

=
{
-
3
,
2
,
1
}, b

=
{
-
12
,
20
,
3
}, c
;

int
i
=
0
,j,k
=
0
,counter
=
15
;

int
main()
{

int
k
=
0
,i
=
0
;

double
num1,E;

float
y
;

do

{
E
=
0
;

for
(j
=
0
;j
<
n;j
++
)
{
c[j]
=
x[j];
}
num1
=
0
;

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

if
(j
!=
i)
num1
=
num1
+
a[i][j]
*
x[j];
}

for
(j
=
0
;j
<
n;j
++
)
{
c[j]
=
x[j];
}

y[i]
=
(b[i]
-
num1)
/
a[i][i];

if
(i
<
n)
{ i
++
; }

else

{

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

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

if
(E
<
(
float
)fabs(c[j]
-
y[j])) E
=
(
float
)fabs(c[j]
-
y[j]);
}

if
(k
>=
7
)
{ cout
<<
"
Iterations is
"
<<
k
+
1
<<
"
,
"
<<
"
Iterative precision is
"
;
printf(
"
%.6lf\n
"
,E);

for
(j
=
0
;j
<
n;j
++
)
printf(
"
x[%d]=%.5lf\t
"
,j,x[j]);
cout
<<
endl
<<
endl;
}
i
=
0
;
k
++
;

if
(E
<
0.001
)
break
;
}
}
while
(k
<=
counter);
cout
<<
"
Iterations is
"
<<
k
<<
endl;
cout
<<
"
Iterative terminating condition
"
;
printf(
"
x[%d]-x[%d]oo<%.6lf\n
"
,k,k
-
1
,E);
cout
<<
"
\t\t---------Solution-----------
"
<<
endl;

for
(i
=
0
;i
<
n;i
++
)
{
printf(
"
x[%d]=%.5lf\n
"
,i,x[i]);
}
}

#include
<
iostream
>

#include
<
math.h
>

using

namespace
std;

int
main()

{

double
a[
3
][
3
]
=
{{
1
,
2
,
-
2
},{
3
,
1
,
-
1
},{
2
,
4
,
1
}},b[
3
]
=
{
10
,
12
,
-
20
};

double
x[
3
]
=
{
0
,
0
,
0
},sum1,sum2;

int
i,j,k,n
=
3
;

for
(k
=
0
;k
<
5
;k
++
)

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

{ sum1
=
0
;sum2
=
0
;

for
(j
=
0
;j
<
i
-
1
;j
++
)

{ sum1
=
sum1
+
a[i][j]
*
x[j];

}

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

{sum2
=
sum2
+
a[i][j]
*
x[j];}

x[i]
=
(b[i]
-
sum1
-
sum2)
/
a[i][i];

}

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

{ printf(
"
x%d=%-15f
"
,i
+
1
,x[i]);}

printf(
"
\n
"
);

}

}