三个矩阵相乘怎么算_两个矩阵的乘积怎么计算?两个矩阵需要符合什么条件才能有乘积( 二 ) _两个矩阵相乘怎么算

for
(j
=
0;
j
<
line_1
*
column_line
+
column_line
*
columns_2;
j++)
scanf("%d",
ptr_memory
+
j);
/*输入这两个矩阵*/
calculate
(line_1,
column_line,
columns_2,
ptr_memory);/*计算并输出结果*/
}
else/*未申请到打印*/
printf("Not
Enough
Memory!
");
free(ptr_memory);/*释放申请的内存*/
}
return
0;
}
/**功能:计算并输出结果
参数:第一个矩阵的行数，列数(第二个矩阵的行数)，第二个矩阵的列数，申请的内存的首地址
返回:无*/
void
calculate
(int
line_1,
int
column_line,
int
columns_2,
int
*ptr_memory)
{
int
i,
j,
k;
int
mid_result1,
mid_result2,
result;/*分别存储中间计算数据1，2，计算结果*/
for
(
i
=
0;
i
<
line_1;
i++)
{
for
(j
=
0;
j
<
line_1;
j++)
{
result
=
0;/*将结果初始化*/
for
(k
=
0;
k
<
column_line;
k++)
{
mid_result1
=
column_line
*
i
+
k;/*第一个乘数相对位置*/
mid_result2
=
line_1
*
column_line
+
columns_2
*
k
+
j;/*第二个乘数相对位置*/
result
+=
*(ptr_memory
+
mid_result1)
*
*(ptr_memory
+
mid_result2);
/**累计计算公式*/
}
if
(j
==
columns_2
-
1)/*如果到了换行的时候换行*/
printf("%d
",result);
else
printf("%d
",result);/*其他情况不换行*/
}
}
}
本题原题:
Input
The
first
line
of
input
is
the
number
of
all
the
test
cases
that
follow.
For
each
test
case,
there
are
three
positive
integer
m,
n,
p
(m,
n,
p
<=
100)
on
the
first
line,
meaning
that
A's
dimension
is
m
rows
by
n
columns
while
B's
is
n
rows
by
p
columns.
Then
comes
the
data
of
matrices
A
and
B.
Matrix
A
is
represented
by
m
lines
of
integers
ranging
in
[-1000,
1000],
with
each
line
containing
n
integers
separated
by
a
space.
A
line
is
corresponding
to
a
row
of
a
matrix.
The
data
format
for
matrix
B
is
much
the
same
except
for
its
potentially
different
dimensions.
Please
refer
to
the
section
Sample
Input.
Output
For
each
test
case,
print
the
product
of
A
multiplied
by
B
in
the
form
described
in
the
section
Input,
but
do
not
include
any
dimension
description.
Sample
Input
2
2
2
2
1
2
3
4
1
2
3
4
1
2
1
1
1
Sample
Output
7
10
15
22
0
Q4：线性代数中，两个矩阵相乘应该怎样计算