c语言程序设计，设计两个矩阵相乘的程序。已知：

#include<stdio.h>

#include<stdlib.h>

#define M 3

 

int main(void)

{

    int i,j,k,matrix1[M][M],matrix2[M][M],row1=M ,col1=M ,row2=M,col2=M,matrix[M][M]   

    /*为需要相乘的两个矩阵赋值：*/ 

    printf("输入第一个矩阵：\n")

    for(i=0i<row1i++){

        for(j=0j<col1j++){

            scanf("%d",&matrix1[i][j]) 

        } 

    } 

    printf("输入第二个矩阵：\n")

    for(i=0i<row2i++){

        for(j=0j<col2j++){

            scanf("%d",&matrix2[i][j]) 

        } 

    }

    /*初始化matrix：*/

    for(i=0i<row1i++){

        for(j=0j<col2j++){

            matrix[i][j]=0 

        } 

    } 

     

    printf("The result:\n") 

    for(i=0i<row1i++){

        for(j=0j<col2j++){

            for(k=0k<col1k++){

                matrix[i][j]=matrix[i][j]+matrix1[i][k]*matrix2[k][j] 

            } 

        } 

    }

 

    for(i=0i<row1i++){

        for(j=0j<col2j++){

            printf("%d ",matrix[i][j]) 

        } 

        printf("\n") 

    } 

    return 0

}

应该是你要的吧。

计算两个矩阵乘积。(附本题原题)

/**计算两个矩阵的乘积

hold

by

cljnnn-hit

date

2006-12-2

13*/

#include

<stdio.h>

#include

<stdlib.h>

void

calculate

(int

line_1,

int

column_line,

int

columns_2,

int

*ptr_memory)/*计算并输出结果*/

int

main()

{

int

*ptr_memory

/*申请内存用，存储此内存的首地址*/

int

times,

line_1,

column_line,

columns_2/*分别存储计算次数，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*/

int

i,

j

scanf("%d",&

times)/*输入计算次数*/

for

(i

=

1

i

<=

times

i++)/*计算times次*/

{

scanf("%d%d%d",

&line_1,

&column_line,

&columns_2)/*输入要计算的两个矩阵的规格*/

ptr_memory

=

(int

*)

malloc((line_1

*

column_line

+

column_line

*

columns_2)

*

sizeof(int))/*申请适当的内存*/

if

(ptr_memory)

/*申请到就进行计算*/

{

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!\n")

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\n",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

0

0

1

Sample

Output

7

10

15

22

0

---