Matrix Multiplication

Problem H: Matrix Multiplication

Time Limit: 1 Sec  Memory Limit: 128 MB
Submit: 84  Solved: 33

Description

Johnny and John are good friends. Johnny is going to take the entrance exams for postgraduate schools. 

Recently, he is reviewing Linear Algebra. Johnny always says that he is very skillful at matrix 

multiplication. As we all know that, if we know   matrix a (m rows and n columns) and b (n rows and t 

columns), the result matrix c (m rows and t columns) can be calculated by expression

c (i, j) = a(i,k)*b(k,j),1<=k<=n.(1<= i <= m, 1 <= j <= t).

But John wants to make things difficult. He wants Johnny to calculate matrix c using expression

c (i, j) = a(i,k)*b(k,j) (1<= i <= m, 1 <= j <= t, i + k is odd and k +  j is even). 

 We consider that 2 is odd and even. At the beginning, c (i, j) =0. 

Input

The first line of input is the number T of test case. 

The first line of each test case contains three integers, m, n, t, indicates the size of the matrix a and b. We 

always assume that the rows of matrix a equals to matrix b's columns. 

The next m lines describe the matrix a. Each line contains n integers. 

Then n lines followed, each line contains t integers, describe the matrix b. 

1<= T <= 10, 1 <= m, n, t <= 100, 0 <= a (i, j) <= 100, 0 <= b (i, j) <= 100. 

Output

For each test case output the matrix c. The numbers are separated by spaces. There is no space at the end of 

each line. 

Sample Input

1 

2 2 2 

1 2 

2 3 

2 3 

3 0

Sample Output

2 0 

4 0

简单题，矩阵相乘运用。

代码：

#include<iostream>#include<stdio.h>#include<cstring>using namespace std;#define MAX 100int a[MAX+5][MAX+5];int b[MAX+5][MAX+5];int c[MAX+5][MAX+5];int main(){    int aa,m,n,t,i,j,k;    scanf("%d",&aa);    while(aa--)    {        scanf("%d%d%d",&m,&n,&t);        memset(a,0,sizeof(a));        memset(b,0,sizeof(b));        memset(c,0,sizeof(c));        for(i=1;i<=m;i++)            for(j=1;j<=n;j++)                scanf("%d",&a[i][j]);        for(i=1;i<=n;i++)            for(j=1;j<=t;j++)                scanf("%d",&b[i][j]);        for(i=1;i<=m;i++)            for(j=1;j<=t;j++)                for(k=1;k<=n;k++)                    if((i+k==2||(i+k)%2==1)&&(j+k==2||(j+k)%2==0))                        c[i][j]+=a[i][k]*b[k][j];        for(i=1;i<=m;i++)        {            for(j=1;j<=t;j++)            {                if(j==1)                    printf("%d",c[i][j]);                else                    printf(" %d",c[i][j]);            }            printf("\n");        }    }    return 0;}

FROM：暑假训练第三场