[第三次训练]Matrix Multiplication

Matrix Multiplication

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

HINT

解题报告

此题是要求输入m,n,t三个数，分别构成m*n与n*t 两个矩阵，然后利用公式的矩阵乘法方式得到新的矩阵m*t，并把它输出。并且规定i+k要为奇数，j+k要为偶数。2即为奇数页为偶数。所以只需判断((i+k)%2==1||i+k==2)&&(j+k)%2==0即可，满足的就加入相应的c(i,j)，然后输出即可（注意输出要按照要求，不要多空格或者换行符，本人因为条件写错，每行多了一个空格，导致PE2次），代码如下：

#include<stdio.h>#include<string.h>long long a[110][110],b[110][110],c[110][110];int main(){    int m,n,t,T,k,i,j;    scanf("%d",&T);    while(T--)    {        memset(c,0,sizeof(c));        scanf("%d%d%d",&m,&n,&t);        for(i=1;i<=m;i++)            for(j=1;j<=n;j++)                scanf("%lld",&a[i][j]);         for(i=1;i<=n;i++)            for(j=1;j<=t;j++)                scanf("%lld",&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==1||i+k==2)&&((j+k)%2==0||j+k==2))                        c[i][j]+=a[i][k]*b[k][j];                }        for(i=1;i<=m;i++)        {            for(j=1;j<=t;j++)            {                printf("%lld",c[i][j]);                if(j!=t)                    printf(" ");                else                    printf("\n");            }        }    }    return 0;}