Problem:
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Matrix Multiplication

Write a program which reads a $n \times m$ matrix $A$ and a $m \times l$ matrix $B$, and prints their product, a $n \times l$ matrix $C$. An element of matrix $C$ is obtained by the following formula:

\[
c_{ij} = \sum_{k=1}^m a_{ik}b_{kj}
\]

where $a_{ij}$, $b_{ij}$ and $c_{ij}$ are elements of $A$, $B$ and $C$ respectively.

Input

In the first line, three integers $n$, $m$ and $l$ are given separated by space characters

In the following lines, the $n \times m$ matrix $A$ and the $m \times l$ matrix $B$ are given.

Output

Print elements of the $n \times l$ matrix $C$ ($c_{ij}$). Print a single space character between adjacent elements.

Constraints

$1 \leq n, m, l \leq 100$

$0 \leq a_{ij}, b_{ij} \leq 10000$

Sample Input

3 2 3
1 2
0 3
4 5
1 2 1
0 3 2

Sample Output

1 8 5
0 9 6
4 23 14

Note

      解説

Solution in C (judge verdict: Time Limit Exceeded)

#include <stdio.h>
#include <string.h>

int main()
{
	int  n,m,l;
	char buff[128];
	char *wk_p;
	int  dt_a[100][100];
	int  dt_b[100][100];
	int  kekka[100][100];
    int  k,i,p,u,a,b,index_x,index_y;
	char *ptr;
	int  *dta_p,*dtb_p,*kekka_p;

	memset(kekka,-1,sizeof(kekka));
	gets(buff);
	sscanf(buff,"%d %d %d",&n,&m,&l);

	for(i=0 ; i<n ; i++) {
		gets(buff);
    	wk_p = strtok(buff," ");
		dta_p = &(dt_a[i][0]);
		for(k=0 ; k<m ; k++,dta_p++) {
			sscanf(wk_p,"%d",dta_p/*&(dt_a[i][k])*/);
			wk_p = strtok(NULL," ");
			if (wk_p == NULL) {
				break;
			}
		}

	}

	for(i=0 ; i<m ; i++) {
		gets(buff);
    	wk_p = strtok(buff," ");
		dtb_p = &(dt_b[i][0]);
		for(k=0 ; k<l ; k++,dtb_p++) {
			sscanf(wk_p,"%d",dtb_p/*&(dt_b[i][k])*/);
			wk_p = strtok(NULL," ");
			if (wk_p == NULL) {
				break;
			}
		}

	}

	for(index_y=0,p=0 ; p<n ; p++,index_y++) {
		for(;;) {
			index_x=0;
			kekka_p=&(kekka[index_y][index_x]);
			for (k=0 ; k<l ; k++,index_x++,kekka_p++) {
				dta_p = &(dt_a[p][0]);
				for (i=0 ; i<m ; i++,dta_p++) {
					if (*kekka_p == -1) {
						*kekka_p = 0;
					}
					a = *dta_p;
					b = dt_b[i][k];
					*kekka_p += b * a;
				}
			}
			kekka_p = &(kekka[index_y][0]);
			for (k=0 ;; k++,kekka_p++) {
				if (*kekka_p!=-1) {

					printf("%d",*kekka_p);
					if (*(kekka_p+1) != -1) {
						printf(" ");
					}
				}
				else {
					printf("\n"); break;
				}
			}
			break;
		}
	}

	return 0;
}