我为矩阵乘法编写了 ac 程序。现在我想将行数分成 5 个 N/5 的块。函数应在第一次迭代中计算第一行块，在第二次迭代中计算第二部分，依此类推。如何在每次迭代中指定起始行和结束行？我在下面试过了。任何人都可以帮我纠正它。
#include <stdio.h>
#include <stdlib.h>
int main()
{
int i, m, n, p, q, c, d, k,g, sum = 0,start=0,end=m/5;
int **first, **second, **multiply;
printf("Enter the number of rows and columns of first matrix\n");
scanf("%d%d", &m, &n);
printf("Value entered %d%d \n",m,n);
first = malloc(m*sizeof(int*));
for ( i =0;i <m; i++)
first[i] =malloc(n*sizeof(int));
printf("Enter the number of rows and columns of second matrix \n");
scanf("%d%d", &p,&q);
printf("value entered %d%d \n",p,q);
second = malloc(p*sizeof(int*));
for( i=0;i<p;i++)
second[i] = malloc(q*sizeof(int));
multiply = malloc(m*sizeof(int*));
for ( i=0;i<m;i++)
multiply[i] = malloc(q*sizeof(int));
printf("Enter the elements of first matrix\n");
for( c = 0 ; c < m ; c++ )
for ( d = 0 ; d < n ; d++ )
scanf("%d", &first[c][d]);
if ( n != p )
printf("Matrices with entered orders can't be multiplied with each other.\n");
else {
printf("Enter the elements of second matrix\n");
for ( c = 0 ; c < p ; c++ ){
for ( d = 0 ; d < q ; d++ )
scanf("%d", &second[c][d]);
}
for(g=0;g<5;g++){
for ( c = start ; c < end ; c++ ) {
for ( d = 0 ; d < q ; d++ ) {
for ( k = 0 ; k < p ; k++ ) {
sum = sum + first[c][k]*second[k][d];
}
multiply[c][d] = sum;
sum = 0;
}
}
start=start+m/5+1;
end=end+m/5;
}
printf("Product of entered matrices:-\n");
for ( c = 0 ; c < m ; c++ ) {
for ( d = 0 ; d < q ; d++ )
printf("%d\t", multiply[c][d]);
printf("\n");
}
for ( i = 0; i < p; i++)
free(second[i]);
free(second);
for ( i = 0; i < m; i++)
free(multiply[i]);
free(multiply);
}
for ( i = 0; i < m; i++)
free(first[i]);
free(first);
return 0;
}