电子科大POJ "任意阶矩阵相乘"

　　　任意阶矩阵的乘法

Time Limit: 3000/1000MS (Java/Others)     Memory Limit: 65535/65535KB (Java/Others)

C-source:

 <span style="color:#333333;">#include<stdio.h>
 #include<malloc.h>
 int main(void)
 {
  int m,g,n;
  int i,j,k;

  int number;
  double **a=NULL,**b=NULL,**result=NULL;
  printf("Please input number:\n");
  scanf("%d",&number);

 while(number!=)
 {
     number--;

     printf("Please input m,g,n:\n");
     scanf("%d%d%d",&m,&g,&n);

     a=(double**)malloc(m*sizeof(double*));
     for(i=;i<m;i++)
         a[i]=(double*)malloc(g*sizeof(double));

     printf("Please input the first jz:\n");
     for(i=;i<m;i++)
         for(j=;j<g;j++)
         scanf("%lf",&a[i][j]);

     b=(double**)malloc(g*sizeof(double *));
     for(i=;i<g;i++)
         b[i]=(double*)malloc(n*sizeof(double));

     printf("Please input the second jz\n");
     for(i=;i<g;i++)
         for(j=;j<n;j++)
             scanf("%lf",&b[i][j]);

     result=(double**)malloc(m*sizeof(double*));
         for(i=;i<m;i++)
             result[i]=(double*)malloc(n*sizeof(double));

     for(i=;i<m;i++)
         for(j=;j<n;j++)
             result[i][j]=;

     for(i=;i<m;i++)
         for(j=;j<n;j++)
             for(k=;k<g;k++)
                 result[i][j]+=a[i][k]*b[k][j];

     printf("answer:\n");
     for(i=;i<m;i++)
     {
         for(j=;j<n;j++)
             printf("%5g",result[i][j]);
             printf("\n");
     }
 }
     for(i=;i<m;i++)
     {
         free(a[i]);
         a[i]=NULL;
         free(result[i]);
         result[i]=NULL;
     }
     free(a);
     a=NULL;
     free(result);
     result=NULL;

     for(i=;i<g;i++)
     {   free(b[i]);
         b[i]=NULL;
     }
     free(b);
     b=NULL;

     return ;
 }