c語言a反

A. 如何用C語言表示：輸入任意二維數組A，求A的逆

下面的代碼是我從fengjian_net 的精彩回答中抄來的，源自《C常用演算法程序集－徐士良》。只是他的代碼在VC2010中要編譯通過，需要修改好幾個地方，增加強制轉換，修改好的代碼如下：

#include <stdlib.h>
#include <math.h>
#include <stdio.h>
int brinv(double a[], int n)
{
int *is,*js,i,j,k,l,u,v;
double d,p;
is=(int*)malloc(n*sizeof(int));
js=(int*)malloc(n*sizeof(int));
for (k=0; k<=n-1; k++)
{
d=0.0;
for (i=k; i<=n-1; i++)
for (j=k; j<=n-1; j++)
{
l=i*n+j; p=fabs(a[l]);
if (p>d) { d=p; is[k]=i; js[k]=j;}
}
if (d+1.0==1.0)
{
free(is); free(js); printf("err**not inv\n");
return(0);
}
if (is[k]!=k)
for (j=0; j<=n-1; j++)
{
u=k*n+j; v=is[k]*n+j;
p=a[u]; a[u]=a[v]; a[v]=p;
}
if (js[k]!=k)
for (i=0; i<=n-1; i++)
{
u=i*n+k; v=i*n+js[k];
p=a[u]; a[u]=a[v]; a[v]=p;
}
l=k*n+k;
a[l]=1.0/a[l];
for (j=0; j<=n-1; j++)
if (j!=k)
{ u=k*n+j; a[u]=a[u]*a[l];}
for (i=0; i<=n-1; i++)
if (i!=k)
for (j=0; j<=n-1; j++)
if (j!=k)
{
u=i*n+j;
a[u]=a[u]-a[i*n+k]*a[k*n+j];
}
for (i=0; i<=n-1; i++)
if (i!=k)
{ u=i*n+k; a[u]=-a[u]*a[l];}
}
for (k=n-1; k>=0; k--)
{
if (js[k]!=k)
for (j=0; j<=n-1; j++)
{
u=k*n+j; v=js[k]*n+j;
p=a[u]; a[u]=a[v]; a[v]=p;
}
if (is[k]!=k)
for (i=0; i<=n-1; i++)
{
u=i*n+k; v=i*n+is[k];
p=a[u]; a[u]=a[v]; a[v]=p;
}
}
free(is); free(js);
return(1);
}
void brmul(double a[], double b[],int m,int n,int k,double c[])
{
int i,j,l,u;
for (i=0; i<=m-1; i++)
for (j=0; j<=k-1; j++)
{
u=i*k+j; c[u]=0.0;
for (l=0; l<=n-1; l++)
c[u]=c[u]+a[i*n+l]*b[l*k+j];
}
return;
}
int main()
{
int i,j;
static double a[4][4]={ {0.2368,0.2471,0.2568,1.2671},
{1.1161,0.1254,0.1397,0.1490},
{0.1582,1.1675,0.1768,0.1871},
{0.1968,0.2071,1.2168,0.2271}};
static double b[4][4],c[4][4];
for (i=0; i<=3; i++)
for (j=0; j<=3; j++)
b[i][j]=a[i][j];
i=brinv((double*)a,4);
if (i!=0)
{
printf("MAT A IS:\n");
for (i=0; i<=3; i++)
{
for (j=0; j<=3; j++)
printf("%13.7e ",b[i][j]);
printf("\n");
}
printf("\n");
printf("MAT A- IS:\n");
for (i=0; i<=3; i++)
{
for (j=0; j<=3; j++)
printf("%13.7e ",a[i][j]);
printf("\n");
}
printf("\n");
printf("MAT AA- IS:\n");
brmul((double*)b,(double*)a,4,4,4,(double*)c);
for (i=0; i<=3; i++)
{
for (j=0; j<=3; j++)
printf("%13.7e ",c[i][j]);
printf("\n");
}
}
}