#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=malloc(n*sizeof(int));
js=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(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(b,a,4,4,4,c);
for (i=0; i<=3; i++)
{ for (j=0; j<=3; j++)
printf("%13.7e ",c[i][j]);
printf("\n");
}
}
}
----------------解决方案--------------------------------------------------------
最怕这样的程序了。。。一个注释没有。。。一行缩进没有。。。又长。。
----------------解决方案--------------------------------------------------------
格式化后看的莫名其妙，编译后出来的打印也看不懂.....
说一说这个是要做什么的吧
#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=malloc(n*sizeof(int));
js=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(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(b,a,4,4,4,c);
for(i=0;i<=3;i++)
{
for(j=0;j<=3;j++)
printf("%13.7e ",c[i][j]);
printf("\n");
}
}
}
----------------解决方案--------------------------------------------------------
MAT A IS --输入矩阵
MAT A- IS --输出逆矩阵
MAT AA- IS ---输出伴随矩阵
----------------解决方案--------------------------------------------------------
请大家帮我再仔细看看好么，有些关键的，请你们帮我注释下，这是求逆矩阵的
----------------解决方案--------------------------------------------------------
晕 我也怕！~~~~~~~~~~~~~~~~~~~~`
----------------解决方案--------------------------------------------------------
格式很乱
----------------解决方案--------------------------------------------------------
我连看都看不懂。嘿鄙
----------------解决方案--------------------------------------------------------
逐行分析程序各个函数的各个算法，较长，最近期末考试，遂无闲时注释,仅提示至此．
----------------解决方案--------------------------------------------------------
跟踪调试....
----------------解决方案--------------------------------------------------------