c语言编写一个子函数求矩阵的逆矩阵

Question

子函数如下

void qiuni(int a[10][10])
{
”补充的部分“
}

Answer 1

#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");
}
}
}

Answer 2

#include<stdio.h>
#define N 10
int getA(int arcs[N][N],int n)//按第一行展开计算|A|
{
if(n==1)
{
return arcs[0][0];
}
int ans = 0;
int temp[N][N];
int i,j,k;
for(i=0;i<n;i++)
{
for(j=0;j<n-1;j++)
{
for(k=0;k<n-1;k++)
{
temp[j][k] = arcs[j+1][(k>=i)?k+1:k];

}
}
int t = getA(temp,n-1);
if(i%2==0)
{
ans += arcs[0][i]*t;
}
else
{
ans -= arcs[0][i]*t;
}
}
return ans;
}
void getAStart(int arcs[N][N],int n,int ans[N][N])//计算每一行每一列的每个元素所对应的余子式，组成A*
{
if(n==1)
{
ans[0][0] = 1;
return;
}
int i,j,k,t;
int temp[N][N];
for(i=0;i<n;i++)
{
for(j=0;j<n;j++)
{
for(k=0;k<n-1;k++)
{
for(t=0;t<n-1;t++)
{
temp[k][t] = arcs[k>=i?k+1:k][t>=j?t+1:t];
}
}

ans[j][i] = getA(temp,n-1);
if((i+j)%2 == 1)
{
ans[j][i] = - ans[j][i];
}
}
}
}

int main()
{
int arcs[N][N];
int astar[N][N];
int i,j;
int n;
while(scanf("%d",&n)!=EOF && n)
{
for(i=0;i<n;i++)
{
for(j=0;j<n;j++)
{
scanf("%d",&arcs[i][j]);
}
}

int a = getA(arcs,n);
if(a==0)
{
printf("can not transform!\n");
}
else
{
getAStart(arcs,n,astar);
for(i=0;i<n;i++)
{
for(j=0;j<n;j++)
{
printf("%.3lf ",(double)astar[i][j]/a);
}
printf("\n");
}
}
printf("\n");

}

return 0;
}