Question

高分求一道C语言程序设计，要求用到函数·选择分支，数组~~~急求，谢谢！！！

谢谢各位大哥大姐了，我就这么多分，全部奉上了~~~~~

Answer 1

求矩阵的逆矩阵~~~~函数·选择分支，数组~~~都用上了
#include <stdlib.h>
#include <math.h>
#include <stdio.h>

double *inv(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 NULL;
}
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 a;
}

void display(double M[],int n) /*显示结果*/
{

int i,j;
double *result=NULL,*A=NULL;

/*显示原矩阵*/
printf("\n\nThe OriginalMatrix is:\n");
for(i=0;i<n;i++)
{
for(j=0;j<n;j++)
{
printf("%10.4f\t", M[i*n+j]);

}
printf("\n");
}

A=(double*)malloc(n*n*sizeof(double));
for(i=0;i<n*n;i++) A[i]=M[i];
result=inv(A,n);
if(result!=NULL)
{
printf("\n\nThe InverseMatrix is:\n");
for(i=0;i<n;i++)
{
for(j=0;j<n;j++)
{
printf("%10.4f\t", result[i*n+j]);

}
printf("\n");
}
}
else
{
printf("The Matrix is singular ! Can't be inverse!\n");
}
free(A);
}

#include <conio.h>
/* Test */
void main()
{
double matrix[3][3]={{1,-5,-2},{-1,3,1},{3,-4,-1}};
double matrix1[3][3]={{0,0,0},{-1,3,1},{3,-4,-1}};
double matrix2[4][4]={{3,2,1,1},{2,3,5,9},{-1,2,5,-2},{1,0,-1,3}};
display(matrix[0],3);
display(matrix1[0],3);
display(matrix2[0],4);
printf("push any key to be end .");
getch();
}

Answer 2

学习中