有多组测试数据。每组测试数据先输入一个整数n，表示方阵的阶。然后下面输入n阶方阵。输出其逆矩阵。若无逆矩阵，则输出No inverse matrix。
#include
#include
#include
using namespace std;
const double eps = 1e-6;
bool is_zero( const double num )
{
return fabs(num) < eps;
}
void create( double ** & matrix, const int n )
{
matrix = new double* [n];
for ( int i = 0; i < n; ++i )
matrix[i] = new double[n];
}
void input ( double ** matrix, const int n )
{
for ( int i = 0; i < n; ++i )
{
for ( int j = 0; j < n; ++ j )
cin >> matrix[i][j];
}
}
bool inverse ( double ** matrix1, double ** matrix2, const int n )
{
int i, j;
for ( i = 0; i < n; ++ i )
{
for ( j = 0; j < n; ++ j )
{
if ( i == j )
matrix2[i][j] = 1;
else
matrix2[i][j] = 0;
}
}
for ( i = 0; i < n; ++i )
{
int rowmaxpos = i;
for ( j = i + 1; j < n; ++j )
{
if ( matrix1[i][j] > matrix1[i][rowmaxpos] )
rowmaxpos = j;
}
for ( j = i; j < n; ++ j )
{
swap( matrix1[j][rowmaxpos], matrix1[j][i]);
swap( matrix2[j][rowmaxpos], matrix2[j][i]);
}
if ( !is_zero(matrix1[i][i]) )
{
int divisor = matrix1[i][i];
for ( j = i; j < n; ++ j )
{
matrix1[i][j] /= divisor;
matrix2[i][j] /= divisor;
}
for ( j = i + 1; j < n; ++ j )
{
int multiple = matrix1[j][i];
for ( int k = i; k < n; ++ k )
{
matrix1[i][j] -= matrix1[i][k] * multiple;
matrix2[i][j] -= matrix2[i][k] * multiple;
}
}
}
else
return false;
}
return true;
}
void output( double ** matrix, const int n )
{
for ( int i = 0; i < n; ++i )
{
for ( int j = 0; j < n; ++ j )
cout << matrix[i][j] << ' ';
cout<
}
}
void destroy( double ** matrix, const int n )
{
for ( int i = 0; i < n; ++ i )
delete [] matrix[i];
delete [] matrix;
}
int main()
{
int n;
double ** matrix1;
double ** matrix2;
while ( cin >> n )
{
create( matrix1, n );
create( matrix2, n );
input( matrix1, n);
if ( inverse(matrix1, matrix2, n) )
output( matrix2, n );
else
cout << "No inverse matrix" << endl;
destroy( matrix1, n );
destroy( matrix2, n );
}
return 0;
}
