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 <iostream>#include <cmath>#include <algorithm>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;elsematrix2[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;}}}elsereturn 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<<endl;}}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 );elsecout << "No inverse matrix" << endl;destroy( matrix1, n );destroy( matrix2, n );}return 0;}
