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; 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<<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 ); else cout << "No inverse matrix" << endl; destroy( matrix1, n ); destroy( matrix2, n ); } return 0; }