以下是我的程序,我遇到了行列式函数的问题。
文件输入为:
2
1 0
0 1
3
8 9 1
3 5 2
-2 3 -1
0
对于第二个矩阵,它为输入文件中矩阵2的行列式函数的结果读取nan,对于我的代码可能出错的任何想法?
#include <iostream>
#include <fstream>
#include <cmath>
using namespace std;
const int maxsize = 10;
ifstream fin;
ofstream fout;
void transpose (double omatrix[][maxsize],double tmatrix [][maxsize], int array_size)
{
for(int i = 0; i < array_size; i++)
{
for(int j = 0; j < array_size; j++)
{
tmatrix[j][i] = omatrix[i][j];
}
}
}
void sub (double omatrix[][maxsize], double smatrix[][maxsize], int array_size, int i, int j)
{
int counter1 = 0, counter2 = 0;
for (int a = 0; a < array_size; a++)
{
if (a != i)
{
for (int b = 0; b < array_size; b++)
{
if (b != j)
{
smatrix[counter1][counter2] = omatrix[a][b];
counter2++;
}
}
counter1++;
}
}
}
double determininant(double original_matrix[][maxsize], int array_size)
{
double D = 0.0, temp[maxsize][maxsize];
if(array_size == 1)
{
return original_matrix[0][0];
}
else if(array_size == 2)
{
return (original_matrix[0][0] * original_matrix[1][1]) - (original_matrix[0][1] * original_matrix[1][0]);
}
for(int i = 0; i < array_size; i++)
{
sub (original_matrix,temp,array_size, 0, i);
D += pow(-1.0,i) * original_matrix[0][i] * determininant(temp, array_size - 1);
}
return D;
}
void inverse(double omatrix[][maxsize], int array_size, double imatrix[][maxsize])
{
double D = determininant(omatrix, array_size);
double c[maxsize][maxsize];
if (D != 0)
{
for(int i = 0; i < array_size; i++)
{
for(int j = 0; j < array_size; j++)
{
sub (omatrix, c, array_size, i, j);
imatrix[j][i] = pow(-1.0, i+j) * determininant( c, array_size - 1) / D;
}
}
}
}
void mult(double omatrix[][maxsize], double imatrix[][maxsize], double pmatrix[][maxsize], int array_size)
{
for(int i = 0; i < array_size; i++)
{
for(int j = 0; j < array_size; j++)
{
pmatrix[i][j] = 0;
for(int k = 0; k < array_size; k++)
{
pmatrix[i][j] += omatrix[i][k] *imatrix[k][j];
}
}
}
}
void print (const double m[][maxsize], int array_size)
{
for(int i = 0; i < array_size; i++)
{
for(int j = 0; j < array_size; j++)
{
fout << m[i][j] << " ";
}
fout << "\n";
}
fout << "\n";
}
int main()
{
double original_matrix[maxsize][maxsize],
inverse_matrix [maxsize][maxsize],
transposed_matrix[maxsize][maxsize],
product_matrix [maxsize][maxsize],
D;
int array_size;
fout.setf(ios::fixed);
fout.precision(2);
char File_Name[10];
cout << "Please enter a name for the output file";
cin >> File_Name;
fin.open ("input.txt");
fout.open (File_Name);
if (fin.fail())
{
cout << "Input file opening failed. \n";
return 0;
}
while(fin >> array_size)
{
if(array_size != 0)
{
for(int i = 0; i < array_size; i++)
{
for(int j = 0; j < array_size; j++)
{
fin >> original_matrix[i][j];
}
}
fout << "THE ORIGIONAL ARRAY! \n";
print (original_matrix, array_size);
transpose (original_matrix, transposed_matrix, array_size);
fout << "THE TRANSPOSED VIRSION! \n";
print (transposed_matrix, array_size);
fout << determininant(original_matrix, array_size) << "\n\n";
inverse(original_matrix, array_size, inverse_matrix);
print (inverse_matrix, array_size);
mult(original_matrix, inverse_matrix, product_matrix, array_size);
print (product_matrix, array_size);
}
}
fin.close();
fout.close();
return 0;
}
输出结果为:
THE ORIGIONAL ARRAY!
1.00 0.00
0.00 1.00
THE TRANSPOSED VIRSION!
1.00 0.00
0.00 1.00
1.00
1.00 -0.00
-0.00 1.00
1.00 0.00
0.00 1.00
THE ORIGIONAL ARRAY!
8.00 9.00 1.00
3.00 5.00 2.00
-2.00 3.00 -1.00
THE TRANSPOSED VIRSION!
8.00 3.00 -2.00
9.00 5.00 3.00
1.00 2.00 -1.00
nan <--------------------- is wrong should be -78.00
which makes the next 2 mess up.
0.00 -0.00 0.00
-0.00 0.00 -0.00
0.00 -0.00 0.00
0.00 0.00 0.00
0.00 0.00 0.00
0.00 0.00 0.00
答案 0 :(得分:2)
您忘记重置counter2
中的sub(...)
。因此,您要复制到结果矩阵中的错误位置,并在计算中使用未初始化的内存。
(故事的寓意是在你真正需要它们的最小范围内声明变量。)