具有几何级数的C ++用户定义的二维数组

Question

I am learning C++ by myself,by solving different problems. I am trying to solve problem which was originally designed for Pascal in C++. It should ask user to input 3 integers M,N and q. Then it should make 2d array of integers with size MxN, where all the elements of (I=I,...M) line will be the members of geometrical progression with first element equal to number of line (I) and denominator q.

I wanted to create a dynamic massive, but I realized that it won't really work with two undefined integers. So, I tried vectors. I guess that I created them in a right way, but I've got no idea how to make a geometrical progression.

Here is my code:

#include <iostream>
#include <vector>
using namespace std;
int main()
{
    int m, n, q;

    cout << "Enter the number for M \n";
    cin >> m;
    if (cin.fail())
    {
        cin.clear();
        cin.ignore();
        cout << "This is not a number! " << endl;
        system("pause");
        return 0;
    }
    cout << "Enter the  number for N \n";
    cin >> n;
    if (cin.fail())
    {
        cin.clear();
        cin.ignore();
        cout << "This is not a number! " << endl;
        system("pause");
        return 0;
    }
    cout << "Enter the number  for Q \n";
    cin >> q;
    if (cin.fail())
    {
        cin.clear();
        cin.ignore();
        cout << "This is not a number! " << endl;
        system("pause");
        return 0;
    }
    int** matrix;
    matrix = new int*[m];
    for (int i = 0; i < m; i++)
        matrix[i] = new int[n];


    for (int i = 0; i < m; i++)
    {
        matrix[i][0] = i + 1;
    }
    for (int i = 0; i < m; i++)
    {
        for (int j = 1; j < n; j++)
        {
            matrix[i][j] = (i + 1)*pow(i, j);
            cout << matrix[i][j];

        }
    }
    system("pause");
    return 0;





}

Answer 1

Note: You can create a two dimensional array of a variable size, although it involves memory allocation and is slightly ugly.

int** matrix;
matrix = new int*[M];
for (int i = 0; i < M; i++)
    matrix[i] = new int[N];

That's the code to create an array of size MxN. Don't forget to deallocate your memory like so:

for (int i = 0; i < M; i++)
    delete matrix[i];
delete matrix;

As far as your question about the geometric progression, I am unsure of what you are asking. When you say geometric progression do you refer to something along the lines of 2 10 50 250 etc.? I am not sure what you mean by "lines" as you don't refer to any such variable in your code.

EDIT

So once the MxN matrix is created, iterate through the rows and initialize the rows like so:

for (int i = 0; i < M; i++)
{
    matrix[i][0] = i+1;
}

This should set the first column of each row to the correct number. Then something along the lines of this should fill out the rest of the geometric progression:

for (int i = 0; i < M; i++)
{
    for (int j = 1; j < N; j++)
    {
        matrix[i][j] = (i+1)*pow(r,j); 
        //note that you'll probably have to do some typecasting
        //p.s. I'm not 100% sure that this is the correct formula
    }
}

I think this is what you are looking for. Let me know if it works because I haven't tested it myself.

Print the matrix like this:

for (int i = 0; i < rows; i++)
{
    for (int j = 0; j < cols; j++)
    {
        std::cout << matrix[i][j] << " ";
    }
    std::cout << "\n";
}