矩阵中独立对角线的总和

Question

我正在学习考试,我正在尝试处理动态矩阵.我遇到了一个关于计算矩阵的每个对角线之和的问题,该矩阵的值和大小由用户选择.我的程序的目的是打印,这要归功于一个函数,其参数是矩阵及其大小,即每个对角线和的值.我将向您展示代码并对其进行深入描述.

----------------
| 52 | 35 | 5  |  Example of matrix.
----------------  Imagine the first diagonal to be the one which goes right-to-left
| 2  | 71 | 1  |  and only consists in the number "47".
----------------  The second diagonal would be the one which goes right-to-left and
| 3  | 60 | 25 |  consists in the number "15" and "79".
----------------  So to get the sum of the second diagonal it would be:
| 79 | 55 | 98 |     
----------------  sum = m[n_rows - 1][diag - 2] + m[n_rows - 2][diag - 1]
| 47 | 15 | 66 | 
----------------  When diag > columns, in order to avoid error regarding matrix size,
                  I should lower the quantity "n_rows - 1" by the quantity "diag - n_columns".

根据我的描述,这就是我的想法:

void diag_matrix(int** m, int righe, int colonne){//righe = rows, colonne = columns. 
//M is the matrix.

// diag is the number of the diagonal I'm considering.
    for(int diag = 1; diag < (righe + colonne); diag++){ 

        int sum = 0;// the sum
        int i = 0;// the counter of the cicle
        int l = 0;// this is the value to riallign the row in case diag > column
        int temp = diag;//I use this variable not to modify the value of diag.

        // What I want is: when the column-index/row-index of the matrix reaches 0, the cicle will interrupt (after final iteration);
        while(righe - i - l - 1 > 0 || diag - 1 - i > 0){

            if (diag > colonne){//this condition changes l-value only if diag value is greater than column. Explanation outside the code
                l = diag - colonne;//this is the value to subtract to row-index
                temp = colonne;//this position is necessary to set column-index to its maxium.
            }

            sum = sum + m[righe - 1 - l - i][temp -1 - i];//pretty clear I think.
            i++;//the i is incremented by one.

        }// end of while-statement

        cout << "Somma Diagonale " << diag << " = " << sum << ".\n";

    }// end of for-statement

}//end of function declaration

显然它不起作用,但我无法弄清楚问题.

Answer 1

您可以简化代码，找到每个对角线的起始位置，然后只要坐标保留在矩阵内即可逐步遍历矩阵。像这样的东西：

#include <iostream>

using namespace std;

void diag_matrix(int** m, int rows, int cols)
{
    for (int diag = 1; diag < rows + cols; diag++)
    {
        int x, y;
        if (diag < rows)
        {
            y = rows - diag;
            x = 0;
        }
        else
        {
            y = 0;
            x = diag - rows;
        }
        int sum = 0;
        cout << "Summing diagonal #" << diag << ":";
        while ((x < cols) && (y < rows))
        {
            sum += m[y][x];
            cout << " " << m[y][x];
            x++;
            y++;
        }
        cout << " result: " << sum << "." << endl;
    }
}

int main(int argc, char* argv[])
{
    int rows = 5, cols = 3;
    int **m = new int*[rows];
    for (int i = 0; i < rows; i++)
        m[i] = new int[cols];
    m[0][0] = 52; m[0][1] = 35; m[0][2] =  5;
    m[1][0] =  2; m[1][1] = 71; m[1][2] =  1;
    m[2][0] =  3; m[2][1] = 60; m[2][2] = 25;
    m[3][0] = 79; m[3][1] = 55; m[3][2] = 98;
    m[4][0] = 47; m[4][1] = 15; m[4][2] = 66;
    diag_matrix(m, rows, cols);
    for (int i = 0; i < rows; i++)
        delete[] m[i];
    delete[] m;
    return 0;
}