C++程序计算矩阵对角线之和

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The utilization of 2-dimensional arrays or matrices is extremely advantageous for several applications. Matrix rows and columns are used to hold numbers. We can define 2D 在C++中使用多维数组来表示矩阵。在本文中，我们将看看如何实现 use C++ to calculate the diagonal sum of a given square matrix.

The matrices have two diagonals, the main diagonal and the secondary diagonal (sometimes referred to as major and minor diagonals). The major diagonal starts from the top-left corner (index [0, 0]) to the bottom-right corner (index [n-1, n-1]) where n is the order of the 正方形矩阵。主对角线从右上角（索引[n-1, 0]）开始，到左下角 corner (index [0, n-1]). Let us see the algorithm to find the sum of the elements along with these two diagonals.

Matrix Diagonal Sum

矩阵对角线之和

$$\begin{bmatrix} 8 & 5& 3\newline 6 & 7& 1\newline 2 & 4& 9\ \end{bmatrix},$$

Sum of all elements in major diagonal: (8 + 7 + 9) = 24
Sum of all elements in minor diagonal: (3 + 7 + 2) = 12

In the previous example, one 3 x 3 matrix was used. We have scanned the diagonals individually and calculated the sum. Let us see the algorithm and implementation for a clear view.

Algorithm

• 读取矩阵 M 作为输入
• 考虑 M 具有 n 行和 n 列
• sum_major := 0
• sum_minor := 0
• 对于i从0到n-1的范围，执行
• for j rangign from 0 to n - 1, do
  • if i and j are the same, then
    • sum_major := sum_major + M[ i ][ j ]
  • end if
  • if (i + j) is same as (N - 1), then
    • sum_minor := sum_minor + M[ i ][ j ]
  • end if
• end for
• end for
• return sum

Example

#include <iostream>
#include <cmath>
#define N 7
using namespace std;
float solve( int M[ N ][ N ] ){
   int sum_major = 0;
   int sum_minor = 0;
   for ( int i = 0; i < N; i++ ) {
      for ( int j = 0; j < N; j++ ) {
         if( i == j ) {
            sum_major = sum_major + M[ i ][ j ];
         }
         if( (i + j) == N - 1) {
            sum_minor = sum_minor + M[ i ][ j ];
         }
      }
   }
   cout << "The sum of major diagonal: " << sum_major << endl;
   cout << "The sum of minor diagonal: " << sum_minor << endl;
}
int main(){
   int mat1[ N ][ N ] = {
      {5, 8, 74, 21, 69, 78, 25},
      {48, 2, 98, 6, 63, 52, 3},
      {85, 12, 10, 6, 9, 47, 21},
      {6, 12, 18, 32, 5, 10, 32},
      {8, 45, 74, 69, 1, 14, 56},
      {7, 69, 17, 25, 89, 23, 47},
      {98, 23, 15, 20, 63, 21, 56},
   };
   cout << "For the first matrix: " << endl;
   solve( mat1 );
   int mat2[ N ][ N ] = {
      {6, 8, 35, 21, 87, 8, 26},
      {99, 2, 36, 326, 25, 24, 56},
      {15, 215, 3, 157, 8, 41, 23},
      {96, 115, 17, 5, 3, 10, 18},
      {56, 4, 78, 5, 10, 22, 58},
      {85, 41, 29, 65, 47, 36, 78},
      {12, 23, 87, 45, 69, 96, 12}
   };
   cout << "\nFor the second matrix: " << endl;
   solve( mat2 );
}

输出

For the first matrix: 
The sum of major diagonal: 129
The sum of minor diagonal: 359

For the second matrix: 
The sum of major diagonal: 74
The sum of minor diagonal: 194

Conclusion

In this article, we have seen how to calculate the diagonal sums of a given square matrix. 主对角线从左上角延伸到右下角，而副对角线则从左下角延伸到右上角 斜线从右上角开始到左下角。要找到这些的总和 diagonal elements, we loop through all elements. When both row and column index values 相同，它表示主对角线元素，当两个索引的和为 与矩阵的阶数n-1相同，它将添加到副对角线上 procedure takes two nested loops and we are traversing through all elements present in the 2D数组。因此，计算两条对角线的和将花费O(n²)的时间 给定的矩阵。

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