有一个 n×n 的方阵，第 1 列、对角线和第 n 列可以构成一个 “N”字，请编程求出这个 “N” 字上所有元素的和。

要解决这个问题，你可以使用两个嵌套的循环来遍历矩阵（二维数组）。首先，从左到右遍历第一列，然后从左上到右下的对角线，再从右到左遍历最后一列。对于每个位置，如果它的值为1（通常表示为白色），则将其添加到总和中。下面是用C++编写的示例代码：

#include <iostream>
using namespace std;

int main() {
    int n, sum = 0; // n is the size of the square matrix, and sum will store the total N shape sum
    cout << "Enter the size of the matrix (n x n): ";
    cin >> n;

    // Create a boolean matrix representing an 'N' shape where 1 represents white (on the pattern) and 0 represents black (not on the pattern)
    bool matrix[n][n];
    for(int i = 0; i < n; ++i) {
        if(i == 0 || i == n - 1) { // First column and last column are part of the 'N'
            matrix[i][i] = true;
        } else if(i % n == 0 || i + 1 == n) { // Diagonal from top-left to bottom-right
            matrix[i][i] = true;
        }
    }

    // Calculate the sum of all 'white' elements in the 'N' shape
    for(int i = 0; i < n; ++i) {
        for(int j = 0; j < n; ++j) {
            if(matrix[i][j]) {
                sum += matrix[i][j]; // Add 1 to the sum if the element is part of the 'N' shape
            }
        }
    }

    cout << "The sum of all elements in the 'N' shape is: " << sum << endl;

    return 0;
}