基础模拟,的确是一堆if-else if嵌套，无技术含量

#include <bits/stdc++.h>
using namespace std;
int n , posx , posy;
int maze[40][40];
int main()
{
    cin >> n;
    posx = 1;
    posy = n / 2 + 1;
    maze[posx][posy] = 1;
    for (int i = 2 ; i <= n * n ; i++)
    {
        if (posx == 1 && posy != n)
        {
            maze[n][posy + 1] = i;
            posx = n;
            posy++;
        }
        else if (posy == n && posx != 1)
        {
            maze[posx - 1][1] = i;
            posx--;
            posy = 1;
        }
        else if (posx == 1 && posy == n)
        {
            maze[posx + 1][posy] = i;
            posx++;
        }
        else
        {
            if (maze[posx - 1][posy + 1] == 0)
            {
                maze[posx - 1][posy + 1] = i;
                posx--;
                posy++;
            }
            else
            {
                maze[posx + 1][posy] = i;
                posx++;
            }
        }
    }
    for (int i = 1 ; i <= n ; i++)
    {
        for (int j = 1 ; j <= n ; j++)
        {
            cout << maze[i][j] << " ";
        }
        cout << endl;
    }
    return 0;
}

按照题目那$4$个要求模拟即可，$a[1][n / 2 + 1] = 1;$ 其余数字就是

int i = 2; i <= n * n; i++

用两个变量$pos_x, pos_y$记录上一个点的$i,j$

初始化$pos_x = 1, pos_y = n / 2 + 1$

然后就是一堆$if-else\ if$嵌套