菱形 曼哈顿距离

#include <iostream>
#include <cmath>
using namespace std;
int main()
{
    int n;
    cin >> n;
    int cx = n / 2, cy = n / 2;
    for (int i = 0; i < n; i++)
    {
        for (int j = 0; j < n; j++)
        {
            if (abs(i - cx) + abs(j - cy) <= n / 2) cout << '*';//距离也要小于n/2
            else cout << ' ';
        }
        cout << endl;//每次换行多要输出回车
    }
    return 0;
}