一层层覆盖应该是最直观的

拿 $n = 5$ 举例。首先计算出需要覆盖 int(0.5*(n+1)) = 3 次。
第一次覆盖，将整个5x5正方形赋值为1；第二次覆盖，把中间3x3的正方形赋值为2......

#include <iostream>
using namespace std;

int main(){
    int n;
    cin >> n;

    while (n != 0){
        int a[n][n];

        for (int t = 0; t < int(0.5*(n+1)); t ++){
            for (int i = t; i < n - t; i ++ ){
                for (int j = t; j < n - t; j ++ ){
                    a[i][j] = t + 1;    
                }
            }
        }

        for (int i = 0; i < n; i ++){
            for (int j = 0; j < n; j ++){
                cout << a[i][j] << " ";
            }
            cout << endl;
        }
        cout << endl;
        cin >> n;
    }

    return 0;
}