方法1：找规律法

#include <iostream>

using namespace std;

int main(void)
{
    int n;
    while (cin >> n) {
        if (n == 0) break;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                cout << min(min(i + 1, j + 1), min(n - i, n - j)) << ' ';
            }
            cout << endl;
        }
        cout <<  endl;
    }
    return 0;
}

方法2：曼哈顿距离法

#include <iostream>

using namespace std;

int main(void)
{
    int n;
    while (cin >> n) {
        if (n == 0) break;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if (n % 2) // 奇数
                    cout << (n + 1) / 2 - max(abs(n / 2 - i), abs(n / 2 - j)) << ' ';
                else 
                    cout << (n + 1) / 2.0 - max(abs((n - 1) / 2.0 - i), abs((n - 1) / 2.0 - j)) << ' ';
            }
            cout << endl;
        }
        cout <<  endl;
    }
    return 0;
}

方法3：蛇形矩阵求解

#include <iostream>
#include <cstring>

using namespace std;

const int N = 100 + 10;
int m[N][N];

int main(void)
{
    int n;
    int dx[4] = {0, 1, 0, -1}, dy[4] = {1, 0, -1, 0};
    while (cin >> n) {
        memset(m, 0, sizeof(m));
        if (n == 0) break;
        int d = 0, x = 0, y = 0;
        int cnt = 0;
        int res = 1;
        for (int i = 0; i < n * n; i++) {
            int a = x + dx[d], b = y + dy[d];
            m[x][y] = res;
            if (a < 0 || a >= n || b < 0 || b >= n || m[a][b]) {
                d = (d + 1) % 4;
                a = x + dx[d], b = y + dy[d];
                cnt++;
                if (!(cnt % 4)) res++;
            }
            x = a, y = b;
        }
        for (int i = 0; i < n; i ++){
            for (int j = 0; j < n; j ++)
                cout << m[i][j] << ' ';
            cout << endl;
        }
        cout << endl;
    }
    return 0;
}