解题思路
M[i][j]的值为到四边距离的最小值+1
时间复杂度:$O(n^2)$
C++ 代码
#include<bits/stdc++.h>
using namespace std;
int main()
{
int x = 1;
while (x != 0)
{
cin >> x;
for (int i = 0;i < x;i++)
{
for (int j = 0;j < x;j++)
{
cout << min(i,min(j,min(x-i-1,x-j-1)))+1;
if (j != x-1) cout << " ";
}
cout << endl;
}
if (x != 0) cout << endl;
}
return 0;
}