陈未一 • 3个月前
using namespace std;
int main() {
int n;
cin >> n;
int a[n + 1][n + 1];
int x = (n + 1) / 2;
for(int k = 1;k <= x;k++)
{
for(int i = x + 1 - k;i <= n - x + k;i++)
{
int t = x - k + 1;
a[n - x + k][i] = t;
a[i][n - x + k] = t;
a[x + 1 - k][i] = t;
a[i][x + 1 - k] = t;
}
}
for(int i = 1;i <= n;i++)
{
for(int j = 1;j <= n;j++)
{
cout << setw(5) << a[i][j];
}
cout << endl;
}
}
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