题目要求：
Input a value n, then print out a n×n matrix.
Example 1: Input 2, output 1 2 4 3 Example2: Input 5, output 1 2 3 4 5 16 17 18 19 6 15 24 25 20 7 14 23 22 21 8 13 12 11 10 9思路：
把该输出划分为多个环，每一个环上有4根线。分别为0,1,2,3号。
最外层的环每根线上有n-1个数字，每靠内一层。线上数字少2个。
计算每一个位置(row, col)相应的环、线、线内索引。进而求出环内索引和全局索引。在该位置打印全局索引就可以。
代码例如以下：
#include <algorithm>
using namespace std;
int get_number( int row, int col, int n)
{
int n_1 = n - 1;
int loop1 = min(row, col);
int loop2 = min(n_1-row, n_1-col);
int loop = min(loop1, loop2);
int line, index_in_line;
if (col == loop && row != loop)
{
line = 3;
index_in_line = (n_1-loop) - row;
}
else if (n_1-row == loop)
{
line = 2;
index_in_line = (n_1-loop) - col;
}
else if (n_1-col == loop)
{
line = 1;
index_in_line = row - loop;
}
else if (row == loop)
{
line = 0;
index_in_line = col - loop;
}
int line_length = n_1 - loop * 2;
int index_in_loop = line * line_length + index_in_line;
int before_loop = 4 * (n_1 - loop + 1) * loop;
return before_loop + index_in_loop + 1;
}
void print_square( int n)
{
for ( int row=0; row<n; ++row)
{
for ( int col=0; col<n; ++col)
{
int number = get_number(row, col, n);
printf( "%4d " , number);
}
printf( "\n" );
}
}
int main () {
for ( int i=1; i<10; ++i)
{
print_square(i);
printf( "\n" );
}
return 0;
}