蛇形填数

这个问题直接用4个while循环就可以写出来，注意一下边界处理就行了

#include <stdio.h>
#include <string.h>
int a[103][103];
int main()
{
    int n;
    while(~scanf("%d", &n))
    {
        memset(a, 0, sizeof(a));
        int num = 1;
        int cnt = n * n;
        int i = 0, j = n - 1;
        a[0][n - 1] = 1;
        while (num < cnt)
        {
            while (i + 1 < n && a[i + 1][j] == 0)//向下填数 
                a[++i][j] = ++num;
            while (j - 1 >= 0 && a[i][j - 1] == 0)//向左填数 
                a[i][--j] = ++num;
            while (i - 1 >= 0 && a[i - 1][j] == 0)//向上填数 
                a[--i][j] = ++num;
            while (j + 1 < n && a[i][j + 1] == 0)//向右填数 
                a[i][++j] = ++num;
            
        }
        for (int i = 0; i < n; i++)
        {
            for (int j = 0; j < n; j++)
                printf("%d ", a[i][j]);
            puts("");
        }
    }
    return 0;

}