HDU 2137 circumgyrate the string

HDU 2137 circumgyrate the string

Problem Description
Give you a string, just circumgyrate. The number N means you just circumgyrate the string N times, and each time you circumgyrate the string for 45 degree anticlockwise.

Input
In each case there is string and a integer N. And the length of the string is always odd, so the center of the string will not be changed, and the string is always horizontal at the beginning. The length of the string will not exceed 80, so we can see the complete result on the screen.

Output
For each case, print the circumgrated string.

Sample Input
asdfass 7
Sample Output

a
 s
  d
   f
    a
     s
      s

简单的字符串处理问题，但是有几个注意点：

1、输入有可能是负数
2、当输入是2或6时，注意前面的空格，要不然会PE的

一共八种情况，一一个一个的写就行了，比较麻烦

代码如下：

#include<iostream>
#include <cstdio>
#include <string.h>
using namespace std;

int main()
{
#ifndef ONLINE_JUDGE
    freopen("1.txt","r",stdin);  
#endif
    char a[85];
    int n, len, i, j;
    while(~scanf("%s%d", a, &n)) 
    {
        while (n < 0)
        {
            n += 8;
        }
        if (n >= 8)
        {
            n %= 8;
        }

        len = strlen(a);
        if (n == 0)
        {
            cout << a << endl;
        }
        else if (n == 1)
        {
            for (i = len - 1; i >= 0; i--)
            {
                for(j = 1; j <= i; j++)
                {
                    printf(" ");
                }
                cout << a[i] << endl;
            }

        }
        else if (n == 2)
        {
            for (i = len - 1; i >= 0; i--)
            {
                for (j = 0; j < len/2; j++)
                    cout << " ";
                printf("%c\n", a[i]);
            }
        }
        else if (n == 3)
        {
            for (i = len - 1; i >= 0; i--)
            {
                for (j = 1; j < len - i; j++)
                {
                    printf(" ");
                }
                cout << a[i];

                cout << endl;
            }
        }
        else if (n == 4)
        {
            for (i = len - 1; i >= 0; i--)
            {
                cout << a[i];
            }
            cout << endl;
        }
        else if (n == 5)
        {
            for (i = 0; i < len; i++)
            {
                for(j = 1; j < len - i; j++ )
                {
                    printf(" ");
                }
                cout << a[i] << endl;
            }
        }
        else if (n == 6)
        {
            for (i = 0; i < len ; i++)
            {
                for(j = 1; j <= len/2; j++)
                {
                    printf(" ");
                }
                printf("%c\n", a[i]);
            }
        }
        else if (n == 7)
        {
            for(i = 0; i < len; i++)
            {
                for(j = 0; j < i; j++)
                {
                    cout << " ";
                }
                cout << a[i] << endl;
            }
        }
    // cout << endl << endl;
    }
    return 0;
}