HDU2137: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
 


 

水题

就是把一个字符串转来转去

 

#include <stdio.h>
#include <string.h>

int main()
{
    char str[100];
    int len,i,n,j;
    while(~scanf("%s%d",str,&n))
    {
        if(n>=8)
        {
            n = n%8;
        }
        else if(n<0)
        {
            n = n%8;
            n = n+8;
            n = n%8;
        }
        len = strlen(str);
        if(n == 0)
            puts(str);
        else if(n == 1)
        {
            for(i = len-1; i>=0; i--)
            {
                for(j = 0; j<len; j++)
                {
                    if(j == i)
                    {
                        printf("%c\n",str[j]);
                        break;
                    }
                    else
                        printf(" ");
                }
            }
        }
        else if(n == 2)
        {
            for(i = len-1; i>=0; i--)
            {
                for(j = 0; j<=len/2; j++)
                {
                    if(j == len/2)
                    {
                        printf("%c\n",str[i]);
                        break;
                    }
                    else
                        printf(" ");
                }
            }
        }
        else if(n == 3)
        {
            for(i = 0; i<len; i++)
            {
                for(j = 0; j<len; j++)
                {
                    if(j == i)
                    {
                        printf("%c\n",str[len-1-i]);
                        break;
                    }
                    else
                        printf(" ");
                }
            }
        }
        else if(n == 4)
        {
            for(i = len-1; i>=0; i--)
            {
                putchar(str[i]);
            }
            printf("\n");
        }
        else if(n == 5)
        {
            for(i = 0; i<len; i++)
            {
                for(j = len-1; j>=0; j--)
                {
                    if(j == i)
                    {
                        printf("%c\n",str[i]);
                        break;
                    }
                    else
                        printf(" ");
                }
            }
        }
        else if(n == 6)
        {
            for(i = 0; i<len; i++)
            {
                for(j = 0; j<=len/2; j++)
                {
                    if(j == len/2)
                    {
                        printf("%c\n",str[i]);
                        break;
                    }
                    else
                        printf(" ");
                }
            }
        }
         else if(n == 7)
        {
            for(i = 0; i<len; i++)
            {
                for(j = 0; j<len; j++)
                {
                    if(j == i)
                    {
                        printf("%c\n",str[i]);
                        break;
                    }
                    else
                        printf(" ");
                }
            }
        }
    }
    return 0;
}


 

 

posted @ 2013-04-11 12:09  javawebsoa  Views(175)  Comments(0Edit  收藏  举报