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;

}