hd2137

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circumgyrate the string

Time Limit: 10000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

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
 
题目大意就是一字符串，嗯，长度总为奇数，以中间的字符为轴逆时针旋转，每次旋转45度。
 
不造有木有可以不用从0到7一个一个写的方法，挺麻烦，
嗯，注意n可以为负
 
<pre name="code" class="html">#include<stdio.h>#include<string.h>int main(){    int n,i,j,lens;    char s[100];    while(~scanf("%s %d",s,&n))    {                n=n%8;        if(n<0)n+=8;        lens=strlen(s);        if(n==0)printf("%s\n",s);        else if(n==1)        {            for(i=0;i<lens;++i)            {                for(j=0;j<lens;++j)                {                    if(j<lens-i-1)printf(" ");                    else if(j==lens-i-1)printf("%c",s[lens-i-1]);                }                printf("\n");            }        }        else if(n==2)        {            for(i=0;i<lens;++i)            {                for(j=0;j<lens;++j)                {                    if(j<(lens-1)/2)printf(" ");                    else if(j==(lens-1)/2)printf("%c",s[lens-1-i]);                }                printf("\n");            }        }        else if(n==3)        {            for(i=0;i<lens;++i)            {                for(j=0;j<lens;++j)                {                    if(j<i)printf(" ");                    else if(j==i)printf("%c",s[lens-i-1]);                }                printf("\n");            }        }        else if(n==4)        {            for(i=lens-1;i>=0;i--)printf("%c",s[i]);            printf("\n");        }        else if(n==5)        {            for(i=0;i<lens;++i)            {                for(j=0;j<lens;++j)                {                    if(j<lens-i-1)printf(" ");                    else if(j==lens-i-1)printf("%c",s[i]);                }                printf("\n");            }        }        else if(n==6)        {            for(i=0;i<lens;++i)            {                for(j=0;j<lens;++j)                {                    if(j<(lens-1)/2)printf(" ");                    else if(j==(lens-1)/2)printf("%c",s[i]);                }                printf("\n");            }        }        else if(n==7)        {            for(i=0;i<lens;++i)            {                for(j=0;j<lens;++j)                {                    if(j<i)printf(" ");                    else if(j==i)printf("%c",s[i]);                }                printf("\n");            }        }    }    return 0;}

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