HDU 2137 circumgyrate the string

http://acm.hdu.edu.cn/showproblem.php?pid=2137

 

circumgyrate the string

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

Total Submission(s):
2164    Accepted
Submission(s): 464


[align=left]Problem Description[/align]
  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.

 

 

[align=left]Input[/align]
  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.

 

 

[align=left]Output[/align]
  For
each case, print the circumgrated string.

 

 

[align=left]Sample Input[/align]

asdfass
7

 

 

[align=left]Sample Output[/align]

a s d f a s
s

 

 

[align=left]Author[/align]
wangye
 

 

[align=left]Source[/align]

HDU 2007-11 Programming Contest_WarmUp
 

 

[align=left]Recommend[/align]
威士忌
 

题意：给一段字符串和数字N，N表示这段字符串逆时针旋转的次数，每次旋转45°，输出旋转后的字符串。

分析：旋转每8次一循环，并且当N的值取0和4；1和5；2和6；3和7时，输出格式一样，字符串的顺序相反。
代码如下：

#include<stdio.h>

#include<string.h>

#include<math.h>

int main()

{

 int n,i,flag,j,l;

 char w[81];

 while(scanf("%s%d",w,&n)!=EOF)

 {

  l=strlen(w);

  if(n%8==0) puts(w);

  if(n%8==4||n%8==-4)

  {

   for(i=0;i<l;i++)

     
printf("%c",w[l-1-i]);

     
printf("\n");

  }

  if(n%8==5||n%8==-3)

       
{

    
 flag=l-1;

  
       
for(i=0;i<l;i++)

    
{

      for(j=0;j<flag;j++)

         
printf(" ");

             
printf("%c\n",w[i]);

             
flag--;

     
} 

       
}

       
if(n%8==1||n%8==-7)

  {

   flag=l-1;

  
       
for(i=0;i<l;i++)

    
{

      for(j=0;j<flag;j++)

         
printf(" ");

             
printf("%c\n",w[l-1-i]);

             
flag--;

     
} 

  }

  if(n%8==6||n%8==-2)

  {

    
flag=l/2;

   for(i=0;i<l;i++)

    
{

      for(j=0;j<flag;j++)

         
printf(" ");

             
printf("%c\n",w[i]);

     
}

  }

  if(n%8==2||n%8==-6)

  {

    
flag=l/2;

   for(i=0;i<l;i++)

    
{