HDU 2137 模拟

circumgyrate the string

Time Limit: 10000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4901    Accepted Submission(s): 1145

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

 

Author

wangye

注意n可以是负的。。。

#include<cstdio>
#include<iostream>
#include<cstring>
char s[1000];
void ok(int n);
using namespace std;
int main()
{
int m=0;
while(scanf("%s %d",s,&m)!=EOF)
{
while(m<0)
m+=8;
getchar();
ok(m%8);
}
return 0;
}
void ok(int n)
{
int k=strlen(s);
if(n==0)
printf("%s\n",s);
else if(n==1)
{
for(int i=k-1;i>=0;i--)
{
for(int j=0;j<i;j++)
printf(" ");
printf("%c\n",s[i]);
}
}
else if(n==2)
{

for(int i=k-1;i>=0;i--)
{
 for(int j=0;j<k/2;j++)
  printf(" ");
printf("%c\n",s[i]);
   }
}
else if(n==3)
{
for(int i=k-1;i>=0;i--)
{
for(int j=0;j<k-1-i;j++)
printf(" ");
printf("%c\n",s[i]);
}   
}
else if(n==4)
{
 for(int i=k-1;i>=0;i--)
  printf("%c",s[i]);
 printf("\n");

}
else if(n==5)
{
  for(int i=0;i<k;i++)
{
for(int j=0;j<k-1-i;j++)
printf(" ");
printf("%c\n",s[i]);
}
}
else if(n==6)
{

for(int i=0;i<k;i++)
{
         for(int j=0;j<k/2;j++)
  printf(" ");
printf("%c\n",s[i]);
  }
}
else if(n==7)
{
for(int i=0;i<k;i++)
{
for(int j=0;j<i;j++)
printf(" ");
printf("%c\n",s[i]);
}
}
}