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;
}