| Cube painting |
We have a machine for painting cubes. It is supplied with three different colors: blue, red and green. Each face of the cube gets one of these colors. The cube‘s faces are numbered as in Figure 1.
Figure 1.
Since a cube has 6 faces, our machine can paint a face-numbered cube in 
different ways. When ignoring the face-numbers, the number of different paintings is much less, because a cube can be rotated. See example below. We denote a painted cube by a string of 6 characters, where each character is a b, r, org. The 
character ( 
) from the left gives the color of face i. For example, Figure 2 is a picture of rbgggr and Figure 3 corresponds to rggbgr. Notice that both cubes are painted in the same way: by rotating it around the vertical axis by 90 
, the one changes into the other.
Input
The input of your program is a textfile that ends with the standard end-of-file marker. Each line is a string of 12 characters. The first 6 characters of this string are the representation of a painted cube, the remaining 6 characters give you the representation of another cube. Your program determines whether these two cubes are painted in the same way, that is, whether by any combination of rotations one can be turned into the other. (Reflections are not allowed.)
Output
The output is a file of boolean. For each line of input, output contains TRUE if the second half can be obtained from the first half by rotation as describes above, FALSE otherwise.
Sample Input
rbgggrrggbgr rrrbbbrrbbbr rbgrbgrrrrrg
Sample Output
TRUE FALSE FALSE 这个方法是参考大神的，实在是不能太6.
#include <cstdio>
#include <cstring>
char str[15];
int vis[3]; /*每两个对面形成一组颜色对， 每个正方体有三个颜色对，
匹配下三个颜色对是否相同。（用vis标记检查顺序）*/
int main()
{
while(scanf("%s", str) != EOF)
{
int cnt = 0;
memset(vis, 0, sizeof(vis));
for(int i = 0; i < 3; i++)
{
for(int j = 0; j < 3; j++)
{
if((str[i] == str[6+j] && str[5-i] == str[11-j]) || (str[i] == str[11-j] && str[5-i] == str[6+j]) && !vis[j])
{
vis[j] = 1;
cnt++;
break;
}
}
}
if(cnt == 3) printf("TRUE\n");
else printf("FALSE\n");
}
}