uva 253 Cube painting

                          uva 253 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, or g. 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

Sample Input

rbgggrrggbgrrrrbbbrrbbbr rbgrbgrrrrrg

Sample Output

TRUEFALSE FALSE

题目大意：判断两个正方题可否旋转得到。

解题思路：只要每两个对面形成一组颜色对， 每个正方体有三个颜色对，匹配下颜色对是否相同。

#include <stdio.h>
#include <string.h>
int main() {
	int vis[5], cnt = 0;
	char pain[20];
	while (scanf ("%s", pain) != EOF) {
		cnt = 0;
		memset (vis, 0, sizeof(vis));
		for (int i = 0; i < 3; i++) {
			for (int j = 0; j < 3; j++) {
				if ((pain[i] == pain[6+j] && pain[5-i] == pain[11-j]) ||(pain[i] == pain[11-j] && pain[5-i] == pain[6+j]) && !vis[j]) {
					cnt++;
					vis[j] = 1;
					break;	
				}
			}
		}
			if (cnt==3)
				printf("TRUE\n");
			else
				printf("FALSE\n");
	}
	return 0;
	
}