HDU 5292 Pocket Cube

Pocket Cube

Problem Description

Pocket Cube is the 2×2×2 equivalent of a Rubik’s Cube(3×3×3). The cube consists of 8 pieces, all corners. (from wiki)

 

The right position rotates in red face clockwisely. 
You can get more details from input case.
w represents white , y represents yellow , o represents orange , r represents red , g represents green , b represents blue. In the right position, white and yellow , orange and red , green and blue are in the opposite face. 

 

Input

The first line of input contains only one integer T(<=10000), the number of test cases. 
Each case contains a Pocket Cube described above. After each case , there is a blacnk line. It guarantees that the corners of the Cube is right.

 

Output

Each case contains only one line. Each line should start with “Case #i: ”,with i implying the case number, followed by “YES” or “NO”,”YES” means you can return it to the right position, otherwise “NO”.

 

Sample Input

2    g y     g y o o w g r r o o w g r r     b w     b w     y b     y b    r w     g b b y o w o r y g y o g b     r w     o y     b r     w g

 

Sample Output

Case #1: YESCase #2: NO

/* * 二阶魔方任何一个块都可以旋转，这里考虑对色面，（黄，白）， （绿， 蓝）， （红， 橙） * 将标准魔方展开 *          0  0 *          0  0 *    1 -1  1  -1  1  -1 *    -1  1  -1  1  -1  1 *          0  0 *          0  0 *          1  -1 *          -1  1 * 只要考虑一组对立面即可，比如（黄， 白），对应到上述展开后，只要最后总和模三等于零就是合理的 */

#include <bits/stdc++.h>#define endl "\n"using namespace std;const int cube[] = {0, 0, 0, 0, 1, -1, 1, -1, 1, -1, -1, 1, -1, 1, -1, 1, 0, 0, 0, 0, 1, -1, -1, 1};int main() {    int T;    cin >> T;    for(int cas = 1; cas <= T; ++cas) {        char color[2];        int ret = 0;        for(int i = 0; i < 24; ++i) {            scanf("%s", color);            if(color[0] == 'w' || color[0] == 'y') ret += cube[i];        }        cout <<  "Case #" << cas << ": " << (ret % 3 ? "NO" : "YES") << endl;    }    return 0;}