2015 Multi-University Training Contest 1 Hdu5292 Pocket Cube

Pocket Cube

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 110    Accepted Submission(s): 57

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)

It was a Pocket Cube. Unfortunately, the Cube fell to the ground and broke. It took you some time to explore the construction of the Cube. Then you assembled the Pocket Cube. Unfortunately, you didn’t assembled it in the right way. So here is the question. You want to know whether it can return to the right position.

The right position means four blocks in each face has the same color. You can only rotate the Cube to return it to the right position.

A Cube is given in its layout.


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

 

#include<iostream>#include<cstdio>;using namespace std;//  不仅要求对面和为0，也要求本面和为0int v[]={1,1,1,1,1,1,0,0,-1,-1,1,1,0,0,-1,-1,        -1,-1,-1,-1,0,0,0,0};   //AC//int v[]={1,-1,-1,1,-1,1,0,0,-1,1,1,-1,0,0,1,-1,//        1,-1,-1,1,0,0,0,0};   //WAint main(){    int T;cin>>T;    for(int ca=1;ca<=T;ca++){        char str[2];        int ans=0;        for(int i=0;i<24;i++){            cin>>str;            if(str[0]=='w'||str[0]=='y')                ans+=v[i];        }        printf("Case #%d: ",ca);        if(ans%3==0)puts("YES");        else puts("NO");    }    return 0;}