UVA - 253 Cube painting

 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 ab, r, or g. The character ( ) from the left gives the color of facei. For example, Figure 2 is a picture of rbgggr and Figure 3 corresponds torggbgr. 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

rbgggrrggbgrrrrbbbrrbbbrrbgrbgrrrrrg

Sample Output

TRUEFALSE
FALSE题意：两个六边形，第一个六边形由前6个字符表示，后一个六边形由后6个字符表示。问第一个六边形与第二个六边形是否相同


#include<string.h>#include<stdio.h>#include<iostream>#define N 200using namespace std;int main(){char a[N];int b[N][N];int c[N][N];int n = 0;while(scanf("%s",a)!= EOF){memset(b,0,sizeof(b));memset(c,0,sizeof(c));for(int i = 0; i < 3;i ++){b[a[i]][a[5-i]] ++ ;b[a[5-i]][a[i]] ++;}int flag  = 0;for(int i = 6 ; i < 9;i ++){c[a[i]][a[17 - i]] ++;c[a[17 - i]][a[i]] ++;}for(int i = 0; i < 3;i ++){if(b[a[i]][a[5-i]] != c[a[i]][a[5-i]]){flag  = 1;printf("FALSE\n");break;}}if(!flag)printf("TRUE\n");}return 0;}