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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.
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.)
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.
rbgggrrggbgr rrrbbbrrbbbr rbgrbgrrrrrg
TRUE
FALSE
FALSE
输入连个骰子，判断是否等价。
1 #include<iostream> 2 3 using namespace std; 4 //列举4*6种可能 5 int a[30][6] = 6 { {}, 7 { 1, 2, 3, 4, 5, 6 }, { 1, 4, 2, 5, 3, 6 }, { 1, 5, 4, 3, 2, 6 }, { 1, 3, 5, 2, 4, 6 }, 8 { 6, 2, 4, 3, 5, 1 }, { 6, 3, 2, 5, 4, 1 }, { 6, 5, 3, 4, 2, 1 }, { 6, 4, 5, 2, 3, 1 }, 9 { 2, 6, 3, 4, 1, 5 }, { 2, 4, 6, 1, 3, 5 }, { 2, 1, 4, 3, 6, 5 }, { 2, 3, 1, 6, 4, 5 }, 10 { 5, 6, 4, 3, 1, 2 }, { 5, 3, 6, 1, 4, 2 }, { 5, 1, 3, 4, 6, 2 }, { 5, 4, 1, 6, 3, 2 }, 11 { 3, 2, 6, 1, 5, 4 }, { 3, 1, 2, 5, 6, 4 }, { 3, 5, 1, 6, 2, 4 }, { 3, 6, 5, 2, 1, 4 }, 12 { 4, 2, 1, 6, 5, 3 }, { 4, 6, 2, 5, 1, 3 }, { 4, 5, 6, 1, 2, 3 }, { 4, 1, 5, 2, 6, 3 } 13 }; 14 15 int main() 16 { 17 int i, j, n, m, s, t; 18 char s1[20], s2[20]; 19 while (cin>>s1) //输入字符串s1共12位 20 { 21 for (i = 6; i <= 12; i++) //将 后六位 赋值给s2 22 { 23 s2[i - 6] = s1[i]; 24 } 25 for (i = 1; i <= 24; i++) 26 { 27 for (j = 0; j <= 5; j++) 28 { 29 30 if (s2[j] == s1[a[i][j] - 1]) 31 { 32 33 } 34 35 else 36 { 37 break; //如果不是一一对应，退出当前循环（最近的那个for循环）
38 } //然后i+1 继续循环
39 } 40 if (j == 6) //j==6说明 这次循环j运行超过六次，存在一一对应。 41 { 42 break; //退出循环 43 } 44 } 45 if (i == 25) //i==25说明循环超过二十四次，未找到一一对应项，False 46 { 47 printf("FALSE\n"); 48 } 49 else 50 { 51 printf("TRUE\n"); 52 } 53 } 54 return 0; 55 }
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原文地址：http://www.cnblogs.com/ghost-song/p/4457666.html
