Each of the six faces on a cube has a different digit (0 to 9) written on it; the same is done to a second cube. By placing the two cubes side-by-side in different positions we can
form a variety of 2-digit numbers.
For example, the square number 64 could be formed:
In fact, by carefully choosing the digits on both cubes it is possible to display all of the square numbers below one-hundred: 01, 04, 09, 16, 25, 36, 49, 64, and 81.
For example, one way this can be achieved is by placing {0, 5, 6, 7, 8, 9} on one cube and {1, 2, 3, 4, 8, 9} on the other cube.
However, for this problem we shall allow the 6 or 9 to be turned upside-down so that an arrangement like {0, 5, 6, 7, 8, 9} and {1, 2, 3, 4, 6, 7} allows for all nine square numbers
to be displayed; otherwise it would be impossible to obtain 09.
In determining a distinct arrangement we are interested in the digits on each cube, not the order.
{1, 2, 3, 4, 5, 6} is equivalent to {3, 6, 4, 1, 2, 5}
{1, 2, 3, 4, 5, 6} is distinct from {1, 2, 3, 4, 5, 9}
But because we are allowing 6 and 9 to be reversed, the two distinct sets in the last example both represent the extended set {1, 2, 3, 4, 5, 6, 9} for the purpose of forming 2-digit
numbers.
How many distinct arrangements of the two cubes allow for all of the square numbers to be displayed?
本来想用C++写，但是set<vector<int>>太磨人了┑(￣Д ￣)┍
还是python好
from itertools import combinations
squares=[(0,1),(0,4),(0,6),(1,6),(2,5),(3,6),(4,6),(8,1)]
dice=list(combinations([0,1,2,3,4,5,6,7,8,6],6))
def valid(c1,c2):
return all(x in c1 and y in c2 or x in c2 and y in c1 for x,y in squares)
ans=sum(1 for i,c1 in enumerate(dice) for c2 in dice[:i] if valid(c1,c2))
print('ans = ',ans)
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