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test1 的判定函数：

'''	@ sum(1,2,3,4) = sum(5,6,7,8) and 	@ sum(1,2,5,6) = sum(3,4,7,8) and	@ sum(1,3,5,7) = sum(2,4,6,8)	@ used for test 1'''def sumEqual(data):	eq1 = sum(data[0:4]) == sum(data[4:8])	eq2 = sum(data[i << 1] for i in range(4)) == \		  sum(data[i << 1 | 1] for i in range(4))	eq3 = sum(data[0:2]) + sum(data[4:6]) == \		  sum(data[2:4]) + sum(data[6:8])	return eq1 and eq2 and eq3

'''	@ for square matrix, the slop is +1 or -1, means	@ abs(p.x - q.x) == abs(p.y - q.y)	@ for data[], we make index i stand for row(coordinate x) i, and 	@ data[i] stand for col(coordinate y) data[i], so the conflict 	@ condition is: abs(i - j) == (data[i] - data[j])	@ used for test 2'''def NotConflict(data):	for i in range(len(data) - 1):		for j in range(i + 1, len(data)):			if abs(i - j) == abs(data[i] - data[j]):				return False	return True

cnt = 0 '''	@ get all the permutition of data and print ones that s.t. cf,'''def permutition(data, start, cf):	if len( data ) <= 1:		return 	if cf(data):		print data		global cnt;		cnt += 1	for i in range(start, len( data ) - 1):		for j in range( i + 1, len( data )):			data[i], data[j] = data[j], data[i]			permutition(data, i + 1, cf)			data[i], data[j] = data[j], data[i]if __name__ == '__main__':	data = [0,1,2,3,4,5,6,7,8]	permutition(data, 0, cf = NotConflict)	print cnt

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