Cube digit pairs
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?
立方體數字對
在一個立方體的六個面上分别标上不同的數字(從0到9),對另一個立方體也如法炮制。将這兩個立方體按不同的方向并排擺放,我們可以得到各種各樣的兩位數。
例如,平方數64可以通過這樣擺放獲得:
事實上,通過仔細地選擇兩個立方體上的數字,我們可以擺放出所有小于100的平方數:01、04、09、16、25、36、49、64和81。
例如,其中一種方式就是在一個立方體上标上{0, 5, 6, 7, 8, 9},在另一個立方體上标上{1, 2, 3, 4, 8, 9}。
在這個問題中,我們允許将标有6或9的面颠倒過來互相表示,隻有這樣,如{0, 5, 6, 7, 8, 9}和{1, 2, 3, 4, 6, 7}這樣本來無法表示09的标法,才能夠擺放出全部九個平方數。
在考慮什麼是不同的标法時,我們關注的是立方體上有哪些數字,而不關心它們的順序。
{1, 2, 3, 4, 5, 6}等價于{3, 6, 4, 1, 2, 5}
{1, 2, 3, 4, 5, 6}不同于{1, 2, 3, 4, 5, 9}
但因為我們允許在擺放兩位數時将6和9颠倒過來互相表示,這個例子中的兩個不同的集合都可以代表拓展集{1, 2, 3, 4, 5, 6, 9}。
對這兩個立方體有多少中不同的标法可以擺放出所有的平方數?
解題
我發現這個翻譯我了解不透
在兩個六面體上面塗:0-9的數字,主要這裡有10個數字,隻用其中6個圖,兩個六面體塗的數字可以不一樣的,6可以當9用,9可以當6用。
兩個六面體上面的數字能組合成:1-9的平方:01 04 09 16 25 36 49 64 81 ,求這樣的塗法有多少種?
骰子說成六面體還吊的。
0-9 十個數 取出6個 就是一個骰子的塗法。
先組合出塗法的種類。
再判斷是否能組成1-9的平方
參考程式
# coding=gbk
import time as time
from itertools import combinations
def run():
dice=list(combinations([0,1,2,3,4,5,6,7,8,6],6))
ans = 0
for i1,d1 in enumerate(dice):
for d2 in dice[i1:]:
if valid(d1,d2) == True:
ans +=1
print ans
def valid(c1,c2):
squares=[(0,1),(0,4),(0,6),(1,6),(2,5),(3,6),(4,6),(8,1)]
return all(x in c1 and y in c2 or x in c2 and y in c1 for x,y in squares)
t0 = time.time()
run()
t1 = time.time()
print "running time=",(t1-t0),"s"
# 1217
# running time= 0.0620000362396 s