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问题：我需要掷 3 个骰子。如果两个（或三个）骰子返回相同的数字，则停止。如果 3 个骰子都是唯一的（例如 2、4 和 6），则再次掷骰子。执行此操作，直到掷出双倍/三倍，或 7 次，以先到者为准。

注意：我是 python 新手。

这是我到目前为止所拥有的，但这实际上是生成了 216 种可能的组合：

import itertools

all_possible = list(itertools.permutations([1,2,3,4,5,6],3))
input = raw_input()

print all_possible

这会产生这种类型的输出：

[(1, 2, 3), (1, 2, 4), (1, 2, 5), (1, 2, 6), (1, 3, 2), (1, 3, 4), (1, 3, 5), (1, 3, 6), (1, 4, 2), (1, 4, 3), (1, 4, 5), (1, 4, 6), (1, 5, 2), (1, 5, 3), (1, 5, 4), (1, 5, 6), (1, 6, 2), (1, 6, 3), (1, 6, 4), (1, 6, 5), (2, 1, 3), (2, 1, 4), (2, 1, 5), (2, 1, 6), (2, 3, 1), (2, 3, 4), (2, 3, 5), (2, 3, 6), (2, 4, 1), (2, 4, 3), (2, 4, 5), (2, 4, 6), (2, 5, 1), (2, 5, 3), (2, 5, 4), (2, 5, 6), (2, 6, 1), (2, 6, 3), (2, 6, 4), (2, 6, 5), (3, 1, 2), (3, 1, 4), (3, 1, 5), (3, 1, 6), (3, 2, 1), (3, 2, 4), (3, 2, 5), (3, 2, 6), (3, 4, 1), (3, 4, 2), (3, 4, 5), (3, 4, 6), (3, 5, 1), (3, 5, 2), (3, 5, 4), (3, 5, 6), (3, 6, 1), (3, 6, 2), (3, 6, 4), (3, 6, 5), (4, 1, 2), (4, 1, 3), (4, 1, 5), (4, 1, 6), (4, 2, 1), (4, 2, 3), (4, 2, 5), (4, 2, 6), (4, 3, 1), (4, 3, 2), (4, 3, 5), (4, 3, 6), (4, 5, 1), (4, 5, 2), (4, 5, 3), (4, 5, 6), (4, 6, 1), (4, 6, 2), (4, 6, 3), (4, 6, 5), (5, 1, 2), (5, 1, 3), (5, 1, 4), (5, 1, 6), (5, 2, 1), (5, 2, 3), (5, 2, 4), (5, 2, 6), (5, 3, 1), (5, 3, 2), (5, 3, 4), (5, 3, 6), (5, 4, 1), (5, 4, 2), (5, 4, 3), (5, 4, 6), (5, 6, 1), (5, 6, 2), (5, 6, 3), (5, 6, 4), (6, 1, 2), (6, 1, 3), (6, 1, 4), (6, 1, 5), (6, 2, 1), (6, 2, 3), (6, 2, 4), (6, 2, 5), (6, 3, 1), (6, 3, 2), (6, 3, 4), (6, 3, 5), (6, 4, 1), (6, 4, 2), (6, 4, 3), (6, 4, 5), (6, 5, 1), (6, 5, 2), (6, 5, 3), (6, 5, 4)]

这也不是很好，因为它只产生没有双倍或三倍 - 据我所知，一切都只是独特的组合。

----------更新----------- 好的--我拿了这个并通过从数组中剥离每个值并将它们求和来稍微扩展它（可能至少可能的有效方式）。它可以工作，如果在休息之前生成了多个集合，它们都会打印出来。我现在要做的是总结。所以：

def gen_random_termagants():
for _ in range(7): 
    # instead of three separate variables, we use a list here
    # the nice thing is, that you can freely vary the number of
    # 'parallel' dice rolls this way 
    dice = [random.randint(1, 6) for _ in range(3)]

    # this is more general and will break as soon as there are
    # duplicate (die) values in the list (meaning, break, if not all elements 
    # are different)
    first_die = dice[0]
    second_die = dice[1]
    third_die = dice[2]
    total_term = first_die + second_die + third_die
    print "Total: %s" % total_term
    if len(dice) > len(set(dice)):
        break

这是输出示例：

How many tervigons? ::>3
Let's calculate some termagants based on 3 tervigons...
You'll get a minimum of 9 termagants per turn.
You'll get a maximum of 54 termagants per turn.
minimums: 5 turns [45] :: 6 turns [54] :: 7 turns [63]
averages: 5 turns [157] :: 6 turns [189] :: 7 turns [220]
maximums: 5 turns [270] :: 6 turns [324] :: 7 turns [378]
Total: 9
Total: 8

所以在这个例子中，我希望它返回 17（即 9 + 8）。

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2 回答 2

7

Python 带有一个很棒的标准库（正如您在使用itertools时可能已经发现的那样），您还可以在其中找到一个随机模块。

您可以使用random.randint来模拟掷骰子。有多种方法可以解决这个问题。第一个代码示例有些有限，第二个更通用。

import random

# '_' (underscore) is used for values that are generated, but that you do not
# care about - here we only want to repeat seven times and do not care about
# the actual loop count 
for _ in range(7): 
    # generate three random numbers between [1 and 6] 
    # and store the values in a, b, c respectively (tuple unpacking)
    a, b, c = (random.randint(1, 6) for _ in range(3))

    # if one of the conditions holds, break out of the loop early
    if a == b or a == c or b == c or a == b == c:
        break

正如@Paulo 指出的那样，您可以使用另一种更简洁的方法来检查n列表（或元组）的元素是否都不同，即将所有元素放在一个集合中：

for _ in range(7): 
    # instead of three separate variables, we use a list here
    # the nice thing is, that you can freely vary the number of
    # 'parallel' dice rolls this way 
    dice = [random.randint(1, 6) for _ in range(3)]

    # this is more general and will break as soon as there are
    # duplicate (die) values in the list (meaning, break, if not all elements 
    # are different)
    if len(dice) > len(set(dice)):
        break

回答您更新的问题，只需使用sum：

    total = sum(dice)
于 2013-09-06T21:24:43.367 回答
1

那是输入排列的正确输出，您要的是笛卡尔积。见产品。

EG 这将打印您期望的输出。

from itertools import product

d = [ x for x in range(1, 7) ]
for r in product(d, d, d):
    print r
于 2013-09-06T21:26:22.133 回答