Question

更新：我已经确定当try函数失败时发生错误，我想如果我修改代码不使用try和box函数返回false，如果它没有工作可能会解决问题

所以我试图弄清楚关闭盒子的胜利概率 http://www.mastersgames.com/rules/shut-box-rules.htm

我已经完成了大部分工作，所以这个功能关闭了＃34;插槽＆＃34;使用尽可能多的策略。但现在我试图通过使用递归来解决每个骰子滚动序列的组合问题。

问题：列表last_current仅在标记为1的点上进行编辑，然后使用标记＆＃34; LCC＆＃34;进行打印，但是稍后在使用标记＆＃34; LC＆＃34打印时;，在第2点，列表中的第一个元素已被删除。

global Total_Count
Total_Count = 0
Shut_Count = 0
Startlist = ["1","2","3","4","5","6","7","8","9"]
shutting = []


def box(n,current):
    if str(n) in current:
        current.remove(str(n))
        for i in range(len(shutting)):
            current.remove(shutting[i])
            shutting.pop()
        return current
    else:
        for i in range(n):
            if str(n-i) in current:
                shutting.append(str(n-i))
                return box(i,current)
                break
            else:
                pass


def recurs(Dice_1,Dice_2,current):
    global Total_Count
    for i in range(Dice_1):
        for j in range(Dice_2):
            print(i,j,current)
       1)   last_current = current
            print("LCC",last_current)
            try:
                n = i + j + 2
                print(n,current)
                new = box(n,current)
                current = new
                print("Try",current)
            except:
       2)       print("LC",last_current)
                current = last_current
                Total_Count += 1
                print("Total",Total_Count,"Current",current)
            else:
                #print(current)
                if range(len(current)) == 0:
                    Total_Count += 1
                    Shut_Count += 1
                    print("Total",Total_Count,"Shut",Shut_Count)
                else:
                    recurs(Dice_1,Dice_2,current)

recurs(6,6,Startlist)
print(Total_Count,Shut_Count)

忽略可怕的编码和使用休息时间，我也不希望别人解决问题只是为什么代码破解的解决方案

关闭Box解决方案，列出神奇地改变 - python

0 个答案: