Question

如果您的第一轮掷骰数是7或11，那么您将获胜。如果是2或3或12，那么您输了。如果是其他任何数字，则必须再次滚动，直到击中该数字或按7或11退出。

我遇到的问题基本上是打了一个数字，但我需要程序来保存随机选择的数字并重新开始。并且如果它重新开始并命中了另一个数字，而没有跳开，则会再次开始直到该数字被击中或突然消失为止。

对理论和基本的基础编程的一些解释确实可以帮助我。我不确定从已经开始的地方去哪里。

Python：

<?xml version="1.0" encoding="utf-8"?>
<android.support.v4.widget.DrawerLayout xmlns:android="http://schemas.android.com/apk/res/android"
    xmlns:app="http://schemas.android.com/apk/res-auto"
    xmlns:tools="http://schemas.android.com/tools"
    android:id="@+id/home_screen_main_drawer_layout"
    android:layout_width="match_parent"
    android:layout_height="match_parent"
    android:fitsSystemWindows="true"
    tools:context=".HomeScreen">

    <RelativeLayout
        android:layout_width="match_parent"
        android:layout_height="match_parent">

        <!--Top Layout-->

        <RelativeLayout
            android:layout_width="match_parent"
            android:layout_height="120dp"
            android:layout_alignParentTop="true"
            android:background="@color/colorPrimary">

            <android.support.v7.widget.Toolbar
                android:id="@+id/drawer_menu_toolbar"
                android:layout_width="match_parent"
                android:layout_height="?attr/actionBarSize"
                android:theme="@style/ThemeOverlay.AppCompat.Dark.ActionBar">

                <FrameLayout
                    android:id="@+id/drawer_menu_fragment_container"
                    android:layout_width="match_parent"
                    android:layout_height="match_parent" />


                <android.support.v7.widget.Toolbar
                    android:id="@+id/search_toolbar"
                    android:layout_width="match_parent"
                    android:layout_height="40dp"
                    android:layout_alignParentBottom="true"
                    android:layout_marginStart="8dp"
                    android:layout_marginEnd="8dp"
                    android:layout_marginBottom="10dp"
                    android:background="@android:"


                <EditText
                    android:id="@+id/searchHere"
                    android:layout_width="match_parent"
                    android:layout_height="40dp"
                    android:background="@drawable/seachbar_homescreen"
                    android:hint="Search Here"
                    android:textSize="15sp" />

            </android.support.v7.widget.Toolbar>
        </RelativeLayout>


    </RelativeLayout>

    <!--<android.support.design.widget.NavigationView
        android:id="@+id/home_screen_navigation_view"
        android:layout_width="wrap_content"
        android:layout_height="match_parent"
        android:layout_alignParentStart="true"
        android:layout_gravity="start"
        android:visibility="gone"
        app:headerLayout="@layout/home_screen_menu_drawer_header"
        app:menu="@menu/home_screen_menu_drawer" />
-->
    <android.support.design.widget.NavigationView
        android:id="@+id/nav_view"
        android:layout_width="wrap_content"
        android:layout_height="match_parent"
        android:layout_gravity="start"
        app:headerLayout="@layout/home_screen_menu_drawer_header"
        app:menu="@menu/home_screen_menu_drawer" />

</android.support.v4.widget.DrawerLayout>

我只希望程序像“好，这是滚动的数字。我要记住这一点，然后再次开始查看它是否可以击中相同的数字。”

Answer 1

您似乎丢失的实际上是在循环中将骰子滚动。

通常的想法是掷骰子一次，并检查立即赢或输。

如果不是瞬间赢/输，然后进入循环，连续掷骰子直到赢或输。< / p>

以下伪代码^（a）（带有注释）显示了如何执行此操作：

# Store first throw.

firstThrow = rnd(1..6) + rnd(1..6)
print "First throw " firstThrow

# Set winner if house or player wins on first throw.

winner = ''
if firstThrow is one of (7, 11):            winner = 'player'
else if firstThrow is one of (2, 3, 12):    winner = 'house'

# Loop until a winner found (may have already happened above).

while winner == '':
    # Get subsequent throw.

    throw = rnd(1..6) + rnd(1..6)
    print "Next throw " throw

    # Set winner (exits loop) if win or lose.

    if throw is one of (7, 11):             winner = 'house'
    else if throw is same as firstThrow:    winner = 'player'

# Have a winner, output it.

print "Winner was " winner

^（a）对于几乎可以肯定是课堂作业的问题，我发现最好提供伪代码，并让提问者将其转换为他们选择的语言，因为这样做往往会使开发人员变得更好。

在这种情况下，转换为Python相对简单，但我敦促您自己尝试一下。如果您为此而苦苦挣扎，我在下面提供了一些Python代码（希望您不会一字不漏地使用它，因为您（或其他人）可能拥有的任何教育者，反正还是个坏主意，能够看到此答案并相应地评分）：

import random

# Store first throw.

firstThrow = random.randint(1, 6) + random.randint(1, 6)
print("First throw", firstThrow)

# Set winner if house or player wins on first throw.

if firstThrow in (7, 11):      winner = 'player'
elif firstThrow in (2, 3, 12): winner = 'house'
else:                          winner = ''

# Loop until a winner found (may have already happened above).

while winner == '':
    # Get subsequent throw.

    throw = random.randint(1, 6) + random.randint(1, 6)
    print("Next throw", throw)

    # Set winner (exits loop) if win or lose condition found.

    if throw in (7, 11):       winner = 'house'
    elif throw == firstThrow:  winner = 'player'

# Have a winner, output it.

print("Winner was",  winner)

这是一些示例运行供确认：

First throw 11
Winner was player

First throw 7
Winner was player

First throw 4
Next throw 7
Winner was house

First throw 9
Next throw 10
Next throw 7
Winner was house

First throw 11
Winner was player
First throw 10
Next throw 5
Next throw 11
Winner was house

First throw 8
Next throw 5
Next throw 11
Winner was house

First throw 8
Next throw 7
Winner was house

Answer 2

import random
numbers_occurred_till_now =[] # here we are going to store all the dice value that will occur.
while True:
    roll = random.randint(1, 6) + random.randint(1, 6)
    numbers_occurred_till_now.append(roll) # when the number is generated, we are appending it in the list.
    if roll == 7 or roll == 11:
        print('nice')
        print(numbers_occurred_till_now)
        break
    elif  roll == 3 or roll == 2 or roll == 12:
        print('crapped out')
        print(numbers_occurred_till_now)
        break
    elif numbers:
        print('your number is:', roll, "roll again...")
        print(numbers_occurred_till_now)

在Python中构建掷骰子表程序

2 个答案: