python——模拟抓狐狸的小游戏

模拟抓狐狸的小游戏

假设一共有一排5个洞口，小狐狸最开始的时候在其中一个洞口，然后人随机打开一个洞口，如果里面有小狐狸就抓到了。如果洞口里没有小狐狸就明天再来抓，但是第二天小狐狸会在有人来抓之前跳到隔壁洞口里。

思路如下：

代码

"""编写程序模拟抓狐狸的小游戏。假设一共有一排5个洞口，小狐狸最开始的时候在其中一个洞口，
然后人随机打开一个洞口，如果里面有小狐狸就抓到了。如果洞口里没有小狐狸就明天再来抓，
但是第二天小狐狸会在有人来抓之前跳到隔壁洞口里。截图上传。"""
import random


def fox_hole(holes):
    fox_locate = random.choice(range(1, 5))  # 狐狸在随机洞中
    holes[fox_locate - 1] = 1  # 狐狸在洞中留下痕迹
    grep(fox_locate, holes)


def grep(fox_locate, all_holes):
    direction = [-1, 1]
    while True:
        prediction = int(input("请随机掏个窝 按（q）退出"))
        while prediction != 'q':
            if fox_locate == prediction - 1:
                print("Bingo，你在第 " + str(prediction) + "个洞抓到了狐狸")
                return False
            else:
                next_direction = random.choice(direction)
                # 狐狸下一步有两个方向-1 和 1
                if next_direction == -1 and fox_locate - 1 != 0:
                    fox_locate = fox_locate - 1
                    print("哎呀没抓到，狐狸在第 " + str(fox_locate) + "个洞")
                    break
                # 如果到最左端，就向右移动
                elif next_direction == -1 and fox_locate - 1 == 0:
                    fox_locate = fox_locate + 1
                    print("哎呀没抓到，狐狸在第 " + str(fox_locate) + "个洞")
                    break
                elif next_direction == 1 and fox_locate + 1 != len(all_holes):
                    fox_locate = fox_locate + 1
                    print("哎呀没抓到，狐狸在第 " + str(fox_locate) + "个洞")
                    break
                elif next_direction == 1 and fox_locate + 1 == len(all_holes):
                    fox_locate = fox_locate - 1
                    print("哎呀没抓到，狐狸在第 " + str(fox_locate) + "个洞")
                    break
        else:
            break


hole = [0] * 5  # 5个洞
fox_hole(hole)