接著昨天的工作，今天來看《Python游戲編程快速上手》的第五章----龍穴探險。
這一章我將不會給出書中的代碼，我會直接給我我修改后的代碼。我增加了一些游戲的規(guī)則，如下：

• 管理員（編程人員）可以調(diào)節(jié)龍穴的數(shù)量，也可以調(diào)節(jié)安全龍穴的數(shù)量，當(dāng)然安全龍穴的數(shù)量必須不大于龍穴的數(shù)量，由于這個是由管理員設(shè)置的，所以無需加判斷

代碼還是非常簡單的，因為還沒有涉及UI，加上UI后會更有可玩性。
代碼如下;

import random import timenum = 3 safenum = 1def storyBegin():print("Now you are in a island full of dragons.")print("In front of you,there are ", num, " caves.")print("In ", safenum, " cave, the dragon is friendly and will share his treasure with you.")print("The others dragon is greedy and hungry,and will eat you on sight.")print()def chooseCave():print("Which one do you choose?(1 to ", num,")")cave = int(input())if cave > 0 and cave <= num:return caveelse:print("You can only choose in 1 to ", num)chooseCave()def checkCave(chooseC):print("You approach the cave...")time.sleep(2)print("It is dark and spooky...")time.sleep(2)print("A large dragon jumps out in front of you!He opens his jaws and ...")time.sleep(2)print()friendCave = random.sample(range(1, num), safenum)if chooseC in friendCave:print("Gives you his treasure!")else:print("Gobbles you down in one bite!")if __name__ == "__main__":while True:storyBegin()chooseC = chooseCave()checkCave(chooseC)print("Do you want to play again?(yes or no)")againP = input()if againP not in ["y", "yes"]:break

加油吧，少年！

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