本文實(shí)例講述了Python實(shí)現(xiàn)的簡(jiǎn)單算術(shù)游戲。分享給大家供大家參考。具體實(shí)現(xiàn)方法如下：
#!/usr/bin/env pythonfrom operator import add, sub from random import randint, choiceops = {'+': add, '-':sub}#定義一個(gè)字典MAXTRIES = 2 def doprob(): op = choice('+-') #用choice從'+-'中隨意選擇操作符 nums = [randint(1,10) for i in range(2)] #用randint(1,10)隨機(jī)生成一個(gè)1到10的數(shù),隨機(jī)兩次使用range(2) nums.sort(reverse=True) #按升序排序 ans = ops[op](*nums) #利用函數(shù) pr = '%d %s %d = ' % (nums[0], op, nums[1]) oops = 0 #oops用來(lái)計(jì)算failure測(cè)試,當(dāng)三次時(shí)自動(dòng)給出答案 while True: try: if int(raw_input(pr)) == ans: print 'correct' break if oops == MAXTRIES: print 'answer/n %s%d' % (pr, ans) break else: print 'incorrect... try again' oops += 1 except (KeyboardInterrupt, EOFError, ValueError): print 'invalid ipnut... try again'def main(): while True: doprob() try: opt = raw_input('Again? [y]').lower() if opt and opt[0] == 'n': break except (KeyboardInterrupt, EOFError): breakif __name__ == '__main__': main()運(yùn)行結(jié)果如下：
8 - 1 = 7correctAgain? [y]y7 - 1 = 6correctAgain? [y]y9 + 4 = 0incorrect... try again9 + 4 =
希望本文所述對(duì)大家的Python程序設(shè)計(jì)有所幫助。
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