#!/usr/bin/env python
from operator import add, sub
from random import randint, choice
ops = {'+': add, '-':sub}
#定义一个字典
MAXTRIES = 2
def doprob():
op = choice('+-')
#用choice从'+-'中随意选择操作符
nums = [randint(1,10) for i in range(2)]
#用randint(1,10)随机生成一个1到10的数,随机两次使用range(2)
nums.sort(reverse=True)
#按升序排序
ans = ops[op](*nums)
#利用函数
pr = '%d %s %d = ' % (nums[0], op, nums[1])
oops = 0
#oops用来计算failure测试,当三次时自动给出答案
while True:
try:
if int(raw_input(pr)) == ans:
print 'correct'
break
if oops == MAXTRIES:
print 'answer
%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):
break
if __name__ == '__main__':
main()
运行结果如下：
8 - 1 = 7 correct Again? [y]y 7 - 1 = 6 correct Again? [y]y 9 + 4 = 0 incorrect... try again 9 + 4 =
希望本文所述对大家的Python程序设计有所帮助。
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