s = (die1==1)+(die2==1)+(die3==1)+(die4==1)+(die5==1)
if s >= 3:
print('yes')
else:
s = (die1==2)+(die2==2)+(die3==2)+(die4==2)+(die5==2)
if s >= 3:
print('yes')
else:
s = (die1==3)+(die2==3)+(die3==3)+(die4==3)+(die5==3)
if s >= 3:
print('yes')
else:
s = (die1==4)+(die2==4)+(die3==4)+(die4==4)+(die5==4)
if s >= 3:
print('yes')
else:
s = (die1==5)+(die2==5)+(die3==5)+(die4==5)+(die5==5)
if s >= 3:
print('yes')
else:
s = (die1==6)+(die2==6)+(die3==6)+(die4==6)+(die5==6)
if s >= 3:
print('yes')
else:
print('no')
快速更新。谢谢大家的快速回答。在另一个论坛上给出我的答案: Counting repeated characters in a string in Python 我是这样做的: DICE值=str(die1)、str(die2)、str(die3)、str(die4)、str(die5) 打印(值)
对于值中的项: 计数=DICE值。计数(项目) 打印(计数)
不带数组:
在找到更好的解决方案后编辑我的帖子:
我知道对一个大于等于3的数字执行循环或显式字符串/整数搜索会更快,但在这种情况下,隐式数组循环是可能的:-)
当然可以。使用
collections.Counter
:严格地说,这确实使用了
list
(我认为这就是您的意思;数组完全是另一回事),因此它取决于您的限制的确切含义。你可以把它改成tuple
。。。你知道吗相关问题 更多 >
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