📅  最后修改于: 2023-12-03 14:58:33.013000             🧑  作者: Mango
这是Sudo GATE 2020 Mock I(2019年12月27日)的第33章“门”的题目。在这道题目中,我们需要实现一个门的开关过程,而门的开关过程依赖于门前的开关状态和门背后的开关状态之间的逻辑关系。
实现一个函数,名为 door
,该函数具有两个形参:
status_front
– 前面的开关状态,可能为字符串 "Open"
或 "Closed"
status_back
– 背面的开关状态,可能为字符串 "Open"
或 "Closed"
函数调用返回值应该是:
"Closed"
,门状态为 "Closed"
。"Open"
,门状态为 "Open"
。"Closed"
,而另一个是 "Open"
,门的状态为 "Open"
。函数应该满足以下额外的约束条件:
"Closed"
。"Open"
。输入格式: 两个字符串:status_front
和 status_back
,分别表示门前和背后的状态。
输出格式: 一个字符串,表示门的状态,可能为 "Open"
或 "Closed"
。
输入:status_front = "Closed", status_back = "Closed"
输出:"Closed"
输入:status_front = "Open", status_back = "Open"
输出:"Open"
输入:status_front = "Open", status_back = "Closed"
输出:"Open"
输入:status_front = "Closed", status_back = "Open"
输出:"Open"
输入:status_front = "Open", status_back = "Open outside disturbance"
输出:"Closed"
输入:status_front = "Closed", status_back = "Closed outside disturbance"
输出:"Open"
实现门开关的逻辑判断需要注意以下几点:
针对上述要求,我们可以使用以下的实现方式:
def door(status_front, status_back):
if status_front == "Closed" and status_back == "Closed":
return "Closed"
elif status_front == "Open" and status_back == "Open":
return "Open"
elif status_front == "Open" or status_back == "Open":
return "Open"
elif status_front == "Open outside disturbance" and status_back == "Open":
return "Closed"
elif status_front == "Closed outside disturbance" and status_back == "Closed":
return "Open"
部分简单到不需要解释,简单说明下其他的几个判断条件:
"Closed"
,而另一个是 "Open"
,门的状态为 "Open"
。"Closed"
。"Open"
。本题需要考虑门开关状态的逻辑关系,需要处理多种开关情况下的门状态变化,要注意针对特殊情况的额外处理。