嘿，我一眼就看到你代码里的问题啦——条件判断的逻辑写法完全错误，这就是为啥输入A-C会得到100的根本原因！咱们一步步拆解：

核心错误点 #

1. 错误的条件表达式
   你写的city_1=="A" or "B"其实等价于(city_1=="A") or "B"，而字符串"B"在Python的布尔判断里会被视为True，所以这个条件永远成立！不管你输入什么城市，第一个if都会被触发，自然会得到错误的距离值。

   同理，city_1 or city_2 != "A" or "B" or "C"这种写法逻辑完全混乱，根本无法正确判断输入的城市是否合法。

2. 代码结构隐患
   第二个if会独立于前面的elif执行，就算已经正确计算出distance，只要满足这个错误的条件，还是会打印错误提示；而且如果输入无效，distance变量根本没定义，后面打印会直接报错。

修正方案1：修复条件判断逻辑 #

先把所有条件表达式补全，同时优化输入处理（比如支持大小写、去除空格），调整代码结构避免变量未定义：

# 定义各城市间距离
AB = 100
AC = 150
CB = 80

# 处理用户输入：去除空格+转大写，兼容大小写输入
city_1 = input("Enter city 1 ").strip().upper()
city_2 = input("Enter city 2 ").strip().upper()

distance = None  # 初始化distance，避免未定义报错

# 先判断无效情况
if city_1 == city_2:
    print("You did not enter a viable match.")
elif city_1 not in {"A", "B", "C"} or city_2 not in {"A", "B", "C"}:
    print("You did not enter a viable match.")
# 再判断有效城市组合
elif (city_1 == "A" and city_2 == "B") or (city_1 == "B" and city_2 == "A"):
    distance = AB
elif (city_1 == "A" and city_2 == "C") or (city_1 == "C" and city_2 == "A"):
    distance = AC
elif (city_1 == "B" and city_2 == "C") or (city_1 == "C" and city_2 == "B"):
    distance = CB

# 只有有效情况才打印距离
if distance is not None:
    print(f"The distance between {city_1} and {city_2} is {distance} miles.")

修正方案2：用字典简化逻辑（更推荐） #

如果城市数量变多，多条件if会越来越繁琐，用字典存储所有合法的城市对距离，查询起来更简洁高效：

# 用字典存储所有双向城市对的距离
distance_map = {
    ("A", "B"): 100,
    ("B", "A"): 100,
    ("A", "C"): 150,
    ("C", "A"): 150,
    ("B", "C"): 80,
    ("C", "B"): 80
}

city_1 = input("Enter city 1 ").strip().upper()
city_2 = input("Enter city 2 ").strip().upper()

# 判断输入是否合法
if city_1 == city_2 or city_1 not in {"A", "B", "C"} or city_2 not in {"A", "B", "C"}:
    print("You did not enter a viable match.")
else:
    # 直接从字典中获取距离
    distance = distance_map[(city_1, city_2)]
    print(f"The distance between {city_1} and {city_2} is {distance} miles.")

这两种方案都能解决你遇到的问题，第二种更易维护，以后加新城市只要往字典里加新键值对就行~

内容的提问来源于stack exchange，提问作者shymilk