Python 3.x Google Kickstart 2020第E轮运行时错误

为什么在GoogleIDE中测试代码时会出现RE（运行时错误）

这是整个pastbin:

下面是我的Python代码的链接，其中包含输入和输出，运行良好：

这里是我的代码的副本，以防你不想麻烦检查一个外部链接

def Solution(cases_nbr):
    ans=[] # Output list
    
    for case in range(cases_nbr):
        dico={}
        input() # this is not needed
        # exmpl tmp_ls=10 7 4 6 8 10 11 will be -> 10 7 4 6 8 11 as int each
        tmp_ls=sorted(set(map(int,input().split()))) 
        for number in tmp_ls:
            # create a new key (number) on dico which holds empty list.
            dico[number]=[] 
            # loop through each uniq number in tmp_ls
            for cmp_n in tmp_ls:
                # check if the current cmp_n is not the current key (number)
                # and also check if its not alreay a key in dico
                if cmp_n!=number and cmp_n not in dico.keys():
                    # that will basically append the list : dico[key]
                    # with the Absolute diffrence between current number and cmp_n
                    dico[number].append(abs(number-cmp_n))
        
        # at the end of this loop we will have something like :
        # since the example i gave was 10 7 4 6 8 11 then the generated dico will be :
        
        #{4: [2, 3, 4, 6, 7], 6: [1, 2, 4, 5], 7: [1, 3, 4], 8: [2, 3], 10: [1], 11: []}
        #That way i will be able to find the longest contiguous arithmetic subarray
        # by simply checking the most found value in each list in dico
        
        tmp=[0,0]
        
        # i take the first list since it has the all the substraction results
        for diff in dico[list(dico.keys())[0]]:
            # then i loop through each value(diff) and check if it exists in the rest of the lists
            #tmp_count will end up having the count of diff in the other lists
            tmp_count=len([1 for k in dico.keys() if diff in dico[k]])
            # this if check if there is a longest count then it stores it in tmp[diff, count_val]
            if tmp[1]<tmp_count:tmp=[diff, tmp_count]
        
        # as we said
        # the resulted dico is :
        #{4: [2, 3, 4, 6, 7], 6: [1, 2, 4, 5], 7: [1, 3, 4], 8: [2, 3], 10: [1], 11: []}
        #basically the result should be len([4 6 8 10]) because thats the longest contiguous arithmetic subarray
        # However tmp now would have only : tmp[2,3] wheres it should have tmp[2,4]
        # that why i loop again through the keys and check if there is no tmp[0] which is 'diff' 
        # in that dico[key] list , for example : 10: [1]
        for _ in dico.keys():
            if tmp[0] not in dico[_]:
                # if so , i re-loop through [10 7 4 6 8 11] and test if  that 
                # abs(x-key) is equal to tmp[0] , in this case :
                # abs(10-6) is indeed == 2 , so i simply increment tmp[1] 
                tmp[1]+=len([_ for x in tmp_ls if x != _ and abs(x-_)==tmp[0] ])
        
        # that way i will end up having the correct result .
        
        # simply i append anser list with the formatted string output they said .
        ans.append(f'Case #{case+1}: {tmp[1]}')
    return ans

if __name__  == '__main__':
    cases_nbr=int(input())
    for ans in Solution(cases_nbr):
        print(ans)
    

def溶液（案例编号）：
ans=[]输出列表
对于范围内的案例（案例编号）：
dico={}
input（）#这是不需要的
#exmpl tmp_ls=10 7 4 6 8 10 11将->10 7 4 6 8 11作为int
tmp_ls=sorted（set（map（int，input（）.split（）））
对于tmp_ls中的编号：
#在保存空列表的dico上创建一个新密钥（编号）。
dico[编号]=[]
#通过tmp_ls中的每个uniq编号循环
对于tmp中的cmp\u n：
#检查当前cmp\n是否不是当前密钥（编号）
#同时也要检查它是否真的是dico中的一个键
如果是的话=数字和cmp\n不在dico.keys（）中：
#这将基本上附加列表：dico[key]
#在当前数量和cmp\n之间存在绝对差异
dico[number].追加（abs（number-cmp\n））
#在这个循环的最后，我们将有如下内容：
#由于我给出的示例是10 7 4 6 8 11，因此生成的dico将是：
#{4: [2, 3, 4, 6, 7], 6: [1, 2, 4, 5], 7: [1, 3, 4], 8: [2, 3], 10: [1], 11: []}
#这样我就可以找到最长的连续算术子数组
#只需检查dico中每个列表中找到最多的值
tmp=[0,0]
#我选择第一个列表，因为它包含所有的减法结果
对于dico[列表（dico.keys（））[0]]中的差异：
#然后我循环遍历每个值（diff），并检查它是否存在于列表的其余部分中
#tmp_count最终会在其他列表中包含diff的计数
tmp_count=len（[1表示dico中的k.keys（），如果dico中的diff[k]]））
#此if检查是否存在最长计数，然后将其存储在tmp[diff，count\u val]
如果tmp[1]，我看不到可能发生错误的明显位置。您是否可以提供更详细的错误消息？您能否提供此代码试图解决的问题的描述？你能用注释来说明你的代码在每一步都做了什么，这样我们在搜索错误的原因时更容易理解吗？@GreenCoveGuy Done（我有时不擅长解释代码），请重新检查。