无法让python中的输出合理化

Question

在CodeJam2020的一个问题中，我得到以下输出

Case #1:    CJC
Case #2: IMPOSSIBLE
Case #3:      CJJCC
Case #4:   CC

同时，实际产出应该是这样的：

Case #1: CJC
Case #2: IMPOSSIBLE
Case #3: JCCJJ
Case #4: CC

有人能告诉我如何纠正这个问题吗？我正在使用Python3。 我的print语句如下：print('Case #{}: {}'.format(count, output))

下面是我的代码

def checkOverlap(task1,task2):
start1 = int(task1[0])
start2 = int(task2[0])
end1 = int(task1[1])
end2 = int(task2[1])

if(end2 > start1 and start2 < end1):
    return True
else:
    return False


T= input()
test = int(T)
countTest = 0
while(countTest<test):
    A = input()
    countA = int(A)
    lim = 0
    activity = []
    while(lim<countA):
        activity.append(input().split(" "))
        lim += 1

    flag = True
    output=' '*countA
    output+='C'
    J=[]
    C=[activity[0]]

   impossible = False
   for i in range(0,countA-1):
       for j in range(i+1,countA):
           if (C.__contains__(activity[i]) and J.__contains__(activity[j])) or \
               (J.__contains__(activity[i]) and C.__contains__(activity[j])):
               continue

           check = checkOverlap(activity[i],activity[j])
           if(check == False):
               if C.__contains__(activity[i]):
                   if not C.__contains__(activity[j]):
                       C.append(activity[j])
               else:
                   if not J.__contains__(activity[j]):
                       J.append(activity[j])
           else:
               if (C.__contains__(activity[i])):
                   if(C.__contains__(activity[j])):
                    impossible = True
                   else:
                       if not J.__contains__(activity[j]):
                           J.append(activity[j])
               else:
                   if J.__contains__(activity[j]):
                       impossible = True
                   else:
                       if not C.__contains__(activity[j]):
                           C.append(activity[j])

   if impossible == True:

       output = 'IMPOSSIBLE'


   else:
       for i in range(1,countA):
           if C.__contains__(activity[i]):
               output += 'C'
           else:
               output += 'J'
   count = (countTest + 1)
   print('Case #{}: {}'.format(count, output))



   countTest += 1

Answer 1

为什么代码中有这一行

    output=' '*countA

如果你摆脱它（换成，比如说，^{），你会更快乐。你要做的是用一堆空格开始字符串，然后打印出来。（您添加的所有其他内容都会被追加，因此最后的字符串将是"    CJJCC"。）这会破坏您想要的对齐方式

Answer 2

开头有output=' '*countA，在CJC和类似的情况下，你只做output += 'C'和output += 'J'，所以你会在字符串的开头找到countA空格