【PAT_Python解】1014 福尔摩斯的约会

原题链接：PTA | 程序设计类实验辅助教学平台

Tips：以下Python代码仅个人理解，非最优算法，仅供参考！

ls=[]   #装输入数据，你也可以S1,S2,S3,S4=input(),···
D,H,M='','',''
dict = {'A':'MON','B':'TUE','C':'WED','D':'THU',
         'E':'FRI','F':'SAT','G':'SUN'}
#用ord编码，有固定逻辑，也可以用下面字典匹配
# dict1 = {'A':10,'B':11,'C':12,'D':13,'E':14,'F':15,
#          'G':16,'H':17,'I':18,'J':19,'K':20,'L':21,
#          'M':22,'N':23,'O':24}
for i in range(4):  #初始4行输入数据
    ls.append(input())
#Day判断
_Dindex=0
ran = min(len(ls[0]),len(ls[1]))
for i in range(ran):
    if ls[0][i]==ls[1][i] and 'A' <= ls[0][i] <= 'G':
        D=dict[ls[0][i]]
        _Dindex=i
        break
#HH判断
for i in range(_Dindex+1,ran):
    if ls[0][i]==ls[1][i] and ('0'<=ls[0][i]<='9' or 'A'<=ls[0][i]<='N'):
            if '0'<=ls[0][i]<='9':
                H=int(ls[0][i])
                break
            elif 'A'<=ls[0][i]<='N':
                H=ord(str(ls[0][i]))-55
                #H=dict1[ls[0][i]]  #配合dict1
                break
#MM判断
ran = min(len(ls[2]),len(ls[3]))
for i in range(ran):
    if ls[2][i]==ls[3][i] and ('a'<=ls[2][i]<='z' or 'A'<=ls[2][i]<='Z'):
        M=i
        break
print('{} {:02d}:{:02d}'.format(D,H,M),end='')