A.charB.intC.floatD.list答案：A 解析：python里面都是字符串

A.*iB.3_1C.AI!D.Templist答案：D  解析：命名规则不能有特殊符号、不能数字开头

A.'abc"ab"B.'abc'ab'C."abc"ab"D."abc\"ab"答案：D  解析：\转义字符

A.dict1={}B.dict2={3:5}C.dict3={[1,2,3]:"uestc"}D.dict4={(1,2,3):"uestc"}答案：C  解析：Python不支持字典的key为list或dict类型list和dict类型是unhashable（不可哈希）的

A.IB.PC.YD.L答案：A  解析：Is[2][-1][1] 找到Is里面的第二个元素【10，"LIST"】-1:代表最后一个，然后找到下标1对应的I

s=input()if s.endswith("er"):    s=s.rstrip("er")#去掉右边的结尾erelif s.endswith("ly"):    s=s.rstrip("ly")elif s.endswith("ing"):#判断结束位置是否是ing    s=s.rstrip("ing")print(s)

n=int(input())cnt=0for i in range(1,n+1):    s=0    for j in range(1,i):        if i%j==0:            s+=j#完成累加功能    if s==i:#如果满足题意则累加和打印输出        cnt+=1        print(i)print("*"+str(cnt))

d=int(input())s=d*d/4print("{:.2f}".format(s))#格式化输出

n=input()d1={"a":"01","b":"02","c":"03","d":"04",    "e":"05","f":"06","g":"07","h":"08",    "i":"09","j":"10","k":"11","l":"12",    "m":"13","n":"14","o":"15","p":"16",    "q":"17","r":"18","s":"19","t":"20",    "u":"21","v":"22","w":"23","x":"24",    "y":"25","z":"26"}d2={"0":"27","1":"28","2":"29","3":"30",    "4":"31","5":"32","6":"33","7":"34",    "8":"35","9":"36"," ":"00"}
s1=""s2=""for i in n:    if i in d1:        s1+=d1[i]    else:        s1+=i#空格情况for i in s1:    if i in d2:        s2+=d2[i]    else:        s2+=i#空格情况print(s2)

#思路：通过对比天数大小判断节气#月份-天数month_days=[31,29,31,30,31,30,31,31,30,31,30,31]#节气jq=["LC","YS","JZ","CF",    "QM","GY","LX","XM",    "MZ","XZ","XS","DS",    "LQ","CS","BL","QF",    "HL","SJ","LD","XX",    "DX","DZ","HX","DH"]#节气-天数days=[35,50,65,80,      95,110,126,141,      157,173,188,204,      220,235,251,266,      282,297,313,328,      343,357,371,386]
#输入字符串s=input()#分割成列表a=s.split("*")#建立空列表t=[]for i in a:    t.append(int(i))#年-月-日year=t[0]month=t[1]date=t[2]
cnt=0#天数计数#统计总天数if year==2020:    for i in range(month-1):#当月之前的统计出来        cnt+=month_days[i]    cnt+=date#累加这个月的天数elif year==2021:    cnt=366+date#因为最多到下年一月份,可以直接统计天数#筛选for i in range(len(days)):    if days[i]==cnt:#匹配        print(jq[i])        break    elif cnt<days[i]:#小于下一个节气，输出下一个节气        print(jq[i])        break