标签:下划线 for 循环 位置 十进制 后退 打印 字符 NPU 不能
---恢复内容开始---
#实现用户输入用户名和密码,当用户名为 seven 或 alex 且 密码为 123 时,显示登陆成功,否则登陆失败,失败时允许重复输入三次 s = 3 while s>0: name = input("请输入用户名:").strip() password = input("请输入密码:").strip() if name == "seven" or name == "alex" and password =="123": print("登录成功") break else: print("登录失败,你还有%s次机会" %s) s -= 1
#使用 while 循环实现输出 2 -3 + 4-5 + 6 ... + 100的和 sum = 0 a = 2 while a<=100: if a%2==0: sum += a else: sum -= a a += 1 print(sum)
#使用 for 循环和 range 实现输的1 -2 + 3-4 + 5-6 ... + 99的和 sum = 0 for i in range(100): if i % 2 == 0: sum -= i else: sum += i print(sum)
#使用 while 循环实现输出 1,2,3,4,5,7,8,9,11,12 a = 1 while a<13: print(a) a += 1
#使用 while 循环实现输出 1-100 内的所有奇数 a = 1 while a <= 100: if a % 2 == 0: pass else: print(a) a += 1
#分别书写数字 5,10,32,7 的二进制表示 for i in [5,10,32,7]: print("{}的二进制数是{}".format(i,bin(i))) print("%s的二进制数是%s" %(i, bin(i)))
#输入一年份,判断该年份是否是闰年并输出结果。(编程题) #注:凡符合下面两个条件之一的年份是闰年。 (1) 能被4整除但不能被100整除。 (2) 能被400整除。 year = input("请输入您需要查询的年份:").strip() year = int(year) if year % 4 == 0 and year % 100 != 0 or year % 400 ==0: print("%d是闰年" %year)
#编写登陆接口 # 基础需求: # 让用户输入用户名密码 # 认证成功后显示欢迎信息 # 输错三次后退出程序 count = 3 while count > 0: name = input("帐号:") password = input("密码:") if name == "long" and password == "123": print("欢迎登录") break else: count -= 1 print("登录失败,你还有%d次机会" %count)
#如有一个变量n1 = 5,请使用int提供的方法,得到该变量最少可以用多少个二进制位表示? n1 = 5 print(int.bit_length(n1))
#判断name变量对应的值是否以"al"开头,并输出结果 name=" aleX" print(name.startswith("al"))
#将name变量对应的值中的"l"替换为"p",并输出结果 name=" aleX" print(name.replace("l","p"))
#请输出name变量对应的值的后2个字符? name=" aleX" print(name[-2:]) #-1代表最后一个字符,-2代表倒数第二个字符
#获取子序列,仅不包含最后一个字符。如:oldboy 则获取oldbo;root则获取roo name=" aleX" print(name.rstrip(name[-1:]))
#利用下划线将列表的每一个元素拼接成字符串,li = "alexericrain" li = "alexericrain" print("_".join(li))
#计算用户输入的内容中有几个十进制数?几个字母? #如: #content=input(‘请输入内容:‘) #如: asduiaf878123jkjsfd-213928 content = input(‘请输入内容:‘) .strip() #一定要记得使用strip去除空格 v,r = content.split("+") print(int(v)+int(r))
#计算用户输入的内容中有几个十进制数?几个字母? # 如: # content=input(‘请输入内容:‘) #如: asduiaf878123jkjsfd-213928 content=input(‘请输入内容:‘).strip() v = 0 s = 0 for i in content: if i.isdecimal(): v += 1 if i.isalpha(): s += 1 else: pass print("数字的个数是%d,字母的个数是%d" %(v,s))
while True: name = input("请输入用户名:") if name == "q" or name == "Q": break if len(name) <= 20: pass else: name = name[:20] password = input("请输入密码:") if password == "q" or password == "Q": break if len(password) <= 20: pass else: password = password[:20] email = input("请输入邮箱:") if email == "q" or email == "Q": break if len(email) <= 20: pass else: email = email[:20] template = "{0}\t{1}\t{2}\n" v = template.format(name,password,email) break print(v.expandtabs(20)) #将\t以及\n转化为空格或者回车
#往names列表里black_girl前面插入一个alex names = ["old_driver","rain","jack","shanshan","peiqi","black_girl"] names.insert(names.index("black_girl"),"alex") #占有原来的位置 print(names)
#把shanshan的名字改成中文,姗姗 names = ["old_driver","rain","jack","shanshan","peiqi","black_girl"] names[names.index("shanshan")] = "姗姗" print(names)
#往names列表里rain的后面插入一个子列表,["oldboy","oldgirl"] names = ["old_driver","rain","jack","shanshan","peiqi","black_girl"] names.insert(names.index("rain")+1,["oldboy","oldgirl"]) print(names)
#创建新列表[1,2,3,4,2,5,6,2],合并入names列表 names = ["old_driver","rain","jack","shanshan","peiqi","black_girl"] names.extend([1,2,3,4,5,6,2]) print(names)
#.取出names列表中索引2-10的元素,步长为2 names = ["old_driver","rain","jack","shanshan","peiqi","black_girl"] print(names[2:10:2])
#循环names列表,打印每个元素的索引值,和元素 names = ["old_driver","rain","jack","shanshan","peiqi","black_girl"] for index,name in enumerate(names): print("%s,%s" %(index,name))
#循环names列表,打印每个元素的索引值,和元素,当索引值为偶数时,把对应的元素改为-1 names = ["old_driver","rain","jack","shanshan","peiqi","black_girl"] for index,name in enumerate(names): if index % 2 == 0: name = "-1" else: pass print(index,name)
#这一题没理解
#names里有3个2,请返回第2个2的索引值。不要人肉数,要动态找(提示,找到第一个2的位置,在次基础上再找第2个) print(names.index(2,names.index(2)+1))
#查找列表(或元祖或字典)中元素,移除每个元素的空格,并查找以 a 或 A 开头 并且以 c 结尾的所有元素 # li = ["alec", " aric", "Alex", "Tony", "rain"] # tu = ("alec", " aric", "Alex", "Tony", "rain") # dic = {‘k1‘: "alex", ‘k2‘: ‘ aric‘,"k3": "Alex", "k4": "Tony"} li = ["alec", " aric", "Alex", "Tony", "rain"] for i in li: i = i.strip() c_a = i.startswith("a") c_A = i.startswith("A") c_c = i.endswith("d") if c_a or c_A and c_c: print(i) tu = ("alec", " aric", "Alex", "Tony", "rain") for i in tu: i = i.strip() c_a = i.startswith("a") c_A = i.startswith("A") c_c = i.endswith("d") if c_a or c_A and c_c: print(i) dic = {‘k1‘: "alex", ‘k2‘: ‘ aric‘,"k3": "Alex", "k4": "Tony"} for i in dic: i = i.strip() c_a = i.startswith("a") c_A = i.startswith("A") c_c = i.endswith("d") if c_a or c_A and c_c: print(i)
#列表中追加元素 “seven”,并输出添加后的列表 li = [‘alex‘,‘eric‘,‘rain‘] li.append("seven") print(li)
#请在列表的第 1 个位置插入元素 “Tony”,并输出添加后的列表 li = [‘alex‘,‘eric‘,‘rain‘] li.insert(0,"Tony") #append是追加 insert是在指定位置添加 print(li)
---恢复内容结束---
标签:下划线 for 循环 位置 十进制 后退 打印 字符 NPU 不能
原文地址:https://www.cnblogs.com/newt/p/9042875.html