f or t or f and t and t or ff or t or f or f答案:ture
-
2)not 2 > 1 and 3 < 4 or 4 > 5 and 2 > 1 and 9 > 8 or 7 < 6
f and t or f and t and t or ff or f or f答案:false -
求出下列逻辑语句的值。
1),8 or 3 and 4 or 2 and 0 or 9 and 7= 8 or (3 and 4) or (2 and 0) or (9 and 7)
= 8 or 4 or 0 or 7
= 8 or 0 or 7
= 8 or 7
= 82),0 or 2 and 3 and 4 or 6 and 0 or 3
= 0 or (2 and 3 and 4) or (6 and 0) or 3
= 0 or (3 and 4) or 0 or 3
= 0 or 4 or 0 or 3
= 4 or 0 or 3
= 4 or 3
= 4 -
下列结果是什么?
1)、6 or 2 > 1
# 2)、3 or 2 > 1
# 3)、0 or 5 < 4
# 4)、5 < 4 or 3
# 5)、2 > 1 or 6
# 6)、3 and 2 > 1
# 7)、0 and 3 > 1
# 8)、2 > 1 and 3
# 9)、3 > 1 and 0
# 10)、3 > 1 and 2 or 2 < 3 and 3 and 4 or 3 > 2
答案
1)、6 or 2 > 1
= 6 or True
= 6
2)、3 or 2 > 1
= 3 or True
=3
3)、0 or 5 < 4
= 0 or False
= False
4)、5 < 4 or 3
= False or 3
=3
5)、2 > 1 or 6
=True or 6
=True
6)、3 and 2 > 1
=3 and True
=True
7)、0 and 3 > 1
=0 and True
=0
8)、2 > 1 and 3
=True and 3
=3
9)、3 > 1 and 0
=True and 0
=0 -
while循环语句基本结构?
while 条件:
循环体 -
利用while语句写出猜大小的游戏:
设定一个理想数字比如:66,让用户输入数字,如果比66大,则显示猜测的结果大了;如果比66小,则显示猜测的结果小了;只有等于66,显示猜测结果正确,然后退出循环。# while True:
# count = 66
# guess_number = int(input("请用户输入数字:"))
# if guess_number > count:
# print("结果大了")
# elif guess_number < count:
# print("结果小了")
# elif guess_number == count:
# print("结果正确")
# break -
在5题的基础上进行升级:
给用户三次猜测机会,如果三次之内猜测对了,则显示猜测正确,退出循环,如果三次之内没有猜测正确,则自动退出循环,并显示‘太笨了你....’。times=1
# while times<=3:
# count = 66
# guess_number = int(input("请用户输入数字:"))
# if guess_number > count:
# print("结果大了")
# elif guess_number < count:
# print("结果小了")
# elif guess_number == count:
# print("结果正确")
# break
# times += 1
# else:
# print("太笨了") -
使用while循环输出 1 2 3 4 5 6 8 9 10
#方法1
# count = 1
# while count<11:
# if count != 7:
# print(count)
# count+=1
#方法2
# count = 1
# while count<11:
# if count == 7:
# print(" ")
# else:
# print(count)
# count += 1
#方法3
# count = 1
# while count<11:
# if count == 7:
# pass
# else:
# print(count)
# count += 1
方法4
count = 1
while count<11:
if count == 7:
count+=1
print(count)
count += 1
方法5
count = 0
while count<10:
count+=1
if count == 7:
continue
print(count) -
求1-100的所有数的和
#count=1
# sum=0
# while count <=100:
# sum+=count
# count+=1
# print(sum) -
输出 1-100 内的所有奇数
count=1
while count <=100:
if count %2!=0:
print(count)
count+=1 -
输出 1-100 内的所有偶数
count=1
while count <=100:
if count %2==0:
print(count)
count+=1 -
求1-2+3-4+5 ... 99的所有数的和
count=1
sum =0
while count <=99:
if count%2==0:
sum -= count
else:
sum += count
count+=1 -
用户登录(三次输错机会)且每次输错误时显示剩余错误次数(提示:使用字符串格式化)
count=1
while count <4:
username = input("用户名")
passsword = input("密码")
if username == "tony" and passsword == "0825":
print("登录成功")
else:
print("用户名或密码错误,还有%s次机会"%(3-count))
count +=1 -
简述ASCII、Unicode、utf-8编码
ASCII:只包含英文字母,数字,特殊字符。均占8位(1字节)
Unicode:万国码,包括世界上所有的文字,不论中文、英文或符号,每个字符均占32位(4字节)。浪费空间,浪费资源。
utf-8:万国码升级版,英文占8位(1字节)、欧洲字符占16位(2字节)、中文字符站24位(3字节) -
简述位和字节的关系?
浙公网安备 33010602011771号