Python之流程控制语句

Posted 2021-01-07 zhangcaiwang

#一.流程控制之if ..else语句
if 条件1：
    pass
elif 条件2：
    pass
elif 条件3：
    pass
else:
    pass


#1.简单的if打印
age_of_girl=31
if age_of_girl > 30:
    print(‘阿姨好‘)

#2.if ..else 单分支
age_of_girl=31
if age_of_girl > 30:
    print(‘阿姨好‘)
else:
    print(‘小姐好‘)

#3.多个条件判断
age_of_girl=18
height=171
weight=99
is_pretty=True
if age_of_girl >= 18 and age_of_girl < 22 and height > 170 weight < 100 and is_pretty == True:
    print(‘表白‘)
else:
    print(‘阿姨好‘)


#4.多分支if..else 
score=input(‘>>: ‘)
score=int(score)
if score >= 90:
    print(‘优秀‘)
elif score >=80:
    print(‘良好‘)
elif
    print(‘普通‘)
else:
    print(‘很差‘)


#二流程控制之while循环
#1.打印0-10
count=0
while count <= 10:
    print(‘loop‘,count)
    count+=1

#2.打印0-10之间的偶数
count=0
while count <= 10:
     if count%2 == 0:
             print(‘loop‘,count)
         count+=1


#3.打印0-10之间的奇数
count=0
while count <= 10:
    if count %2 == 1:
            print(‘loop‘,count)
        count+=1

#break用于退出本层循环
while True:
    print "123"
    break
    print "456"

#continue用于退出本次循环，继续下一次循环
while True:
    print "123"
    continue
    print "456"

#4.使用while循环输出1 2 3 4  5 6 8 9 10
count=1
while count <= 10:
    if count == 7:
        count+=1
        continue
    print(count)
    count+=1

#5.求100内的所有数的和
count=0
while count <= 100:
    res+=count
    count+=1
print(res)

#6.输出1-100内的所有的奇数
count=0
while count <= 100:
    if count%2 != 0:
        print(count)
    count+=1
#7.输出1-100内的所有的偶数
count=0
while count <= 100:
    if count%2 = 0:
        print(count)
    count+=1
#8.求1-2+3-4+5....99的所有数的和
res=0
count=1
while count <= 5:
    if count%2 == 0
        res-=count
    else:
        res+=count
    count+=1
print(res)
#9 用户登陆（三次机会重试）
count=0
while count < 3:
    name=input(‘请输入用户名:‘)
    password=input(‘请输入密码:‘)
    if name == ‘egon‘ and password == ‘123‘:
        print(‘login succcess‘)
    else:
        print(‘用户名或者密码错误‘)
        count+=1

#10.猜年龄游戏
#允许用户最多尝试3次 每尝试3次后，如果还没猜对，就问用户是否还想继续玩，如果回答Y或y, 就继续让其猜3次，以此往复，如果回答N或n，就退出程序
#如果猜对了，就直接退出 
age=73
count=0
while True:
    if count == 3:
        choice=input(‘继续Y/N‘)
        if choice == ‘Y‘ or choice == y:
                count=0
        else：
            break
guess=int(input(‘>>>‘))
if guess == age：
    print(‘you  got it‘)
    break
count+=1


#三.流程控制之for循环
打印金字塔
max_level=5
for current_level in range(1,max_level+1):
    for i in range(max_level-current_level):
        print(‘ ‘,end=‘‘) #在一行中连续打印多个空格
    for j in range(2*current_level-1):
        print(‘*‘,end=‘‘) #在一行中连续打印多个空格
    print()