PYTHON基础之条件语句

了解到一些Python基础语句 那就反手敲一敲!!!

1,关于个税算法

   基本算法:

扣除三险一金后月收入 = 工资 - 三险一金

应纳税所得额=扣除三险一金后月收入-扣除标准

应纳个人所得税税额=全月应纳税所得额×适用税率-速算扣除数

个税 

基本条件: 扣除标准3500元/月三险一金缴纳 的基准工资上限为7662 ;超过,就按照7662缴纳;

gongzi = int(input('请输入工资:'))
sanxian = 0
if gongzi >=7662:
    sanxian= 7662*0.225
else:
    snaxian=gongzi*0.225
suode = gongzi - sanxian -3500
shuilu = 0
kouchu = 0
if suode<=0:
    print('loser')
else:

    if suode<=1500:
        shuilu = 0.03
    elif suode<=4500:
        shuilu =0.1
        kouchu =105
    elif suode <= 9000:
        shuilu = 0.2
        kouchu = 555
    elif suode <= 35000:
        shuilu = 0.25
        kouchu = 1005
    elif suode <= 55000:
        shuilu = 0.3
        kouchu = 2755
    elif suode <= 80000:
        shuilu = 0.35
        kouchu = 5505
    geshui = suode*shuilu-kouchu
    print(geshui)

2,数字类型操作

  运用所学的if for  条件语句  简单的运用小操作 如:数字之间的加减乘除  及and or  在语句上的运用

# 字符串, 数字, 布尔,列表,字典
# 问题1 怎么判断输入一个整数为整数
num = input('请输入一个整数:')
num = int(num)
if num % 2 == 0:
     print('偶数')
else:
     print('奇数')


# 怎么判断一个数既是三的倍数也是4的倍数

num = int(input('请输入一个整数'))
if num %12 ==0 :
    print('既是三的倍数又是四的倍数')
else:
    print('不是三和四的倍数')

# 问题三怎么判断这个数是五的倍数还是六的倍数

num = int(input('请输入一个整数'))
if num %5 == 0:
    if num %6 == 0:
        print('既是5的倍数有事6的倍数')
    else:
        print('是五的倍数不是六的倍数')
else:
    if num %6 ==0:
        print('只是留的倍数')
    else:
        print('几部是五的倍数又不是六的倍数')

 # and和or用于多个条件需要判断的情况下.一两个条件举例

 # and: 只要一个条件为假,最终结果为假
 # 条件1  真  条件2  真   结果  真
 #       假         真        假
#        假         假        假
 # or : 只要有一个条件为真,最终结果为真
#        真          真       真
 #       真          假       真
 #      假          假       假

#  框架, 工具 ,包  ,插件
#从random包中引出随机整数的方法
from random import  randint
# 计算机当中的随机数也称之为 ‘伪随机数’
# 计算机找随机数是根据计算机内部的算法来找到的
number =randint(0,3)
print(number)


user_num = input('请输入你要猜的数字')
user_num = int(user_num)
computer_num = randint(0,3)
print(computer_num)
if(user_num == computer_num):
    print('you win')
else:
    print('you lose')

3.使用条件语句实行石头剪刀布


from random import randint
# 0石头  1剪刀   2布
randint (0,2)
user_num = input('请输入你要猜的数字')
# user_num = int(user_num)
computer_num = randint(0,2)
# print(computer_num)
if user_num.isdigit():
    if 0<user_num>=2:
        if user_num-computer_num==-1 or user_num-computer_num== 2:
            print('you win')
        elif user_num - computer_num ==0:
            print('deuce')
        else :
            print('you lose')
    else:
        print('输入的数值大于有效范围')
else:
    print('输入的雷荣格式错误,请输入0-2之间的数')
你以为只有一种吗 ?   不存在的!  一个循环语句将它带走!

# 石头剪刀布
from random import  randint

user_win = 0
computer_win= 0
deuce = 0
# index代表标号  value代表值
for index,value in enumerate(range(3)):
    user_num = int(input('请输入数字'))
    computer_num = randint(0,2)
    if user_num-computer_num == -1 or user_num-computer_num == 2:
        print('第{}局玩家胜'. format(index +1))
        user_win +=1
    elif user_num-computer_num == 0:
        print('第{}局平局'.format(index +1))
        deuce += 1
    else:
        print('第{}局电脑胜'.format(index +1))
        computer_win += 1
    print('-----------第{}局结束-----------'.format(index +1))

    if computer_win == 2:




        print('电脑胜')
        break
    elif user_win == 2:
        print('玩家胜')
        break
    else:
        if deuce == 1 and computer_win - user_win ==0 and index ==2:
              print('平局')
        elif deuce == 3:
            print('平局')
        elif deuce == 2 and index == 2:
            if computer_win - user_win == 1:
                print('电脑胜')
            else:
                print('玩家胜')

4. For循环语句

   运用for循环做一些简单的小问题

# for(int index = 0 ;  index <10 ; index++)
for index in range (10):
    print('woshinibaba')
    print(index)


for index in range(50,100):
    print(index)

for index in range(50,100,5):
    print(index)
#1~100的和
sum = 0
for x in range(1,101):
    sum = sum + x
    print('x=%s'%x)
    print(sum)

#问题2:获取1~100之间所有奇数之和与5的倍数之和的差
sum1 = 0
for x in range(1,101,2):
    sum1 = sum1 + x
sum2 =0
for x in range(0,101,5):
    sum2 = sum2 + x
sum3 = sum1 - sum2
print(sum3)
   今天了解及学习的内容大概如此!  慢慢来吧。。。

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转载自blog.csdn.net/donquixote_/article/details/80827520