练习  输入三条边长如果能构成三角形就计算周长和面积

import math
a,b,c = input().split(',')
a = float(a)
b = float(b)
c = float(c)
if a + b > c and a + c > b and b + c > a:
print('周长%f'%(a + b + c))
p = (a + b + c)/2 area = math.sqrt( p * ( p - a ) * ( p - b )*( p - c ))
print('面积%f'%(area))
else:
print('不能构成三角形')

  练习 求平均值

flag = 0
sum = 0
while 1:
print('Enter an integer, the input ends if it is 0: ')
shu = input()
if shu != '0' :
sum = sum + float(shu)
flag +=1
else:
break
ave = float(sum/flag)
print('平均值是%f' %ave)

  练习 个人所得税计算器

salary = float(input('本月收入: '))
insurance = float(input('五险一金: '))
diff = salary - insurance - 3500
if diff <= 0:
rate = 0
deduction = 0
elif diff < 1500:
rate = 0.03
deduction = 0
elif diff < 4500:
rate = 0.1
deduction = 105
elif diff < 9000:
rate = 0.2
deduction = 555
elif diff < 35000:
rate = 0.25
deduction = 1005
elif diff < 55000:
rate = 0.3
deduction = 2755
elif diff < 80000:
rate = 0.35
deduction = 5505
else:
rate = 0.45
deduction = 13505
tax = abs(diff * rate - deduction)
print('个人所得税: ¥%.2f元' % tax)
print('实际到手收入: ¥%.2f元' % (diff + 3500 - tax))

  生成随机验证码

验证码
import random
import numpy as np
import string
print('开始生成验证码')
s = string.ascii_lowercase
str1 = ""
for i in range (3):
for i in range(0,4):
a = random.choice(s)
b = np.random.choice([1,2,3,4,5,6,7,8,9,0])
c = random.choice([a,b])
print(c,end="")
str1 = str(str1)+str(c)
print('')
shuru = input('请输入验证码')
if shuru == str1 :
print('成功')
break
else:
print('再试一次')

  密码爆破

import itertools
username = 'admin'
print('输入一个6位以内纯数字密码')
password = int(input())
print('开始爆破')
for i in range (0,999999):
print(i)
if int(i) == password:
print('爆破成功,密码%s' %i)
break
else:
continue

  石头剪子布

import numpy as np
res = np.random.choice(['0','1','2'])
x=input('输入0、剪刀 1、石头 2、布')
if res=='0':
if x=='0':
print("tie")
elif x=='1':
print('loser')
else:
print('win')
elif res=='1':
if x=='2':
print("win")
elif x=='0':
print('tie')
else:
print('loser')
else:
if x=='2':
print("loser")
elif x=='1':
print('win')
else:
print('tie')

  

2019_7_31python的更多相关文章

随机推荐

  1. fixture实战---通过fixure,解决方法依赖逻辑

    import pytest@pytest.fixture()def login(): print('输入用户名密码登陆') def test_cart(login): print('用例1,登陆后执行 ...

  2. react踩坑

    1.列表渲染一定要注意key的唯一性,不然会导致页面只更新部分的bug graphVertexConfigList.data.map((item, index) => { return < ...

  3. KiCAD差分布线

    KiCAD差分布线方法 KiCAD在进行差分布线的时候,会自动按照网路名称生成差分对,所以差分对的名称必须是以_P_N或+/-结束,这样才能找到一对差分对,比如说CAN网络,可以定义为CAN_P/CA ...

  4. 使用openntpd替换ntpd

    系统自带的ntp服务太难用,systemd启动几次没启动起来,懒得折腾,换了openntpd一次成功.

  5. Oracle数据库中文乱码问题解决

    1.查看服务器端编码select userenv('language') from dual;我实际查到的结果为:AMERICAN_AMERICA.ZHS16GBK 2.执行语句 select * f ...

  6. CF566E Restoring Map

    题意:乱序给你树上的每一个节点与之相距<=2的节点集合(并不知道这具体是哪个节点). 还原整棵树. 标程: #include<bits/stdc++.h> #define P pai ...

  7. swiper轮播箭头垂直居中

    取消懒加载 for (var i in $('.p01-s9 .lazyload')) { $('.p01-s9 .lazyload').eq(i).attr('src',$('.p01-s9 .la ...

  8. 【LeetCode 21】合并两个有序链表

    题目链接 [题解] 就是归并排序的一趟合并操作. 新建个链表加在上面就好.(用原来的链表的头结点也没问题) 加个头结点会比较好操作一点. 返回的时候返回头结点的next域就行 [代码] /** * D ...

  9. springboot之停止与启动服务的脚本

    最近要开始自动化部署了,我们要编写杀死服务的脚本. 我在windows写好的脚本拷贝到linux,就是不行,好像是因为转义字符吧. 然后,我就手敲了这个下面脚本: pid=`ps -ef|grep / ...

  10. tomcat源码分析一之getCanonicalFile和getAbsolutePath的区别

    最近在看tomcat源码 1.getPath(): 返回定义时的路径,(就是你写什么路径,他就返回什么路径) 2.getAbsolutePath(): 返回绝对路径,但不会处理“.”和“..”的情况 ...