• Python


    问题:输入出生日期和当前的日期,输出活了多少天

    举例:你是昨天出生的,那么输出就为1

    分三种情况讨论:

    1、年份和月份都相同

    2、年份相同月份不同,先计算出生当天是当年的第几天,后计算当前为当年的第几天,相减

    3、年份不同,还是先计算出生当天为当年的第几天,后计算当前为当年的第几天,做闰年判断,逐一相加

    闰年为一下两种情况

    1、能被400整除

    2、能被4整除但不能被100整除

    、、、、、、、、、、、、、、、

    本题来自Udacity的计算机科学导论课程,用来做Python入门

    Python语言兼具一般高级语言和脚本语言的特点,在官网下了一个东东,只会做脚本,函数现在只会一行一行往里敲,然后运行,无法调试,好像是需要找一个开发环境,有空弄

    附代码

    # By Websten from forums
    #
    # Given your birthday and the current date, calculate your age in days. 
    # Account for leap days. 
    #
    # Assume that the birthday and current date are correct dates (and no 
    # time travel). 
    #
    
    
    def is_leap(year):
        result = False
        if year % 400 == 0:
            result = True
        if year % 4 == 0 and year % 100 != 0:
            result = True
        return result
    
    def daysBetweenDates(year1, month1, day1, year2, month2, day2):
        daysOfMonths = [ 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]
        if year1 == year2 and month1 == month2:
            days = day2 - day1
        if year1 == year2:
            days1 = 0
            i = 0
            while i < month1 - 1:
                days1 = days1 + daysOfMonths[i]
                i = i + 1
            days1 = days1 + day1
            if is_leap(year1) and month1 > 2:
                days1 = days1 + 1
            days2 = 0
            i = 0
            while i < month2 - 1:
                days2 = days2 + daysOfMonths[i]
                i = i + 1
            days2 = days2 + day2
            if is_leap(year2) and month2 > 2:
                days2 = days2 + 1
            days = days2 - days1
        else:
            days1 = 0
            i = 0
            while i < month1 - 1:
                days1 = days1 + daysOfMonths[i]
                i = i + 1
            days1 = days1 + day1
            if is_leap(year1) and month1 > 2:
                days1 = days1 + 1
            days2 = 0
            i = 0
            while i < month2 - 1:
                days2 = days2 + daysOfMonths[i]
                i = i + 1
            days2 = days2 + day2
            if is_leap(year2) and month2 > 2:
                days2 = days2 + 1
            days = 365 - days1 + days2
            if is_leap(year1):
                days = days + 1
            year1 = year1 + 1
            while year1 < year2:
                days = days + 365
                year1 = year1 + 1
                if is_leap(year1):
                    days = days + 1
        return days
            
            
            
    
    
    # Test routine
    
    def test():
        test_cases = [((2012,1,1,2012,2,28), 58), 
                      ((2012,1,1,2012,3,1), 60),
                      ((2011,6,30,2012,6,30), 366),
                      ((2011,1,1,2012,8,8), 585 ),
                      ((1900,1,1,1999,12,31), 36523)]
        for (args, answer) in test_cases:
            result = daysBetweenDates(*args)
            if result != answer:
                print "Test with data:", args, "failed"
            else:
                print "Test case passed!"
    
    test()
    View Code
  • 相关阅读:
    备份服务-Rsync
    mysql查询语句,int类型加引号居然也能查
    echo显示带颜色
    python读文件
    python发送邮件
    javadoc源码获取
    mac修改终端用户名
    [tomcat]-tomcat8启动时SessionIdGeneratorBase.createSecureRandom耗时
    shell运算符
    shell中$*和$@ 两个都区别
  • 原文地址:https://www.cnblogs.com/qingkai/p/8667501.html
Copyright © 2020-2023  润新知