Python - 按天算年龄,,问题:输入出生日期和


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

技术分享图片
# 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 resultdef 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 routinedef 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()
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Python - 按天算年龄

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