問題：輸入出生日期和當前的日期，輸出活了多少天

舉例：你是昨天出生的，那麽輸出就為1

分三種情況討論：

1、年份和月份都相同

2、年份相同月份不同，先計算出生當天是當年的第幾天，後計算當前為當年的第幾天，相減

3、年份不同，還是先計算出生當天為當年的第幾天，後計算當前為當年的第幾天，做閏年判斷，逐一相加

閏年為一下兩種情況

1、能被400整除

2、能被4整除但不能被100整除

、、、、、、、、、、、、、、、

本題來自Udacity的計算機科學導論課程，用來做Python入門

Python語言兼具一般高級語言和腳本語言的特點，在官網下了一個東東，只會做腳本，函數現在只會一行一行往裏敲，然後運行，無法調試，好像是需要找一個開發環境，有空弄

附代碼

# 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()

Python - 按天算年齡