Question

写一个函数[day] = weekday(M, D, Y)，它告诉你日期M / D / Y的星期几。告诉您的用户接受的日期范围是什么。请勿使用任何特定于日历的模块或功能。

我找到了这两段代码，但我不理解它们背后的逻辑。更具体地说，我不明白偏移数字的来源。

def weekday(M,D,Y):
    offset = [0, 31, 59, 90, 120, 151, 181, 212, 243, 273, 304, 334]
    week   = ['Sunday', 
              'Monday', 
              'Tuesday', 
              'Wednesday', 
              'Thursday',
              'Friday', 
              'Saturday']
    afterFeb = 1
    if M > 2: afterFeb = 0
    aux = Y - 1700 - afterFeb
    # dayOfWeek for 1700/1/1 = 5, Friday
    dayOfWeek  = 5
    # partial sum of days betweem current date and 1700/1/1
    dayOfWeek += (aux + afterFeb) * 365
    # leap year correction
    dayOfWeek += aux / 4 - aux / 100 + (aux + 100) / 400
    # sum monthly and day offsets
    dayOfWeek += offset[M - 1] + (D - 1)
    dayOfWeek %= 7
    return dayOfWeek, week[dayOfWeek]

def weekday(m, d, y):
    # following numbers in the array are suggested by Sakamoto, Lachman, 
    Keith and Craver
    # one for each month
    # and the remaining logic as well
    t = [0, 3, 2, 5, 0, 3, 5, 1, 4, 6, 2, 4]
    y = y - (m<3)
    day = (y + y/4 - y/100 + y/400 + t[m-1] + d)%7;
    # from 0 to 6, weekdays
    if day == 0:
    print "Sunday"
    elif day == 1:
    print "Monday"
    elif day == 2:
    print "Tuesday"
    elif day == 3:
    print "Wednesday"
    elif day == 4:
    print "Thursday"
    elif day == 5:
    print "Friday"
    elif day == 6:
    print "Saturday"

Answer 1

该程序将开始日期定为1700年1月1日星期五。它计算年，月和日的差异。抵消是这几个月的累积数字。对于Jan，不需要添加任何数字。仅添加0。对于2月，人们必须增加31天的1月，对于3月，一个人增加了31天的jan和28天的feb（总共59天）。等

python中没有日历模块的日期中的星期几

1 个答案: