20221230星期五:
输入指定格式得日期，然后判断这一天是一年中得第几天：
# 方式一：datetime函数：from datetime import *da = input("请输入年月日，输入格式为xxxx-xx-xx：")da1 = datetime.strptime(da[:4] + '-1-1','%Y-%m-%d')da2 = datetime.strptime(da,'%Y-%m-%d')print(type(da1),da1) # <class 'datetime.datetime'> 2022-01-01 00:00:00print(type(da2),da2) # <class 'datetime.datetime'> 2022-12-22 00:00:00print("今天是一年中得第 {} 天".format((da2-da1).days+1),type(da2))# 方式二：datetime.date函数：from datetime import *def findDays(): year = int(input("请输入年份：")) month = int(input("请输入月份：")) day = int(input("请输入日期：")) da1 = date(year,month,day) da2 = date(year,month=1,day=1) count = (da1-da2).days+1 print(count)findDays()# 方式三：比较繁琐得办法：''' 以3月5日为例，应该先把前两个月得加起来，然后再加上5天即本年得第几天，特殊 情况，闰年且输入月份大于3时需考虑多加一天'''da = input("请输入日期，格式为XXXX-XX-XX：")date = da.split('-')year = int(date[0])month = int(date[1])day = int(date[2])list_month = [31,28,31,30,31,30,31,31,30,31,30,31]print('计算前：平年闰年：',list_month[1])count_day = 0if((year % 4 == 0 and year % 100 != 0) or (year % 400 == 0)): list_month[1] = 29print('平年闰年：',list_month[1])for i in range(12): if month > i + 1: count_day += list_month[i] else: count_day += day breakprint("这一天是 {} 年得第 {} 天".format(year,count_day))
