4）：题目：输入某年某月某日，判断这一天是这一年的第几天？
程序分析：以3月5日为例，应该先把前两个月的加起来，然后再加上5天即本年的第几天，特殊情况，闰年且输入月份大于2时需考虑多加一天。
程序源代码：
#!/usr/bin/python
# -*- coding: UTF-8 -*- year = int(raw_input('year:\n')) month = int(raw_input('month:\n')) day = int(raw_input('day:\n')) months = (0,31,59,90,120,151,181,212,243,273,304,334) if 0 < month <= 12: sum = months[month - 1] else: print 'data error' sum += day leap = 0 if (year % 400 == 0) or ((year % 4 == 0) and (year % 100 != 0)): leap = 1 if (leap == 1) and (month > 2): sum += 1 print 'it is the %dth day.' % sum
以上实例输出结果为：
year: 2015 month: 6 day: 7 it is the 158th day.
看另外一个案例：
#!/usr/bin/python # -*- coding: UTF-8 -*- year=int(raw_input("年：\n")) month=int(raw_input("月：\n")) day=int(raw_input("日：\n")) months1=[0,31,60,91,121,152,182,213,244,274,305,335,366] #闰年 months2=[0,31,59,90,120,151,181,212,243,273,304,334,365] #平年 if ((year%4==0)and(year%100!=0)) or((year%100==0)and(year%400==0)): Dth=months1[month-1]+day else: Dth=months2[month-1]+day print "是该年的第%d天"%Dth
闰年需要同时满足以下条件：
- 1、年份能被4整除；
- 2、年份若是 100 的整数倍的话需被400整除，否则是平年。
#!/usr/bin/python # -*- coding: UTF-8 -*- # 输入任意年月日，知道是改年第几天 p = [31,28,31,30,31,30,31,31,30,31,30,31] # 平年 w = [31,29,31,30,31,30,31,31,30,31,30,31] # 闰年 year =int(raw_input("请输入年："+'\n')) month =int(raw_input("请输入月："+'\n')) day=int(raw_input("请输入日："+'\n')) arr=[31,28,31,30,31,30,31,31,30,31,30,31] sum=day for i in range(0,month-1): sum+=arr[i] if year%4==0: if year%100==0 and year%400!=0: print "这是今年的第%d天" % sum else: sum=sum+1 print "这是今年的第%d天" % sum else: print "这是今年的第%d天" % sum
再来参考一个：
#!/usr/bin/python # -*- coding: UTF-8 -*- # 输入任意年月日，知道是改年第几天 p = [31,28,31,30,31,30,31,31,30,31,30,31] # 平年 w = [31,29,31,30,31,30,31,31,30,31,30,31] # 闰年 year =int(raw_input("年:\n")) month =int(raw_input("月:\n")) day =int(raw_input("日:\n")) # 判断闰年，平年 if year % 100 == 0: if year % 400 == 0: d=w else: d=p else: if year % 4 == 0: d=w else: d=p # 计算天数 days = sum(d[0:month-1])+day print "%d.%d.%d是%d年的第%s天。"%(year,month,day,year,days)
还有哦：
#!/usr/bin/python # -*- coding: UTF-8 -*- year = int(input('请输入年份：')) month = int(input('请输入月份：')) day = int(input('请输入日期：')) total = 0 if year%4 == 0: days = 29 else: days = 28 if year%4 == 0: print year, '年是润年，2月有', days, '天！' else: print year, '年是平年，2月有', days, '天！' if month <= 1: total = day elif month == 2: total = 31 + day elif month == 9 or month == 11: total = (month - 2) * 30 + days + month/2 + day + 1 else: total = (month - 2) * 30 + days + month/2 + day print year, '年',month, '月', day, '日,是这一年的第', total, '天！'
完事再来看一个：
#!/usr/bin/python def isLeapYear(a): if (0 == a%4 or 0 == a%400) and 0 != a%100 : return 1 else: return 0 dict = {1: 31, 2: 28, 3: 31, 4: 30, 5: 31, 6: 30, 7: 31, 8: 31, 9: 30, 10: 31, 11: 30, 12: 31} a = int(input("Input year:")) b = int(input("Input month:")) c = int(input("Input day:")) m = 0 k = isLeapYear(a) for i in range(1,b): m = m + dict[i] m = m + isLeapYear(a) + c print(m)
#!/usr/bin/env python # -*- coding: utf-8 -*- year = int(raw_input('year:\n')) month = int(raw_input('month:\n')) day = int(raw_input('day:\n')) days = [31,28,31,30,31,30,31,31,30,31,30,31] if year % 400 == 0 or (year % 4 == 0 and year % 100 != 0): days[2] += 1 now = sum(days[0:month-1])+day print 'it is the %dth day.' %now
#!/usr/bin/env python #coding:utf-8 import time import sys reload(sys) sys.setdefaultencoding('utf-8') # 设置 'utf-8' a = raw_input("输入时间（格式如：2017-04-04）:") t = time.strptime(a,"%Y-%m-%d") print time.strftime("今年的第%j天",t).decode('utf-8')
#! /usr/bin/env python # coding:utf-8 import time D=raw_input("请输入年份，格式如XXXX-XX-XX：") d=time.strptime( D,'%Y-%m-%d').tm_yday print "the {} day of this year!" .format(d)
Python3 参考解法：
#!/usr/bin/python3 date = input("输入年月日(yyyy-mm-dd):") y,m,d = (int(i) for i in date.split('-')) sum=0 special = (1,3,5,7,8,10) for i in range(1,int(m)): if i == 2: if y%400==0 or (y%100!=0 and y%4==0): sum+=29 else: sum+=28 elif(i in special): sum+=31 else: sum+=30 sum+=d print("这一天是一年中的第%d天"%sum)
再来看：
#!/usr/bin/python # -*- coding: UTF-8 -*- i= int(input('请输入年份：')) j= int(input('请输入月份：')) k= int(input('请输入天数：')) month = [31,28,31,30,31,30,31,31,30,31,30,31] day = 0 if i%4 == 0 or i %400 == 0 and i%100 != 0: #闰年 month = month[:j-1] if j >2: #大于2月份 day = sum(month)+1+k elif j== 2 and k ==29: #刚好在闰年闰天 day = sum(month)+1+k else: day = sum(month)+k else: #平年 month = month[:j-1] day = sum(month)+k print('这一天是这一年的第{}天'.format(day))
#!/usr/bin/env python #-*- coding:utf-8 -*- import calendar year = int(input("year: ")) month = int(input("month: ")) day = int(input("day: ")) days = 0 if 0 < month <= 12: _, month_day = calendar.monthrange(year, month) if day <= month_day: for m in range(1, month): _, month_day = calendar.monthrange(year, m) days += month_day days += day print days else: print "input day error" else: print "input month error"
通过计算输入的日期与相应年份1月1日相差的秒数，然后除以每天的秒数3600*24，即可得到输入日期的天数：
#!/usr/bin/env python3 import time def datecount(date): date0=date[0:4]+"-01-01" datet=time.strptime(date,"%Y-%m-%d") #将输入的字符串转化为时间元组 date0t=time.strptime(date0,"%Y-%m-%d") dates=time.mktime(datet) #将时间元组转化为时间戳 date0s=time.mktime(date0t) count=(dates-date0s)/(3600*24) #输入日期的时间戳减当前年份0101的时间戳除以每天秒数 return count+1 a=input("请输入日期：格式如2017-06-16\n") print("{}是{}年第{}天".format(a,a[0:4],int(datecount(a))))
考虑实际的情况，比如输入月份为13月或输入天数为65天时候报错（日期仅校对0-31天，未按照实际日期校对）：
#!/usr/bin/env python #-*- coding:utf-8 -*- print('输入年月日以查看某一日期是当年第几天\n') year = int(input('请输入年份：\n')) month = int(input('请输入月份:\n')) day = int(input('请输入日期:\n')) months = [31,28,31,30,31,30,31,31,30,31,30,31] d = 0 if 0<month<=12: if 0<day<=31: d = d + day if month > 2: if (year % 400 == 0) or ((year % 4 == 0) and (year % 100 != 0)): d = d + 1 for i in range(12): if (i+1) < month: d = d + months[i] i = i + 1 print(d) else: print('输入的日期有错误') else: print('输入的月份有错误')
通过输入时间点的unix时间戳和输入年份首日的Unix时间戳之间的差，来计算经过的时间：
#coding=utf-8 import time print "Please Enter full number just like 02 01 03" y = int(raw_input('Enter the year:')) #分别输入年月日 m = int(raw_input('Enter the month:')) d = int(raw_input('Enter the day:')) a = (y,m,d,00,00,00,00,00,00) #要求长度为9 b = (y,01,01,00,00,00,00,00,00) #输入年份的第一天 timestampa = time.mktime(a) #两个都转换为Unix时间戳，即1970.1.1到现在经过的秒数 timestampb = time.mktime(b) timec = int((timestampa - timestampb)/(3600*24)) #输入的时间戳减去年份首天的时间戳等于经过的秒数，再换算成天，取整 print("There are {} days goes by!".format(timec))
python3 利用time模块，简洁写法：
import time print(time.strptime('2017-9-20', '%Y-%m-%d')[7])
Python2.x 与 Python3.x 兼容：
#!/usr/bin/python # -*- coding: UTF-8 -*- # 觉得自己的逻辑看起来更顺眼，嘻嘻！ days = [0, 31, 59, 90, 120, 151, 181, 212, 243, 273, 304, 334] yy = int(input('请输入年份：')) if yy >= 0: mm = int(input('请输入月份：')) if 0 < mm < 13: dd = int(input('请输入日期：')) if ((yy % 4 == 0) and (yy % 100 != 0)) or (yy % 400 == 0): if mm <= 2: sum = days[mm - 1] + dd else: sum = days[mm - 1] + 1 + dd else: sum = days[mm - 1] + dd print('您输入的时间在这一年的第%d天' % sum) else: print('您输入的月分不正确') else: print('请输入正确的公元年')
#!/usr/bin/python # -*- coding: UTF-8 -*- from functools import reduce year = int(input('请输入年（如：2017）：')) month = int(input('请输入月（如：3）：')) day = int(input('请输入日（如：16）：')) mday = [0,31,28 if year%4 else 29 if year%100 else 28 if year%4 else 29,31,30,31,30,31,31,30,31,30,31] print('{}年{}月{}日是当年的第{}天'.format(year, month, day, reduce(lambda x,y:x+y, mday[:month])+day))
加入异常处理，确保日期输入格式正确：
import time while 1: try: a=input('请输入日期yyyy-mm-dd:') b=time.strptime(a,'%Y-%m-%d') #按输入格式转化成时间格式 local变量 except ValueError: print('请输入正确的日期格式！') else: b=time.strptime(a,'%Y-%m-%d') #按输入格式转化成时间格式 global变量 dd=time.strftime('%j',b) #返回一年中的第几天 yy=time.strftime('%Y',b) #返回年份 print('输入的日期是%s年的第%s天'%(yy,dd)) break
基本上就是各种各样的案例。。。可见诸位的大才。。。
如果感觉不错的话，请多多点赞支持哦。。。
原文链接：https://blog.csdn.net/luyaran/article/details/80016648