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:
月份天数:
月份 | 天数 |
2 | 平年28天,闰年29天 |
1,3,5,7,8,10,12 | 31 |
4,6,9,11 | 30 |
闰年:
1、非整百年:能被4整除的为闰年。(如2004年就是闰年,2100年不是闰年)
>>> L=[31,31,30,31,30,31,31,30,31,30,31] >>> def caldate(a,b,c): s=0 if(a%100!=0 and a%4==0 or a%100==0 and a%400==0): L.insert(1,29) else: L.insert(1,28) for i in range(b-1): s=s+L[i] return s+c >>> caldate(2016,1,2) 2 >>> caldate(2016,3,2) 62
改进版:考虑了月份和天数的有效性(哈哈,对比网上的答案,一看自己的代码就像是菜鸟级的,还有很多需要学习的地方)
def caldate(a,b,c): L=[31,31,30,31,30,31,31,30,31,30,31] s=0 Leap=0 if(a%100!=0 and a%4==0 or a%100==0 and a%400==0): L.insert(1,29) Leap=1 else: L.insert(1,28) if 0<b<=12: if 0<c<=31: if((b==2)and(Leap==1)and(c<=29)or((b==2)and(Leap==0)and(c<=28))): for i in range(b-1): s=s+L[i] return s+c else: print 'The February should not greater than 28 or 29' else: print 'The date is error' else: print 'The month is invalid'
输出:
>>> caldate(2016,4,33) The date is error >>> caldate(2017,2,30) The February should not greater than 28 or 29 >>> caldate(2017,2,28) 59 >>> caldate(2017,2,29) The February should not greater than 28 or 29 >>> caldate(2017,13,29) The month is invalid
网上答案:
#!/usr/bin/python # -*- coding: UTF-8 -*- year = int(raw_input('year: ')) month = int(raw_input('month: ')) day = int(raw_input('day: ')) 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
输出结果:
year: 2015 month: 6 day: 7 it is the 158th day.