第四题

题目：输入某年某月某日，判断这一天是这一年的第几天？

自己的代码如下：

# -*- coding: utf-8 -*-
"""
Created on Fri Oct  4 22:08:42 2019

@author: Franz
"""

def leap(year):
    index = False
    if year % 400 == 0:
        index = True
    else:
        if year % 4 == 0 and year % 100 !=0:
            index = True
    return index

year = int(input('please input the year you want to check: '));
month = int(input('please input the month you want to check: '));
day = int(input('please input the day you want to check: '))

mon_days = [0,31,59,90,120,151,181,212,243,273,304,334];
if not leap(year):
    days = mon_days[month-1]+day;
else:
    if month >= 2:
        days = month_days[month-1]+1+day
    else:
        days = mon_days[month-1]+day;

print('it is the %dth day.' % days)

分析：代码写的很渣，其中很多条件判断可以合到一个if语句中进行判断，对于代码优先级还欠考虑。优点在于领悟到了闰年月份与非闰年月份天数的差异，并根据一个数组实现月份的查询。

如下是标准答案（python-2.7x）。

#!/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