这本来是一个悲剧的故事，昨天想了很久都无法实现，仅仅判断个平闰年和月份就已经错误百出，今年参考了Snow__同学的万年历文章，总算是慢慢能够消化下来。
巨赞无比的思路，特别合适我这种小白
- 首先判断年份是否为平闰年
- 判断月份的天数
- 根据平闰年和月份的天数计算某年某月距离1990有多少平闰年，某月有多少天
程序如下：
# Insure whether Year is Leap year
def is_leap_year(year):
if(year%4 == 0 and year%100 != 0) or year % 400 == 0:
return True
else:
return False
# Calculate how many days in the month
def get_month_day(year, month):
days = 31
if month == 2:
days = 29 if is_leap_year(year) else 28
elif month in [4, 6, 9, 11]:
days = 30
return days
# Calculate how many days before
def get_today(year, month):
total = 0
for i in range(1990, year):
if is_leap_year(year):
total += 366
else:
total += 365
for i in range(1, month):
total += get_month_day(year, i)
return total
if __name__ == '__main__':
while True:
year = input('请输入年份(如：1990)：')
month = input('请输入月份(如：1)：')
try:
year = int(year)
month = int(month)
if month < 1 or month > 12:
print('年份或月份输入错误，请重新输入')
continue
except:
print('年份或月份输入错误，请重新输入')
continue
break
print('日\t一\t二\t三\t四\t五\t六')
count = 0
for i in range((get_today(year, month) % 7) + 1):
print('\t', end = '')
count += 1
for i in range(1, get_month_day(year, month) + 1):
print(i, end = '')
print('\t', end = '')
count += 1
if count %7 == 0:
print('\n')
很清爽的思路风格相比我自己之前的半成品而言，泥石流是什么就显而易见了。
Practice makes perfect
一起好好学习呀
