Python：每日一题 4

新手·ing

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

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冬雪雪冬

def leapyear(n):
    return True if (n % 4 == 0 and n % 100 != 0) or n % 400 == 0 else False

days = [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30 ]
year, month, day = [int(x) for x in input('input year/month/day: ').split('/')]
day2 = sum(days[:month - 1]) + day
if leapyear(year) and month > 2:
    day2 += 1
print(day2)

jerryxjr1220

datetime不是可以直接计算的？算两个日期间的差
import datetime as dt
print dt.datetime(2017,3,5)-dt.datetime(2016,12,31)

新手·ing

冬雪雪冬 发表于 2017-3-25 23:04

感谢版主支持！

新手·ing

jerryxjr1220 发表于 2017-3-25 23:27
datetime不是可以直接计算的？算两个日期间的差
import datetime as dt
print dt.datetime(2017,3,5)-dt. ...

不是的
是随便输入日期，判断

新手·ing

year = int(input('year:'))
month = int(input('month:'))
day = int(input('day:'))
months = (0,31,59,90,120,151,181,212,243,273,304,334)
if 0 < month and month <= 12:
    sum = months[month - 1]
else:
    print('亲，一年只有12个月！')
sum += day
l = 0
if year % 4 == 0 or year % 400 == 0:
    l = 1
if l == 1 and month > 2:
    sum += 1
print('这是 %s 年的第 %s 天！' % (year,sum))

我的解答！

jerryxjr1220

是随便输入日期啊，只要减前一年的12月31不就是当年的第几天？

新手·ing

jerryxjr1220 发表于 2017-3-26 12:02
是随便输入日期啊，只要减前一年的12月31不就是当年的第几天？

大佬...

jerryxjr1220

新手·ing 发表于 2017-3-26 12:32
大佬...

python的强大就在于有非常多好用的库，有现成的解决方案就不要重复造轮子了

新手·ing

jerryxjr1220 发表于 2017-3-26 12:55
python的强大就在于有非常多好用的库，有现成的解决方案就不要重复造轮子了

学习了

gopythoner

我只知道datetime能计算，但是不熟，查了半天，写了个：

def test4(date):
    '''日期的输入方式使用2017-4-1这样的'''
    the_date = datetime.datetime.strptime(date,"%Y-%m-%d")
    the_year = datetime.datetime.strftime(the_date,"%Y")
    start_date = str(the_year)+"-1-1"
    start_date = datetime.datetime.strptime(start_date,"%Y-%m-%d")
    dates = the_date - start_date + datetime.timedelta(days=1)
    print (dates)

test4('2017-4-1')
>>>91 days, 0:00:00

ButcherRabbit

，不用模板的话闰年要判断，填写的月份要判断，填写的日子又要判断，例如，填写的4月31号，四月没有31号想了好久。。感觉我写的好丑好长

余欲渔

def ts(n=2017,y=4,r=11):
    day=0
    ey=29 if n%4==0 and n%100!=0 else 28
    for i in range(1,y):
        if i in[1,3,5,7,8,10,12]: day+=31
        elif i==2:                day+=ey
        else :                    day+=30
    print('%d月%d号是%d年的第%d天'%(y,r,n,(day+r)))


ts()

源稚空

year = int(input('year = '))
month = int(input('month = '))
day  = int(input('day = '))
today = [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30]
if year % 4 == 0 or (year % 400 == 0 and year%100!=0):
     today[1] = 29
i = 0
d = 0
for i in range(month) :
     d = d + today[i]
     i = i + 1
print(d)

seasonjj

day = str(input('输入年月日（列：20170419）：'))
y = int(day[:4])
m = int(day[4:6])
d = int(day[6:8])
md = int(day[4:])
mlist = [0,31,28,31,30,31,30,31,31,30,31,30,31]
i = list(range(m))
days = 0

if y%400 == 0 or (y%4 == 0 and y%100 != 0):
    if md > 228:
        days = 1
    else:
        days = 0
    for each in i:
        days += mlist[each]
    days += d
    print(days)
else:
    for each in i:
        days += mlist[each]
    days += d
    print(days)


好臭好长的感觉

solomonxian

看起来大家的想法差别不大，基本编程还没学会，库什么的还是留后面吧，记得有这么回事就是了

years = {'common_year': [31,28,31,30,31,30,31,31,30,31,30,31],\
         'leap_year': [31,29,31,30,31,30,31,31,30,31,30,31]}
judge = lambda y: 'leap_year' if ((y%4==0 and y%100 !=0) or y%400==0) else 'common_year'
try:
    year = int(input("请输入年份（正整数）："))   
    month = int(input("请输入月份（正整数1-12）："))
    day = int(input("请输入日期（正整数）："))
    months = years[judge(year)] # 年份判断
    
    if year>=0 and 12>=month>0 and months[month-1]>=day>0:
        num = sum(months[:month-1])+ day
        print("这是%d年第 %d 天" %(year, num))
    else:
        print("输入超出合理范围，请重新输入")
        
except ValueError:
    print("输入格式有误")
    

渡漫

year=int(input('输入年份：'))
mouth=int(input('输入月份：'))
day=int(input('输入日期：'))
sumDay=0    #总天数即第几天
isLeapYear=False    
perDay=[0,31,28,31,30,31,30,31,31,30,31,30]     #以列表的方式存储每个月的天数
if (year%4==0 and year%100!=0) or year%400==0:  #是否为闰年
    isLeapYear=True     
for i in range(mouth):      #以循环的方式叠加天数
    sumDay+=perDay[i]
if isLeapYear==True and mouth>2:
    sumDay+=1
sumDay+=day
print('这是一年中的第{0}天'.format(sumDay))

看了一下楼上的大佬，月份判断，日期判断。。。好烦，果然我还是太年轻了

Wofficre

print("请输入某年某月某日")
a=int(input())
b=int(input())
c=int(input())
day = [0,31,59,90,120,151,181,212,243,273,304,334]
m=0
result=0
if((a%4==0 and a%100!=0) or a%400==0):
      m=1
for n in range(1,b):
      result=result+day[n]
result=result+m+c
print ('这是一年的第%d天' %result )

NwkerWang

代码多了点，但是排除了所有输入错误的情况，所有输入错的数值都要重新输入

a = input ("please enter the year:")
while a.isdigit()!= True:
        a = input ("ERROR! Please enter the year again:")
yy = int (a)
if yy % 4 != 0:
        r = 1
elif (yy % 400 == 0) or (yy % 100 != 0):
        r = 2
elif yy % 100 == 0:
        r = 1
b = input ("please enter the month:")
b1 = ['1','2','3','4','5','6','7','8','9','10','11','12']
while b not in b1:
    b = input ("ERROR! please enter the month again:")
mm = int (b)
if r == 1:
    month = [31,28,31,30,31,30,31,31,30,31,30,31]
else:
    month = [31,29,31,30,31,30,31,31,30,31,30,31]
day = month[mm-1]
i = 1
day1 = []
while i<=day:
    temp = str(i)
    day1.append (temp)
    i += 1
c = input ("please enter the day:")
while c not in day1:
    c = input ("ERROR! please enter the day again:")
dd = int (c)
n = 0
while mm != 1:
    n += month[mm-2]
    mm -= 1
n += dd
print (n)

抑痒指

def fuction1(a):
    #该函数判断是否为闰年，是返回1，否则返回0.
    if(a%4 ==0 and a%100 != 0) or a % 400 == 0:
        return 1
    return 0

def fuction2(riqi):
    #通过年月日计算第几天
    runnian = fuction1(riqi[0])
    if riqi[1] == 1:
        return riqi[2]
    elif riqi[1] == 2:
        return 31 + riqi[2]
    elif riqi[1] == 3:
        return runnian + 59 + riqi[2]
    elif riqi[1] == 4:
        return runnian + 90 + riqi[2]
    elif riqi[1] == 5:
        return runnian + riqi[2] + 120
    elif riqi[1] == 6:
        return runnian + riqi[2] + 151
    elif riqi[1] == 7:
        return runnian + riqi[2] + 181
    elif riqi[1] == 8:
        return runnian + riqi[2] + 212
    elif riqi[1] == 9:
        return runnian + riqi[2] + 243
    elif riqi[1] == 10:
        return runnian + riqi[2] + 273
    elif riqi[1] == 11:
        return runnian + riqi[2] + 304
    elif riqi[1] == 12:
        return runnian + riqi[2] + 334

def fuction3(riqi):
    #用户输入年月日
    riqi[0] = int(input('请输入年份，并按回车结束,输入0可结束程序'))
    if riqi[0] != 0:
        riqi[1] = int(input('再输入月份，并按回车结束'))
        riqi[2] = int(input('最后输入几号，并按回车结束'))

'''
fuction3的测试程序
riqi = [0,0,0,0]
fuction3(riqi)
print(riqi)
'''


riqi = [1,0,0,0]
while riqi[0]:
    fuction3(riqi)
    if riqi[0] == 0:
        break
    riqi[3] = fuction2(riqi)
    print('您输入的日期是该年的第',riqi[3],'天','\n')