如何计算某人生日后的第 10000 天是什么时候
How to calculate when one's 10000 day after his or her birthday will be
我想知道如何使用基本的 Python 解决这个问题（不使用任何库）：我如何计算一个人生日后的 10000 天是什么时候（/将是）？
例如，给定星期一 19/05/2008，所需的日期是星期五 05/10/2035（根据 https://www.durrans.com/projects/calc/10000/index.html?dob=19%2F5%2F2008&e=mc2）
到目前为止我已经完成了以下脚本：
years = range(2000, 2050)
lst_days = []
count = 0
tot_days = 0
for year in years:
if((year % 400 == 0) or (year % 100 != 0) and (year % 4 == 0)):
lst_days.append(366)
else:
lst_days.append(365)
while tot_days <= 10000:
tot_days = tot_days + lst_days[count]
count = count+1
print(count)
估计此人从生日算起 10,000 天后的年龄（适用于 2000 年之后出生的人）。但是我该如何继续呢？
仅使用基础 Python 包
基于“无特殊包”意味着您只能使用基础 Python 包，您可以使用 datetime.timedelta 解决此类问题：
import datetime
start_date = datetime.datetime(year=2008, month=5, day=19)
end_date = start_date + datetime.timedelta(days=10000)
print(end_date.date())
没有任何基础包（并且正在解决问题）
Side-stepping 甚至基础 Python 包，并将问题向前推进，以下内容应该有所帮助（我希望！）。
首先定义一个确定年份是否为闰年的函数：
def is_it_a_leap_year(year) -> bool:
"""
Determine if a year is a leap year
Args:
year: int
Extended Summary:
According to:
https://airandspace.si.edu/stories/editorial/science-leap-year
The rule is that if the year is divisible by 100 and not divisible by
400, leap year is skipped. The year 2000 was a leap year, for example,
but the years 1700, 1800, and 1900 were not. The next time a leap year
will be skipped is the year 2100.
"""
if year % 4 != 0:
return False
if year % 100 == 0 and year % 400 != 0:
return False
return True
然后定义一个判断人年龄的函数（利用上面的判断闰年）：
def age_after_n_days(start_year: int,
start_month: int,
start_day: int,
n_days: int) -> tuple:
"""
Calculate an approximate age of a person after a given number of days,
attempting to take into account leap years appropriately.
Return the number of days left until their next birthday
Args:
start_year (int): year of the start date
start_month (int): month of the start date
start_day (int): day of the start date
n_days (int): number of days to elapse
"""
# Check if the start date happens on a leap year and occurs before the
# 29 February (additional leap year day)
start_pre_leap = (is_it_a_leap_year(start_year) and start_month < 3)
# Account for the edge case where you start exactly on the 29 February
if start_month == 2 and start_day == 29:
start_pre_leap = False
# Keep a running counter of age
age = 0
# Store the "current year" whilst iterating through the days
current_year = start_year
# Count the number of days left
days_left = n_days
# While there is at least one year left to elapse...
while days_left > 364:
# Is it a leap year?
if is_it_a_leap_year(current_year):
# If not the first year
if age > 0:
days_left -= 366
# If the first year is a leap year but starting after the 29 Feb...
elif age == 0 and not start_pre_leap:
days_left -= 365
else:
days_left -= 366
# If not a leap year...
else:
days_left -= 365
# If the number of days left hasn't dropped below zero
if days_left >= 0:
# Increment age
age += 1
# Increment year
current_year += 1
return age, days_left
使用您的示例，您可以使用以下方法测试函数：
age, remaining_days = age_after_n_days(start_year=2000, start_month=5, start_day=19, n_days=10000)
现在你有完整的年数和剩余天数
然后您可以使用 remaining_days 计算出准确的日期。
如果导入库日期时间
import datetime
your_date = "01/05/2000"
(day, month, years) = your_date.split("/")
date = datetime.date(int(years), int(month), int(day))
date_10000 = date+datetime.timedelta(days=10000)
print(date_10000)
没有库脚本
your_date = "20/05/2000"
(day, month, year) = your_date.split("/")
days = 10000
year = int(year)
month = int(month)
day = int(day)
end=False
#m1,m3,m5,m7,m8,m10,m12=31
#m2=28
#m4,m6,m9,m11=30
m=[31,28,31,30,31,30,31,31,30,31,30,31]
while end!=True:
if(((year % 400 == 0) or (year % 100 != 0) and (year % 4 == 0)) and(days-366>=0)):
days-=366
year+=1
elif(((year % 400 != 0) or (year % 100 != 0) and (year % 4 != 0)) and(days-366>=0)):
days-=365
year+=1
else:
end=True
end=False
if(((year % 400 == 0) or (year % 100 != 0) and (year % 4 == 0))):
m[1]=29
else:
m[1]=28
while end!=True:
if(days-m[month]>=0):
days-=m[month]
if(month+1!=12):
month+=1
else:
year+=1
if(((year % 400 == 0) or (year % 100 != 0) and (year % 4 == 0))):
m[1]=29
else:
m[1]=28
month=0
else:
end=True
if(day+days>m[month]):
day=day+days-m[month]+1
if(month+1!=12):
month+=1
else:
year+=1
if(((year % 400 == 0) or (year % 100 != 0) and (year % 4 == 0))):
m[1]=29
else:
m[1]=28
month=0
else:
day=day+days
print(day,"/",month,"/",year)
我已经更新了闰年和闰月的代码。这是我正在使用的代码：
n = input("Enter your DOB:(dd/mm/yyyy)")
d,m,y = n.split('/') #Splitting the DOB
d,m,y = int(d), int(m), int(y)
def if_leap(year): # Checking for leap year.
if year % 4 != 0:
return False
elif year % 100 == 0 and year % 400 != 0:
return False
else:
return True
target = 10000
while target > 364: # getting no.of years
if if_leap(y):
target -= 366
y += 1
else:
target -= 365
y += 1
while target > 27: # getting no. of months
if m == 2 :
if if_leap(y):
target -= 29
m += 1
if m >= 12: # Resetting the month to 1 if it's value is greater than 12
y += 1
m -= 12
else:
target -= 28
m += 1
if m >= 12:
y += 1
m -= 12
elif m in [1, 3, 5, 7, 8, 10, 12]:
target -= 31
m += 1
if m >= 12:
y += 1
m -= 12
elif m in [4, 6, 9, 11]:
target -= 30
m += 1
if m >= 12:
y += 1
m -= 12
d = d + target # getting the no. of days
if d > 27:
if m == 2:
if if_leap(y):
d -= 29
m += 1
else:
d -= 28
m += 1
elif m in [1, 3, 5, 7, 8, 10, 12]:
d -= 31
m += 1
else:
d -= 30
m += 1
print(f"The 10000th date will be {d}/{m}/{y}")
输出：
Enter your DOB:(dd/mm/yyyy): 06/01/2006
The 10000th date will be 24/5/2033
P.S：我在查看该网站时得到了一些略有不同的输出。任何人都可以找出代码中的 bug/mistake 吗？这真的很有帮助。
例如。对于日期 08/12/2004，它应该是 25/4/2032 但我的输出显示 24/4/2032.
这是我想出的解决方案，它不涉及库或包，只涉及循环和条件（考虑闰年）：
def isLeapYear(years):
if years % 4 == 0:
if years % 100 == 0:
if years % 400 == 0:
return True
else:
return False
else:
return True
else:
return False
monthDays = [31,28,31,30,31,30,31,31,30,31,30,31]
sum = 0
sumDays = []
for i in monthDays:
sumDays.append(365 - sum)
sum += i
timeInp = input("Please enter your birthdate in the format dd/mm/yyyy\n")
timeInp = timeInp.split("/")
days = int(timeInp[0])
months = int(timeInp[1])
years = int(timeInp[2])
totDays = 10000
if totDays > 366:
if isLeapYear(years):
if months == 1 or months == 2:
totDays -= (sumDays[months - 1] + 1 - days) + 1
else:
totDays -= (sumDays[months - 1] - days) + 1
else:
totDays -= (sumDays[months - 1] - days) + 1
months = 1
days = 1
years += 1
while totDays > 366:
if isLeapYear(years):
totDays -= 366
else:
totDays -= 365
years += 1
i = 0
while totDays != 0:
if isLeapYear(years):
monthDays[1] = 29
else:
monthDays[1] = 28
if totDays >= monthDays[i]:
months += 1
totDays -= monthDays[i]
elif totDays == monthDays[i]:
months += 1
totDays = 0
else:
days += totDays
if days % (monthDays[i] + 1)!= days:
days %= monthDays[i] + 1
months += 1
totDays = 0
if months == 13:
months = 1
years += 1
i += 1
if i == 12:
i = 0
print(str(days) + "/" + str(months) + "/" + str(years))
顾名思义，isLeapYear()接受一个参数years，returns一个布尔值。
我们解决这个问题的第一步是先将我们的日期“翻译”到下一年，以使其更容易。这使我们以后的计算更加容易。为此，我们可以定义一个数组 sumDays 来存储每个月结束一年（进入新的一年）所需的天数。然后，我们从 totDays 中减去这个数额，考虑闰年，并更新我们的变量。
接下来是简单的部分，只要我们有足够的天数来完成一整年，就向前跳几年。
一旦我们无法再增加一整年，我们就逐月进行，直到 运行 天数已满。
示例测试用例：
输入#1：
19/05/2008
输出#1：
5/10/2035
输入#2：
05/05/2020
输出#2：
21/9/2047
输入#3：
29/02/2020
输出#3：
17/7/2047
我在这个网站上检查了我的大部分解决方案：https://www.countcalculate.com/calendar/birthday-in-days/result
我想知道如何使用基本的 Python 解决这个问题（不使用任何库）：我如何计算一个人生日后的 10000 天是什么时候（/将是）？
例如，给定星期一 19/05/2008，所需的日期是星期五 05/10/2035（根据 https://www.durrans.com/projects/calc/10000/index.html?dob=19%2F5%2F2008&e=mc2）
到目前为止我已经完成了以下脚本：
years = range(2000, 2050)
lst_days = []
count = 0
tot_days = 0
for year in years:
if((year % 400 == 0) or (year % 100 != 0) and (year % 4 == 0)):
lst_days.append(366)
else:
lst_days.append(365)
while tot_days <= 10000:
tot_days = tot_days + lst_days[count]
count = count+1
print(count)
估计此人从生日算起 10,000 天后的年龄（适用于 2000 年之后出生的人）。但是我该如何继续呢？
仅使用基础 Python 包
基于“无特殊包”意味着您只能使用基础 Python 包，您可以使用 datetime.timedelta 解决此类问题：
import datetime
start_date = datetime.datetime(year=2008, month=5, day=19)
end_date = start_date + datetime.timedelta(days=10000)
print(end_date.date())
没有任何基础包（并且正在解决问题）
Side-stepping 甚至基础 Python 包，并将问题向前推进，以下内容应该有所帮助（我希望！）。
首先定义一个确定年份是否为闰年的函数：
def is_it_a_leap_year(year) -> bool:
"""
Determine if a year is a leap year
Args:
year: int
Extended Summary:
According to:
https://airandspace.si.edu/stories/editorial/science-leap-year
The rule is that if the year is divisible by 100 and not divisible by
400, leap year is skipped. The year 2000 was a leap year, for example,
but the years 1700, 1800, and 1900 were not. The next time a leap year
will be skipped is the year 2100.
"""
if year % 4 != 0:
return False
if year % 100 == 0 and year % 400 != 0:
return False
return True
然后定义一个判断人年龄的函数（利用上面的判断闰年）：
def age_after_n_days(start_year: int,
start_month: int,
start_day: int,
n_days: int) -> tuple:
"""
Calculate an approximate age of a person after a given number of days,
attempting to take into account leap years appropriately.
Return the number of days left until their next birthday
Args:
start_year (int): year of the start date
start_month (int): month of the start date
start_day (int): day of the start date
n_days (int): number of days to elapse
"""
# Check if the start date happens on a leap year and occurs before the
# 29 February (additional leap year day)
start_pre_leap = (is_it_a_leap_year(start_year) and start_month < 3)
# Account for the edge case where you start exactly on the 29 February
if start_month == 2 and start_day == 29:
start_pre_leap = False
# Keep a running counter of age
age = 0
# Store the "current year" whilst iterating through the days
current_year = start_year
# Count the number of days left
days_left = n_days
# While there is at least one year left to elapse...
while days_left > 364:
# Is it a leap year?
if is_it_a_leap_year(current_year):
# If not the first year
if age > 0:
days_left -= 366
# If the first year is a leap year but starting after the 29 Feb...
elif age == 0 and not start_pre_leap:
days_left -= 365
else:
days_left -= 366
# If not a leap year...
else:
days_left -= 365
# If the number of days left hasn't dropped below zero
if days_left >= 0:
# Increment age
age += 1
# Increment year
current_year += 1
return age, days_left
使用您的示例，您可以使用以下方法测试函数：
age, remaining_days = age_after_n_days(start_year=2000, start_month=5, start_day=19, n_days=10000)
现在你有完整的年数和剩余天数
然后您可以使用 remaining_days 计算出准确的日期。
如果导入库日期时间
import datetime
your_date = "01/05/2000"
(day, month, years) = your_date.split("/")
date = datetime.date(int(years), int(month), int(day))
date_10000 = date+datetime.timedelta(days=10000)
print(date_10000)
没有库脚本
your_date = "20/05/2000"
(day, month, year) = your_date.split("/")
days = 10000
year = int(year)
month = int(month)
day = int(day)
end=False
#m1,m3,m5,m7,m8,m10,m12=31
#m2=28
#m4,m6,m9,m11=30
m=[31,28,31,30,31,30,31,31,30,31,30,31]
while end!=True:
if(((year % 400 == 0) or (year % 100 != 0) and (year % 4 == 0)) and(days-366>=0)):
days-=366
year+=1
elif(((year % 400 != 0) or (year % 100 != 0) and (year % 4 != 0)) and(days-366>=0)):
days-=365
year+=1
else:
end=True
end=False
if(((year % 400 == 0) or (year % 100 != 0) and (year % 4 == 0))):
m[1]=29
else:
m[1]=28
while end!=True:
if(days-m[month]>=0):
days-=m[month]
if(month+1!=12):
month+=1
else:
year+=1
if(((year % 400 == 0) or (year % 100 != 0) and (year % 4 == 0))):
m[1]=29
else:
m[1]=28
month=0
else:
end=True
if(day+days>m[month]):
day=day+days-m[month]+1
if(month+1!=12):
month+=1
else:
year+=1
if(((year % 400 == 0) or (year % 100 != 0) and (year % 4 == 0))):
m[1]=29
else:
m[1]=28
month=0
else:
day=day+days
print(day,"/",month,"/",year)
我已经更新了闰年和闰月的代码。这是我正在使用的代码：
n = input("Enter your DOB:(dd/mm/yyyy)")
d,m,y = n.split('/') #Splitting the DOB
d,m,y = int(d), int(m), int(y)
def if_leap(year): # Checking for leap year.
if year % 4 != 0:
return False
elif year % 100 == 0 and year % 400 != 0:
return False
else:
return True
target = 10000
while target > 364: # getting no.of years
if if_leap(y):
target -= 366
y += 1
else:
target -= 365
y += 1
while target > 27: # getting no. of months
if m == 2 :
if if_leap(y):
target -= 29
m += 1
if m >= 12: # Resetting the month to 1 if it's value is greater than 12
y += 1
m -= 12
else:
target -= 28
m += 1
if m >= 12:
y += 1
m -= 12
elif m in [1, 3, 5, 7, 8, 10, 12]:
target -= 31
m += 1
if m >= 12:
y += 1
m -= 12
elif m in [4, 6, 9, 11]:
target -= 30
m += 1
if m >= 12:
y += 1
m -= 12
d = d + target # getting the no. of days
if d > 27:
if m == 2:
if if_leap(y):
d -= 29
m += 1
else:
d -= 28
m += 1
elif m in [1, 3, 5, 7, 8, 10, 12]:
d -= 31
m += 1
else:
d -= 30
m += 1
print(f"The 10000th date will be {d}/{m}/{y}")
输出：
Enter your DOB:(dd/mm/yyyy): 06/01/2006
The 10000th date will be 24/5/2033
P.S：我在查看该网站时得到了一些略有不同的输出。任何人都可以找出代码中的 bug/mistake 吗？这真的很有帮助。
例如。对于日期 08/12/2004，它应该是 25/4/2032 但我的输出显示 24/4/2032.
这是我想出的解决方案，它不涉及库或包，只涉及循环和条件（考虑闰年）：
def isLeapYear(years):
if years % 4 == 0:
if years % 100 == 0:
if years % 400 == 0:
return True
else:
return False
else:
return True
else:
return False
monthDays = [31,28,31,30,31,30,31,31,30,31,30,31]
sum = 0
sumDays = []
for i in monthDays:
sumDays.append(365 - sum)
sum += i
timeInp = input("Please enter your birthdate in the format dd/mm/yyyy\n")
timeInp = timeInp.split("/")
days = int(timeInp[0])
months = int(timeInp[1])
years = int(timeInp[2])
totDays = 10000
if totDays > 366:
if isLeapYear(years):
if months == 1 or months == 2:
totDays -= (sumDays[months - 1] + 1 - days) + 1
else:
totDays -= (sumDays[months - 1] - days) + 1
else:
totDays -= (sumDays[months - 1] - days) + 1
months = 1
days = 1
years += 1
while totDays > 366:
if isLeapYear(years):
totDays -= 366
else:
totDays -= 365
years += 1
i = 0
while totDays != 0:
if isLeapYear(years):
monthDays[1] = 29
else:
monthDays[1] = 28
if totDays >= monthDays[i]:
months += 1
totDays -= monthDays[i]
elif totDays == monthDays[i]:
months += 1
totDays = 0
else:
days += totDays
if days % (monthDays[i] + 1)!= days:
days %= monthDays[i] + 1
months += 1
totDays = 0
if months == 13:
months = 1
years += 1
i += 1
if i == 12:
i = 0
print(str(days) + "/" + str(months) + "/" + str(years))
顾名思义，isLeapYear()接受一个参数years，returns一个布尔值。
我们解决这个问题的第一步是先将我们的日期“翻译”到下一年，以使其更容易。这使我们以后的计算更加容易。为此，我们可以定义一个数组 sumDays 来存储每个月结束一年（进入新的一年）所需的天数。然后，我们从 totDays 中减去这个数额，考虑闰年，并更新我们的变量。
接下来是简单的部分，只要我们有足够的天数来完成一整年，就向前跳几年。
一旦我们无法再增加一整年，我们就逐月进行，直到 运行 天数已满。
示例测试用例：
输入#1：
19/05/2008
输出#1：
5/10/2035
输入#2：
05/05/2020
输出#2：
21/9/2047
输入#3：
29/02/2020
输出#3：
17/7/2047
我在这个网站上检查了我的大部分解决方案：https://www.countcalculate.com/calendar/birthday-in-days/result