如何打印给定年份下15个闰年的列表？

Question

编写一个Python程序来生成从给定年份开始的下15个闰年。将闰年填充到列表中并显示该列表。还要编写pytest测试用例来测试程序。

def find_leap_years(given_year):
    list_of_leap_years=[0]*15

    # Write your logic here
    if given_year%100==0 and given_year%400!=0:
        for i in range(0,15):
             temp=given_year+(4*(i+1))
             list_of_leap_years[i]=temp
    elif given_year%400==0:
        for i in range(0,15):
            temp=given_year+(4*i)
            list_of_leap_years[i]=temp
    elif given_year%4==0:
        for i in range(0,15):
            temp=given_year+(4*i)
            list_of_leap_years[i]=temp
    elif given_year%4==1:
        for i in range(0,15):
            temp=given_year+3+(4*i)
            list_of_leap_years[i]=temp
    elif given_year%4==2:
        for i in range(0,15):
            temp=given_year+2+(4*i)
            list_of_leap_years[i]=temp
    elif given_year%4==3:
        for i in range(0,15):
            temp=given_year+1+(4*i)
            list_of_leap_years[i]=temp

    return list_of_leap_years

next_leap_years=find_leap_years(1684)
print(next_leap_years)

当我将给定年份设为1684年时，测试用例失败了，因为我的程序在闰年列表中打印1700，但1700不是闰年。

Answer 1

我会尝试一些更简单的方法，并使用datetime模块。当构造2月29日的日期失败时，抛出异常-我们可以捕获它：

from datetime import date

def find_leap_years(year, num=15):
    count = 0
    while count < num:
        try:
            d = date(year, 2, 29)
            yield year
            count += 1
        except ValueError:
            continue
        finally:
            year += 1

for i, years in enumerate( find_leap_years(1690, 15), 1 ):
    print(i, years)

打印(注意，1700不在返回值中)：

1 1692
2 1696
3 1704
4 1708
5 1712
6 1716
7 1720
8 1724
9 1728
10 1732
11 1736
12 1740
13 1744
14 1748
15 1752

Answer 2

def find_leaps(year, counts=15):
     cnt = 0
     leaps = []
     while cnt != counts:
         if is_leap(year):
             leaps.append(year)
             cnt += 1
         year += 1
     return leaps

和

 def is_leap(year):
     return (year%4 == 0 and year%100 != 0) or (year % 400 == 0)

所以

In [3]: find_leaps(1684)
Out[3]:
[1684,
 1688,
 1692,
 1696,
 1704,
 1708,
 1712,
 1716,
 1720,
 1724,
 1728,
 1732,
 1736,
 1740,
 1744]

Answer 3

import calendar

def Leap_Year(Test_input):
    count = 0
    while count < 15:
        if calendar.isleap(Test_input):
            print(Test_input)
            count += 1
        Test_input += 1


Test_input = int(input("Enter Year: "))
Leap_Year(Test_input)