Question

我必须编写一个程序，我输入一个数字，程序会回复该数字是否为闰年。我是python的新手所以我不确定从哪里开始。到目前为止，我知道闰年是4可以整除的任何数字，但不能被100整除（除非它也可以被400整除）。

Answer 1

也许这会有所帮助。

year = int(input("Input year: "))

if year % 4 == 0:
    print("Year is leap.")
    if year % 100 == 0 and year % 400 != 0:
        print("Year is common.")

else:
    print("Year is common.")

Answer 2

您可以做的是在代码顶部插入此功能：

def leapyear(year, querytype='is'):
    import calendar
    querytype == case(querytype, 'lowercase')
    if querytype == 'is':
        return calendar.isleap(year)
    elif querytype == 'closest':
        return year % 4

然后，要检查一年是否为闰年，请键入leapyear(THE_YEAR)’ and it will return True or False。这是如何使用它的另一个例子

def leapyear(year, querytype='is'):
    import calendar
    querytype == querytype.lower()
    if querytype == 'is':
        return calendar.isleap(year)
    elif querytype == 'closest':
        return year % 4
yeartocheck = input('Enter A Year. Your Answer: ')
if leapyear(yeartocheck) == True:
    print('It Is A Leap Year')
else:
    print('It Is Not A Leap Year')

通过我在此答案顶部提供的功能，您还可以通过键入leapyear(THE_YEAR, 'closest')

找到距离最近的闰年多少年

Answer 3

我也是python初学者。我尽力了，看看是否行得通。

def isLeap(year):
    tyear = year
    if tyear % 4 == 0:
        tyear = year
        if tyear % 100 == 0:
            tyear = year
            if tyear % 400 == 0:
                print('Leap Year')
            else:
                print('Not a Leap Year')
        else:
            print('Leap Year')
    else:
        print('Not a Leap Year')

isLeap(2004) # Returns Leap Year
isLeap(2005) # Returns Not a Leap Year

Answer 4

以前的大多数答案都是正确的，这是给猫皮的另一种方法：

数学逻辑基于此full credit

year = 2024
if year % 4 != 0:
    return False
elif year % 100 != 0:
    return True
elif year % 400 != 0:
    return False
else:
    return True

这是测试用例

def checkit(year):
    if year % 4 != 0:
        return False
    elif year % 100 != 0:
        return True
    elif year % 400 != 0:
        return False
    else:
        return True

x = range(2013,2027)
for n in x:    
    print(n,checkit(n))

Answer 5

这就是闰年的方法:)

if year % 4 == 0 and year % 100 != 0 or year % 400 == 0:
print(year, "is leap!")

其他：打印（年，“不是闰年！”）

Answer 6

如果你知道什么是闰年，那么你就知道如何用Python写这个。

你说的是闰年？给定一年，如果它可以被4整除，而不是100，那么这是真的。你如何检查一个数字是否可以被另一个数字整除？模块化算术。 x％y给出整数除法的余数（想想三等算术）。

所以对于一个int，这可以检查它是否可被4整除：

<b>Soaking time</b>:&nbsp; 3 hours <br>
<b>Preparation time:</b> 15 mins <br>
<b>Fermenting time:</b>&nbsp; 8 hours <br>
<span><b>Cooking time:</b> 45 mins<br><br></span>
1.&nbsp;&nbsp;&nbsp;&nbsp; Soak the urad dal (split black lentils)&nbsp;and fenugreek seeds (methi)&nbsp;together in water for 3 hours and drain. <br>
2.&nbsp;&nbsp;&nbsp;&nbsp; Soak the parboiled rice in water for 3 hours and drain. <br>
3.&nbsp;&nbsp;&nbsp;&nbsp; Blend urad dal and fenugreek seeds together in a mixer till smooth and frothy (add water little by little as required) remove and keep aside. <br>
4.&nbsp;&nbsp;&nbsp;&nbsp; Blend the rice in a mixer till smooth. Remove and keep aside. <br>
5.&nbsp;&nbsp;&nbsp;&nbsp; Combine the urad dal paste and rice paste together in a bowl and keep aside to ferment overnight. <br>
6.&nbsp;&nbsp;&nbsp;&nbsp; Once the batter is fermented, add salt to the batter and mix well. <br>
7.&nbsp;&nbsp;&nbsp;&nbsp; Put spoonfuls of the batter into greased idli moulds and steam for 10 to 12 minutes. <br>
8.&nbsp;&nbsp;&nbsp;&nbsp; Serve at room temperature. <br>

除此之外，还应该对闰年进行全面检查。希望这足以让你开始。

Python：确定闰年

6 个答案: