What is the python leap year determination code?

Python leap year determination code: Use inline if statements to implement, the code is [if (year % 4) == 0:if (year % 100)== 0:if (year % 400) == 0:print("{0} is a leap year".format(year))].

The operating environment of this tutorial: Windows 7 system, python version 3.9, DELL G3 computer.

python leap year determination code:

Leap year: leap year: every four years, no leap year, and leap year again every four hundred years.

For example, 2000 is a leap year and 2100 is a normal year

year = int(input("输入一个年份: "))
if (year % 4) == 0:
   if (year % 100) == 0:
       if (year % 400) == 0:
           print("{0} 是闰年".format(year))   # 整百年能被400整除的是闰年
       else:
           print("{0} 不是闰年".format(year))
   else:
       print("{0} 是闰年".format(year))       # 非整百年能被4整除的为闰年
else:
   print("{0} 不是闰年".format(year))

We can also use embedded if statements to achieve:

The output result of executing the above code is:

输入一个年份: 2000
2000 是闰年

rrree

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