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
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