一马当先

时间就是金钱

365 Or 366？

y = int(year) if y % 4 == 0 and y % 100 != 0: print(366) elif y % 400 == 0: print(366) else: print(365)

傻瓜算法

y=int(year) if y%100==0: if y%400==0: print(366) else: print(365) else: if y%4==0: print(366) else: print(365)

好好好

y=int(year);print(365+(y%400==0 or (y%4==0 and y%100!=0)))优先级不好记，括号就保留吧

one line

year = int(year) # 四年一闰，百年不闰，四百年再闰 if year % 4 == 0 and year % 100 != 0 or year % 400 == 0: print(366) else: print(365)

感谢，我还以为只有四年一闰呢

四年一闰，百年不闰，四百年再闰

print(366 if (int(year)%4==0 and int(year)%100!=0) or int(year)%400==0 else 365)

print( 366 if (y:=int(year)) % 4 == 0 and y %100!=0 or y %400 == 0 else 365)


一行

year=int(year) print(365+int(year%4==0 and year%100!=0 or year%400==0))

year = int(year)
if year%100 == 0:
    if year%400 == 0:
        print(366)
    else :print(365)
else:
    if year%4 == 0:
        print(366)
    else:print(365)

不使用if的解法

if int(year)%100==0: print( 366 if int (year)%400==0 else 365) else: print( 366 if int(year)%4==0 else 365)

悄悄学习

import time #year=&quot;2000&quot; a=time.strptime(str(int(year)+1),&quot;%Y&quot;)   #格式化year+1年的时间戳 b=time.strptime(year,&quot;%Y&quot;)                 #格式化year年的时间戳 c=(time.mktime(a)-time.mktime(b))/86400 #计算year年的天数 print(int(c))

直接用时间戳计算，不需要考虑闰年问题

#判断闰平年 if year[2:]==&#x27;00&#x27;: if int(year)%400 == 0: print(366) else: print(365)     else: if int(year)%4 == 0: print(366) else: print(365)

清晰

if int(year)%100==0: print( 366 if int(year)%400==0 else 365) else: print( 366 if int(year)%4==0 else 365)或者是简化为一行代码： print( 366 if int(year)%400==0 or int(year)%4==0 and int(year)%100!=0 else 365)

简单易懂，三目运算

调用内置函数

<pre><code>year=eval(year) if year%100!=0 and year%4==0 or year%400==0: print(&quot;366&quot;) else: print(&quot;365&quot;)</code></pre>

解：

year = int(year) if year % 400 == 0 or (year % 4 == 0 and year % 100 != 0): print(366) else: print(365)

随便看看

y=int(year) if (y&lt;100 and y%4==0)or y%400==0: print(366) else: print(365)

哪里的问题呢

什么是闰年？ 1、能被4整除，但不能被100整除；2、能被400整除；

year1=int(year) if year1%400==0: &nbsp; &nbsp; print ('366') elif year1%4==0 and year1%4!=0: &nbsp; &nbsp; print ('366') else : &nbsp; &nbsp; print ('365')

print(366 if (int(year)%4==0 and int(year)%100!=0)or int(year)%400==0 else 365)

一行代码

if int(year)%400 == 0: &nbsp; &nbsp; print 366 else: &nbsp; &nbsp; if int(year)%4 == 0 and int(year)%100!=0: &nbsp; &nbsp; &nbsp; &nbsp; print 366 &nbsp; &nbsp; else: &nbsp; &nbsp; &nbsp; &nbsp; print 365

四年一闰，百年不闰，四百年再一闰年

why?

y = int(year) if y%400==0: &nbsp; &nbsp; print 366 elif y%4==0 and y%100!=0: &nbsp; &nbsp; print 366 else: &nbsp; &nbsp; print 365

if判断实现

哈哈哈，咱俩思路一样。没啥毛病，就是不太简洁

if int(year)%400==0: &nbsp; &nbsp; print 366 elif int(year)%4 == 0 and int(year)%100 != 0: &nbsp; &nbsp; print 366 else: &nbsp; &nbsp; print 365 满足1.被400整除2.被4整除，但不能被100整除

365,366

shell里都试过了，在这上面就过不了。。。 无语

第23题：365 Or 366？

吃糖果

乘法运算

百度时间

平分果子