Question

在模块操作员的剩余零件工作方面,我遇到一些困难。

基本上,这是我法典的一条线。

fday= (day+24) % 30

我只想增加24天,用其余时间。然而,如果一天用户进入6个用户,结果就等于我不想要的零。

如果进入了6个行动,我如何能够使行动返回6个? 是否有更好的办法这样做?

<>Update:

I m试图采用以前确定的变量(day),并将这一数值增加24天。

然而,如果用户进入该月的第25天,那么我就增加了24天,但每月有49天。

因此,我试图利用单元行动给我剩余的东西,因为它在一个月里工作了30天。

另一个例子:

如果是5个投入,则为5+24,为29个,然后是29%30 = 29。因此,5个工厂为我试图做的工作(这只是增加价值24天,使产出保持在30天以下(因为每月只有30天时间用于我做的事情)。

Answer 1

You say in the comments that for 5 you want 29, for 6 you want 6, and for 7 you want 1. That doesn t make sense. The correct value for 6 is 30 not 6 or 0 when you re calculating a day modulo 30.

In math, division by 30 gives a remainder between 0 and 29 -- that s what modulo 30 is. What you want is a number between 1 and 30.

因此,你需要花一天时间:subtract 1。 (到0和29而不是1和30>之间,再加24,再加< <>30>, 即<>addcode>1 (在1和30之间回过一天)。

result = ((day - 1 + 24) % 30) + 1

That way you ll always get the correct number between 1 and 30 and you don t have to think of 6 as a special case.

Answer 2

Consider using the functionality in the datetime module instead of trying to reinvent it:

>>> from datetime import date, timedelta
>>> for month in (4, 5):
...     for day in (5, 6, 7, 8):
...         start = date(2011, month, day)
...         later = start + timedelta(days=24)
...         print str(start), str(later)
...
2011-04-05 2011-04-29
2011-04-06 2011-04-30
2011-04-07 2011-05-01
2011-04-08 2011-05-02
2011-05-05 2011-05-29
2011-05-06 2011-05-30
2011-05-07 2011-05-31
2011-05-08 2011-06-01

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