假设有三个区间：（1，100），（101，300），(301,999)
给定一个整数n，请问有什么简便有效的方法可以判断n落在哪个区间内呢？

1£¬100£¬101£¬300£¬301£¬999²ÉÓÃ2·Ö·¨£¬ºÙºÙ¡£

ÔÚ07-6-12£¬Shuning Hong <hongshuning在gmail.com> дµÀ£º
>
> ¼ÙÉèÓÐÈý¸öÇø¼ä£º£¨1£¬100£©£¬£¨101£¬300£©£¬(301,999)
> ¸ø¶¨Ò»¸öÕûÊýn£¬ÇëÎÊÓÐʲô¼ò±ãÓÐЧµÄ·½·¨¿ÉÒÔÅжÏnÂäÔÚÄĸöÇø¼äÄÚÄØ£¿
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On 6/12/07, Shuning Hong <hongshuning在gmail.com> wrote:
> 假设有三个区间：（1，100），（101，300），(301,999)
> 给定一个整数n，请问有什么简便有效的方法可以判断n落在哪个区间内呢？

a= 112
In [4]:  a in range(300,400)
   ...:
Out[4]: False
In [5]:  a in range(101,300)
   ...:
Out[5]: True

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import bisect

r = bisect.bisect([1,101,301,1000], n)

if r == 1:
 print (1, 100)
elif r == 2:
 print (101, 300)
elif r == 3:
 print (301, 999)
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>>> l = [[1, 100], [101, 300], [301, 999]]
>>> def locate(v, l):
...     for x in l:
...             if ( v>=x[0] and v<=x[1]):
...                     return x
...
>>> locate(30, l)
[1, 100]
>>> locate(300, l)
[101, 300]
>>> locate(305, l)
[301, 999]

个人以为pan的方法不错.

2007/6/13, pan <nirvana117 at gmail.com>:
>
> import bisect
>
> r = bisect.bisect([1,101,301,1000], n)
>
> if r == 1:
>  print (1, 100)
> elif r == 2:
>  print (101, 300)
> elif r == 3:
>  print (301, 999)
>
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On 6/13/07, 东子/hydon <hydonlee在gmail.com> wrote:
> >>> l = [[1, 100], [101, 300], [301, 999]]
> >>> def locate(v, l):
> ...     for x in l:
> ...             if ( v>=x[0] and v<=x[1]):
> ...                     return x
> ...

在python中，v>=x[0] and v<=x[1]可以写成x[0] <= v <= x[1]


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