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n031k1V2EX member #183126, joined on 2016-07-22 01:37:38 +08:00 |
| 如何从 虾米音乐 分享到微信的链接中的 uxid 中找到相应用户？
1 问与答 • n031k1 • Jun 26, 2019 • Lastly replied by n031k1
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1 |
| 最短包含所有四位数的切分字段
1 问与答 • n031k1 • Dec 28, 2018 • Lastly replied by n031k1
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3 |
n031k1's recent replies
Jun 26, 2019 Replied to a topic by n031k1 › 问与答 › 如何从 虾米音乐 分享到微信的链接中的 uxid 中找到相应用户？ |
1
Dec 28, 2018 Replied to a topic by n031k1 › 问与答 › 最短包含所有四位数的切分字段 |
有没有同学救救伸手党！不行我就要动脑了！
Dec 27, 2018 Replied to a topic by n031k1 › 问与答 › 最短包含所有四位数的切分字段 |
import sys
sys.setrecursionlimit(10001)
v = ['%04d' % a for a in range(0,9999)]
e = dict([(a,[(a[1:]+b) for b in '0123456789' if a[1:] + b != a]) for a in v])
def run(n, e,used, count):
if count + 1 == len(e):
return n
used.add(n)
for i in e[n]:
if i not in used:
r = run(i, e, used, count + 1)
if r is not None:
return n[0] + r
used.remove(n)
return None
print run('0000',e,set(),0)
sys.setrecursionlimit(10001)
v = ['%04d' % a for a in range(0,9999)]
e = dict([(a,[(a[1:]+b) for b in '0123456789' if a[1:] + b != a]) for a in v])
def run(n, e,used, count):
if count + 1 == len(e):
return n
used.add(n)
for i in e[n]:
if i not in used:
r = run(i, e, used, count + 1)
if r is not None:
return n[0] + r
used.remove(n)
return None
print run('0000',e,set(),0)
May 13, 2017 Replied to a topic by mokeyjay › 程序员 › 突然想试一下前阵子流传的那种代餐粉，有 V 友吃过吗？介绍下？ |
记得有人好像试过一个月，最后好像只撑了半个月，总结：食之无味
所以生在天朝，请好好珍惜~
所以生在天朝，请好好珍惜~
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