Dear all,
It would be great if someone helps me with this challenge :
http://uva.onlinejudge.org/index.php?op ... oblem=3191
When the number M cannot "overlap" itself, it is not a big deal to compute it recursively.
By "overlapping itself", here is an example :
If M=1212, then the number "121212" contains M both at its beginning and at its end. If M=12312, then "12312312" is another string containing twice M overlapping itself. The overlap is the substring "12" in the middle. M=123 is a case where M cannot overlap itself.
The tricky cases are when M can overlap itself.
I have looked for similar situations on other topics, like 10722 - Super lucky numbers and 10712 - Count the numbers:
http://online-judge.uva.es/board/viewto ... ky+numbers
Any little hint will allow me to move forward.
Thanks in advance for your help !
12040 - Again Lucky Numbers
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Re: 12040 - Again lucky numbers
this problem can be solved using dp + some prefix matching techniquebrainless_the_swiss wrote:Dear all,
It would be great if someone helps me with this challenge :
http://uva.onlinejudge.org/index.php?op ... oblem=3191
When the number M cannot "overlap" itself, it is not a big deal to compute it recursively.
By "overlapping itself", here is an example :
If M=1212, then the number "121212" contains M both at its beginning and at its end. If M=12312, then "12312312" is another string containing twice M overlapping itself. The overlap is the substring "12" in the middle. M=123 is a case where M cannot overlap itself.
The tricky cases are when M can overlap itself.
I have looked for similar situations on other topics, like 10722 - Super lucky numbers and 10712 - Count the numbers:
http://online-judge.uva.es/board/viewto ... ky+numbers
Any little hint will allow me to move forward.
Thanks in advance for your help !