Submit | All submissions | Best solutions | Back to list |
ACODE - Alphacode |
Alice and Bob need to send secret messages to each other and are discussing ways to encode their messages:
Alice: “Let’s just use a very simple code: We’ll assign ‘A’ the code word 1, ‘B’ will be 2, and so on down to ‘Z’ being assigned 26.”
Bob: “That’s a stupid code, Alice. Suppose I send you the word ‘BEAN’ encoded as 25114. You could decode that in many different ways!”
Alice: “Sure you could, but what words would you get? Other than ‘BEAN’, you’d get ‘BEAAD’, ‘YAAD’, ‘YAN’, ‘YKD’ and ‘BEKD’. I think you would be able to figure out the correct decoding. And why would you send me the word ‘BEAN’ anyway?”
Bob: “OK, maybe that’s a bad example, but I bet you that if you got a string of length 5000 there would be tons of different decodings and with that many you would find at least two different ones that would make sense.”
Alice: “How many different decodings?”
Bob: “Jillions!”
For some reason, Alice is still unconvinced by Bob’s argument, so she requires a program that will determine how many decodings there can be for a given string using her code.
Input
Input will consist of multiple input sets. Each set will consist of a single line of at most 5000 digits representing a valid encryption (for example, no line will begin with a 0). There will be no spaces between the digits. An input line of ‘0’ will terminate the input and should not be processed.
Output
For each input set, output the number of possible decodings for the input string. All answers will be within the range of a 64 bit signed integer.
Example
Input: 25114 1111111111 3333333333 0 Output: 6 89 1
Added by: | Adrian Kuegel |
Date: | 2005-07-09 |
Time limit: | 0.5s |
Source limit: | 50000B |
Memory limit: | 1536MB |
Cluster: | Cube (Intel G860) |
Languages: | All |
Resource: | ACM East Central North America Regional Programming Contest 2004 |
hide comments
|
||||||||||||||
2020-07-12 17:42:35
Solved in 2nd Attempt. |
||||||||||||||
2020-07-03 07:48:36
try 113011 1010 10110 226210 310 Last edit: 2020-07-03 08:46:10 |
||||||||||||||
2020-05-19 04:49:53
This was not a good question, the boundary cases I got to know from the comments. So please check the comments. Also see my submission for more clarity : <snip> Last edit: 2023-06-07 12:15:04 |
||||||||||||||
2020-05-18 19:03:47
I think It should be mentioned that a substring 02 will not be counted valid |
||||||||||||||
2020-05-10 09:05:36
what should be the output for 20201 will 2|020|1 count as a valid partition |
||||||||||||||
2020-05-05 10:11:55
the problem is easy except corner cases like 109, 10009, 20110 etc. |
||||||||||||||
2020-04-29 17:57:40
getting WA .. try Test Cases in the comments |
||||||||||||||
2020-04-29 12:05:03
In 108, will 08 be treated as H or only 10 will be treated? |
||||||||||||||
2020-04-28 17:40:23
Turned such a pretty question into something so ugly thanks to 10 and 20... |
||||||||||||||
2020-04-18 16:05:18
I think this problem is not so much about main logic but about handling corner cases |