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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 |
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2016-12-27 18:43:21
kk Last edit: 2016-12-27 19:12:20 |
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2016-12-24 00:26:07
For a sequence of 1's (or 2's or 12's) the problem is basically asking to calculate the n'th Fibonacci number. 5000'th Fibonacci is kind of a big number. UPD: fortunately, the judge only checks "reasonable" strings so "long long" is enough to get AC... Last edit: 2016-12-24 00:53:41 |
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2016-12-21 18:00:27
AC in 0.00!!! Bottom up approach!!!! Last edit: 2016-12-21 18:01:02 |
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2016-12-06 13:51:26
handling 0's was the tricky part |
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2016-12-04 17:59:54
How could the output be 89 for second input.Any intuitive idea? |
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2016-11-30 23:03:45
0's matter |
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2016-10-19 20:00:10
Highly recommended problem! Must try! The answer for some test cases at SPOJ TOOLKIT are wrong for this problem. If getting WA, try some of the following test cases. 226210 -> 3 301 -> 0 50 -> 0 1020 -> 1 2002-> 0 In short, make sure you don't fail at any inputs with zeros in it. They are the most tricky ones! |
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2016-10-17 14:54:57
@urohit011 use more readable variable names, so that others can help. |
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2016-10-11 13:27:13
solved it in O(n) .tried recursion but failed ..apoorv gaurav singh tera 4k wala t-Shirt aya? |
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2016-10-04 08:05:00
Unclear statement |