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

hide comments
Utkarsh: 2015-12-17 15:37:58

my accepted solution gives 1 as the answer for 100 or 1000..idk why
the correct ans should be 0 for both the testcases

Junaid: 2015-12-17 14:02:34

Finally after working on this problem for more than a day......AC in one go...:)...Thanks for the testcases......

rahul: 2015-12-09 23:59:44

if using python and top down approach
import sys
sys.setrecursionlimit(10000)

sushmitkumar: 2015-12-07 08:56:15

Any reason as to why same JAVA code would take 314M one time and 1341M the next in SPOJ?

mjguru: 2015-11-28 10:17:49

No need for an array. This problem can be solved even without memoization (full on memoization).

Priyank: 2015-11-15 06:19:12

Just remember few imp. cases like 10, 20.

python_user: 2015-10-27 13:41:06

If you believe that your code is correct but is not passing some test cases, then look for ALL of the test cases mentioned in the comment section (till the last!!).

Last edit: 2015-10-27 13:41:21
Heartbreak_Kid: 2015-10-18 16:00:44

what will be the answer for "1101"?

Alex Anderson: 2015-10-11 00:35:50

Try http://www.spoj.com/problems/PONY2/ for a more advanced version of this problem.

Dushyant Singh: 2015-10-07 18:51:34

@ senshiya: Because only 1,10 is only valid and 11,0 is not.


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