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
shehar48: 2019-05-18 21:00:48

AC at first go !

i_master: 2019-05-12 22:32:24

@get_right_jr how is 200202 => 1? Two consecutive 0's are bound to make an invalid testcase.

2_0_0_2_0_2 -> invalid
20_0_2_0_2 -> invalid
2_00_2_0_2 -> invalid
2_0_02_0_2 -> invalid
2_0_0_20_2 -> invalid
2_0_0_2_02 -> invalid
...

Find me a single case where it would be valid.

i_master: 2019-05-12 06:47:46

@amanharitsh because in 110, 11 mapped to K leaves 0 alone, 11_0 and 0 alone is not a valid letter. The reason why 110 is 1 is because its interpreted as 1_10 aka AJ, while 260 cannot be interpreted in any way. If you try 26_0 you get invalid because 0 alone is not valid. If you try 2_60 once again its not valid because 60 is past letter z. Hope this helps.

hetp111: 2019-04-22 21:27:08

Beware of zeros...!

Last edit: 2019-04-22 22:27:41
rohitkk074: 2019-04-15 17:26:44

will there be any 0 s in the input string.
like 20002

mshafquat: 2019-03-27 21:51:09

Update : got it! :)

Last edit: 2019-03-27 22:10:22
mjyoneman: 2019-03-27 14:29:55


i have written a DP solution using java but its getting TLE top-down approach ? any help
I think initializing dp[5001][5001] with all -1 is taking time ?

(edit)- Solved Yeah ! 0.28sec in java :-)

Last edit: 2019-03-27 14:40:19
anadi_s: 2019-03-23 17:11:25

Thank you for the test cases :)

nilesh8757: 2019-03-08 20:36:36

spoj toolkit is broken i guess:
1230
0

it gives 2 for above case.

suraj1611: 2019-02-10 21:16:40

Good one!


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