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
kaizukizio: 2019-11-23 08:22:24

wow

Last edit: 2019-11-23 08:23:37
scolar_fuad: 2019-11-22 06:36:14

easy dp problem

akp_27_ics: 2019-11-11 15:06:31

Ac in 0.00s. First, hack the logic of dynamic programming then take care about zero and you are done.

phantom_w025: 2019-11-03 09:49:14

A good problem, specially for ones who have just started learning dp.

emilio_cosy: 2019-09-14 05:44:13

firstly, check if current character not equal to '0', then use substr to get next 2 element, judget if it smaller than 27, dp + memo

a_itachi: 2019-09-07 06:44:14

how to take input in this problem

vritta: 2019-08-20 02:55:32

/*spoiler alert(contains logic)*/
here we have taken the max combination till [i-1] & saw that if n[i] can be attached with n[i-1] ie. in 1212 at i=3(last) we saw if 2 can be attached with 1, then max combinations till i=1 will be added to max combi. till i=2. And see the case of 0 as no. after 0 cannot be attached ie. in 102, 02 cannot be taken as no.

Last edit: 2019-08-20 02:57:07
rameshus12: 2019-08-16 08:56:24

I'm new to this blog. My answer to this question is not accepted as it says wrong answer.

I have no clue which test case is not working. Is there a place where it shows the submitted code fails for a certain test cases?

Thanks,
Ramesh

np_soo_hard: 2019-08-14 18:26:41

Note that:- All number formed starting with 0 is invalid.
So answer for:-
- 2002 is 0
- 101 is 1 (Letter with 10 and 1)

chirayu_555: 2019-08-10 07:10:43

AC in second go. Nice problem.


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