POJ 2033 Alphacode

Alphacode

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 11666		Accepted: 3564

Description

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 500 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 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 long variable.

Sample Input

25114
1111111111
3333333333
0

Sample Output

6
89
1
题目大意：将一个仅由大写字母组成的字符串，以每个字母的编号代替原字母，组成一个由0~9组成的数字序列。即1代替A，2代替B，10代替J等等。但是，将这个数字序列进行相同方式的解密，却有不止一个解。现给定一个某单词加密后的数字序列，问将这个数字序列解密后，会得到几种不同的单词。

#include <stdio.h>
#include <iostream>
#include <string.h>
using namespace std;

int main()
{
    char str[];
    int dp[];
    while(scanf("%s", str) != EOF && str[] != '')
    {
        memset(dp, , sizeof(dp));
        dp[] = dp[] = ;
        for (int i = ; i < strlen(str); i++)
        {
            if (str[i] == '')
            {
                dp[i + ] = dp[i - ];
            }
            else
            {
                if (str[i] - '' + (str[i - ] - '') *  <=  && str[i - ] != '')
                {
                    dp[i + ] = dp[i] + dp[i - ];
                }
                else
                {
                    dp[i + ] = dp[i];
                }
            }
        }
        printf("%d\n", dp[strlen(str)]);
    }
    return ;
}