Sicily 1001. Alphacode

Constraints

Time Limit: 1 secs, Memory Limit: 32 MB

Description

Alice and Bob need to send secret messages to each other and are discussing ways to encode theirmessages: 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 willdetermine 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 avalid 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 bewithin the range of a long variable.

Sample Input

25114111111111133333333330

Sample Output

6891

思路：

动态规划思想，如果当前字符为‘0’，则dp[i]=dp[i-2],如果当前字符不为‘0’，且与前面的数字组合能构成英文字母，那么dp[i]=dp[i-1]+dp[i-2]。这两种情况中，i=1的时候另外处理。

代码：

#include <iostream>#include <cstring>#include <memory.h>using namespace std;int main(){string s;while(cin>>s && s!="0"){int l = s.length();long dp[l];memset(dp,0,sizeof(dp)); dp[0] = 1;for(int i = 1; i < l; i++){if(s[i] == '0'){if(i ==1)dp[i] = dp[i-1];elsedp[i] = dp[i-2];}else{if((s[i-1] == '1' && s[i] <= '9') || (s[i-1] == '2' && s[i]<='6')){if(i != 1)dp[i] = dp[i-1] + dp[i-2];elsedp[i] = dp[i-1] + 1;}elsedp[i] = dp[i-1];}      }cout<<dp[l-1]<<endl;}return 0;}