1002.Alphacode

sicily 1001

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 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

tips: 一开始我想到的是递归，以字符串为参数，进行判断，后来发现挺麻烦的

      去看网上的解答，发现用dp就挺好的

      昨天有点晚，早上来理一下思路

      我们用dp思想去做，则让dp[i]等于字符串的前i+1个字符不同组合的个数

      首先，输入保证合法，即若字符串中有‘0’，则必与前一个数字搭配，换句话说

      if (dp[i]=='0') dp[i]=dp[i-2]  //因为dp[i]与dp[i-1]形成了固定搭配，相当于在dp[i-2]所具有的所有组合后加上一个两位数而已，不增加任何组合数

      其次，对于每个为‘0’的字符，他们自身必定合法

      if (dp[i]!='0') dp[i]=dp[i-1]   //这里其实是不对的，但是思路要一步一步来才清楚

      这里其实跟为‘0’的解释有相近之处，不过，有一种情况未指出

      if (dp[i]!='0'&&dp[i-1]!='0'&&(dp[i-1]-'0')*10+dp[i]-'0'<=26) dp[i]=dp[i-1]+dp[i-2]   //就是当i位与i-1位的字符为合法组合的情况

      这里其实想一下就懂了，不过做做笔记总是好的

      因为本身合法，故可以认为在dp[i-1]的所有组合上加了一个字符，又因为与前个字符组合亦合法，可以认为在dp[i-2]的所有组合上加了一个字符

      最后，因为有i-2的存在，需要去注意下标的合法

      //代码来源于网络

#include
#include
#include
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];
                          else
                           dp[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];
                                   else
                                   dp[i] = dp[i-1] + 1;
                       }
                      else
                       dp[i] = dp[i-1];
                  }
                  
          }
          cout< 
  
  return 0;
 
  
}