Sicily 1001.Alphacode | 动态规划

1001. Alphacode 

Time Limit: 1sec Memory Limit:32MB 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

刚开始我是用递归写的，发现超时了，后来去搜了一下，才发现是要用动态规划，果然是too young too simple呀！还得努力才行！

递归版本：

#include<iostream>#include<cstdio>using namespace std;int f(string n,int pos){if(pos==n.length()-1) return 1;else if(pos==n.length()-2&&n[pos]>='1'&&n[pos]<='2'&&n[pos+1]>='0'&&n[pos+1]<='6') return 2;else if(pos==n.length()-2) return 1;if(n[pos]>='1'&&n[pos]<='2'&&n[pos+1]>='0'&&n[pos+1]<='6') return f(n,pos+2)+f(n,pos+1);else return f(n,pos+1);}char n[1000];int main(){while(scanf("%s",&n)&&n[0]!='0'){string temp;temp+=n;printf("%d\n",f(temp,0));}} 

动态规划版本：

#include<iostream>#include<cstring>using namespace std;int main(){    string s;    while(cin>>s && s!="0")    {          int len = s.length();          long dp[len];          memset(dp,0,sizeof(dp));           dp[0] = 1;          for(int i = 1; i < len; i++)          {                  if(s[i] == '0')                  {                    if(i ==1)                        dp[i] = dp[i-1];                    else                        dp[i] = dp[i-2];//因为0必须和上面一位组成数字，所以dp[i]就等于上移1+1位                  }                  else                  {                      if((s[i-1] == '1' && s[i] <= '9') || (s[i-1] == '2' && s[i]<='6'))//若符合条件                       {                        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<<dp[len-1]<<endl;    }   // system("pause");}

该段代码参考自：mrlaker‘s blog，本人添加了一些注释以及修改了一些变量名。