1001. Alphacode （sicily）

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

Alice和Bob需要互相传送秘密消息，他们正在讨论怎样为他们的消息编码：

Alice：“我们就用一种非常简单的编码吧：我们将’A’视为1，’B’为2，以此类推，’Z’为26。”

Bob：“那是种愚蠢的编码，Alice。假如我发送’BEAN’，编码为25114。你可以用许多种不同的方式去译码！”

Alice：“你当然可以用不同方式译码，但是你将会得到什么单词？除了’BEAN’你会得到’BEAAD’，’YAAD’，’YAN’，’YKD’和’BEKD’。我认为你能够找出正确的译码。那么为什么不能发送’BEAN’呢？”

Bob：“好吧，也许那不是一个好例子，但是我敢打赌，如果你收到一个长度为500的编码，那么将会有上吨不同的译码，那样你可能不只找出一种有意义的译码了。”

Alice：“多少种不同的译码？”

Bob：“数不胜数！”

由于一些原因，Alice并未被Bob说服，所以她需要一个程序来决定对于一个给定的编码，有多少种译码方式。

    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.

输入样例

    25114

    1111111111

    3333333333

    0

输出样例

    6

    89

    1

 代码实现（由于字符串进行了处理，所以时间不满足1s的要求）

  1 #include <stdlib.h>
  2 #include <iostream>
  3 #include <string>
  4 using namespace std;
  5 #define MAX 10000
  6 int totalCount(string a);
  7 int main()
  8 {
  9     string temp;
 10     long sets[MAX]={0};
 11     int point=0;
 12     while(cin>>temp&&temp!="0")
 13     {
 14          int result=0;
 15          //划分子集——第一位和后面几位、第一、二位和后面几位
 16          if(temp.length()==1)
 17          {
 18              sets[point]=1;
 19              point++;
 20              continue;
 21          }
 22          else
 23          {
 24              string abc1="";
 25              string abc2="";
 26              for(int i=1;i<temp.length();i++)
 27              {
 28                      abc1+=temp[i];
 29                      if(i!=1)
 30                      {
 31                          abc2+=temp[i];   
 32                      }     
 33              }
 34              
 35              int ii=(temp[0]-'0')*10+(temp[1]-'0');
 36              if(ii!=10)
 37              {
 38                  result+=totalCount(abc1);
 39              }
 40              //cout<<"输出 "<<temp[0]<<temp[1]<<ii<<endl;
 41              if(ii>9&&ii<27)
 42              {
 43                   result+=totalCount(abc2);
 44              }
 45              sets[point]=result;
 46              point++;
 47          }
 48          
 49           
 50     }
 51     bool flag=true;
 52     for(int i=0;i<MAX&&flag;i++)
 53     {
 54          if(sets[i]!=0)
 55          {
 56               cout<<sets[i]<<endl;
 57               
 58          }
 59          else
 60          {
 61              flag=false;
 62          }
 63     }
 64     system("pause");
 65     return 0;
 66 }
 67 int totalCount(string a)
 68 {
 69     if(a==""||a.length()==1)
 70     {
 71         return 1;               
 72     }
 73     int count=0;
 74 
 75     int b=(a[0]-'0')*10+(a[1]-'0');
 76     if(b!=10)
 77     {
 78         string abc="";
 79         for(int i=1;i<a.length();i++)
 80         {
 81                 abc+=a[i];
 82         }
 83         count+=totalCount(abc);
 84     }
 85     if(b!=10)
 86     {
 87     }
 88     //cout<<"输出 "<<b<<endl;
 89     if(b<27&&b>9)
 90     {
 91         string abc="";
 92         for(int i=2;i<a.length();i++)
 93         {
 94                 abc+=a[i];
 95         }
 96         count+=totalCount(abc);
 97     }
 98     //cout<<"输出 "<<count<<endl;
 99     return count;
100 }

 

 提交版本

 1 //#include <stdlib.h>
 2 #include <iostream>
 3 #include <string>
 4 #define MAX 100000
 5 using namespace std;
 6 int main()
 7 {
 8     long long array[MAX];
 9     string str;
10     int size;
11     while(cin>>str&&str!="0")
12     {
13         size=str.length();
14         if(size==1)
15         {
16             cout << "1" << endl; 
17             continue;
18         } 
19         else
20         {
21             array[0]=1; 
22             for(int i=1;i<size;i++)
23             {
24                  int flag=(str[i-1]-'0')*10+(str[i]-'0');
25                  if(i==1)
26                  {
27                      if(str[i]=='0'||flag<10||flag>26)
28                      {
29                           array[i]=array[i-1];
30                      }
31                      else
32                      {
33                          array[i]=2;
34                      }
35                  }
36                  else
37                  {
38                      
39                      if(str[i]=='0')
40                      {
41                           array[i]=array[i-2];
42                           array[i-1]=array[i-2];
43                      }
44                      else if(flag<10||flag>26)
45                      {
46                          array[i]=array[i-1];
47                      }
48                      else
49                      {
50                          array[i]=array[i-1]+array[i-2];
51                      }
52                  }
53             }
54         }
55         cout << array[size-1] << endl; 
56     }
57     return 0;
58 }