SPOJ 394. Alphacode 种类数

ACODE - Alphacode

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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 5000 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 at most 5000 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 64 bit signed integer.

Example

Input:25114111111111133333333330Output:6891

          

  用d[i-1]表示前i-1个字母可以转化成的字符串的种类数，那么d[i]一定是》=d[i-1]的，因为如果str[i]是1——9，那么d[i]=d[i-1],如果str[i-1]和str[i]构成一个1-26之间的数（并且str[i+1]!='0',因为如果str[i+1]=='0'，那么它一定是和str[i]一起组成了一个数），d[i]=d[i-1]+d[i-2].另外还要避免1205这样的串中05不能算一个数，0只能和2构成20.

      这个题用cin读20ms,用scanf读是10ms......................

#include<cstdio>#include<iostream>#include<cstring>#define LL long longusing namespace std;char str[5010];LL d[5010];int main(){    //freopen("in.txt","r",stdin);    int i;    // while(cin>>str&&strcmp(str,"0")!=0)    while(scanf("%s",str)!=EOF&&strcmp(str,"0")!=0)    {        memset(d,0,sizeof(d));        d[0]=d[1]=1;        int num=0;        num=(str[0]-'0')*10+str[1]-'0';        if(num>0&&num<=26&&str[2]!='0'&&str[1]!='0')  d[1]++;        for(i=2; str[i]!='\0'; i++)        {            d[i]=d[i-1];            num=(str[i-1]-'0')*10+str[i]-'0';            if(num>0&&num<=26&&str[i+1]!='0'&&str[i]!='0'&&num!=str[i]-'0')                d[i]+=d[i-2];            //cout<<i<<": "<<d[i]<<endl;        }        cout<<d[strlen(str)-1]<<endl;    }    return 0;}