poj2033 简单dp

Alphacode

Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 12136 Accepted: 3698

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

25114111111111133333333330

Sample Output

6891

Source

East Central North America 2004

直接对数字为数进行dp

#include<stdio.h>#include<iostream>#include<algorithm>#include<string.h>using namespace std;char c[100005];long long dp[100005];int main(){    int i,t;    while(1)    {        cin>>c;        t=(int)strlen(c);        if(c[0]=='0')            break;        memset(dp,0,sizeof(dp));        for(i=0;i<t;i++)        {            if(i==0)            {                dp[i]=1;            }            else            {                if(c[i]=='0')                {                    if(i>=2)                    dp[i]=dp[i-2];                    else                    dp[i]=1;                    continue;                }                if(c[i-1]=='0')                {                    dp[i]=dp[i-1];                    continue;                }                if(c[i-1]>'2')                {                    dp[i]=dp[i-1];                }                else                {                    if(c[i-1]=='2'&&c[i]>'6')                    {                        dp[i]=dp[i-1];                        continue;                    }                    if(i>=2)                    dp[i]=dp[i-1]+dp[i-2];                    else                        dp[i]+=dp[i-1]+1;                }            }        }        cout<<dp[t-1]<<endl;    }    return 0;}