Hrbust 1242 Alphacode【dp】好水啊这个dp

Alphacode

Time Limit: 1000 MS

Memory Limit: 30000 K

Total Submit: 67(13 users)

Total Accepted: 39(9 users)

Rating:

Special Judge: No

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

Sample Output

Source

East Central North America 2004

Recommend

程宪庆

题目大意：

给你一个数字字符串，其中我们将A-Z可以设定为1-26的数字（比如A设定为24），问有多少种设定方案。

思路：

计数问题，考虑dp，设定dp【i】表示长度为i的方案数。

那么其状态转移方程：

一共有两种状态，要么紧接着上一个数两个数组成一个数，要么自己独立出来。

①dp【i】=dp【i-1】（a【i】！=‘0’）；独立出来，自己作为一个数的方案递推。

②dp【i】+=dp【i-2】（当前位和上一位组成的数字要在1-26之间，并且不能有前导0）；和上一个数组成一个数的方案递推。

Ac代码：

#include<stdio.h>
#include<string.h>
#define ll long long int
ll dp[2000006];
char a[2000000];
int main()
{
    while(~scanf("%s",a))
    {
        memset(dp,0,sizeof(dp));
        if(strcmp(a,"0")==0)break;
        dp[0]=1;
        dp[1]=1;
        int n=strlen(a);
        for(int i=2;i<=n;i++)
        {
            if(a[i-1]!='0')
            dp[i]=dp[i-1];
            if(a[i-2]!='0'&&(a[i-2]-'0')*10+a[i-1]-'0'<=26&&(a[i-2]-'0')*10+a[i-1]-'0'>=1)
            {
                dp[i]+=dp[i-2];
            }
        }
        printf("%lld\n",dp[n]);
    }
}