2013
12-12

# Alphacode

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 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

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.

25114
1111111111
3333333333
0

6
89
1

HintHint
It can be confirmed that no zero in the head and no trailing zeros in the integer.
Please pay special notice on the position of zero.

#include<stdio.h>
#include<string.h>
#define N 1000005
typedef long long LL;
char s[N];
int a[N],ln;
LL dp[N];
LL dfs(int x)
{
if(x>=ln) return 1;
if(dp[x]>0) return dp[x];
if(a[x+1]!=0) dp[x]+=dfs(x+1);
if(a[x]*10+a[x+1]<=26&&a[x+2]!=0) dp[x]+=dfs(x+2);
return dp[x];
}
int main()
{
while(gets(s+1))
{
if(s[1]=='0') break;
ln=strlen(s+1);
int i;
for(i=1; i<=ln; i++) a[i]=s[i]-'0';
a[i]=1;
memset(dp,0,sizeof(dp));
dp[1]=dfs(1);
printf("%I64d\n",dp[1]);
}
return 0;
}

1. 第一句可以忽略不计了吧。从第二句开始分析，说明这个花色下的所有牌都会在其它里面出现，那么还剩下♠️和♦️。第三句，可以排除2和7，因为在两种花色里有。现在是第四句，因为♠️还剩下多个，只有是♦️B才能知道答案。

2. 您没有考虑 树的根节点是负数的情况， 若树的根节点是个很大的负数，那么就要考虑过不过另外一边子树了