题目描述
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.
输入要求
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
输出要求
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.
假如输入
25114
1111111111
3333333333
0
应当输出
6
89
1
题目大意很简单，用1-26来代表A-Z，比如‘BEAN’代表25114，但25114并不一定是‘BEAN’，你的目的就是为了求出这个数字串可以代表多少个字母串。一开始用的dfs，同时我也注意到了有10，和20要处理，但会超时，之后想了想可以用dp，就和走楼梯差不多，是一步走两阶还是一步走一阶，是不是很像啊。
#include<iostream>
#include<algorithm>
#include <vector>
#include<string.h>
#include<string>
#include<ctype.h>
#include<math.h>
#include <map>
using namespace std;
void solve();
int main()
{
solve();
return 0;
}
void solve()
{
char str[10000];
int ans[10000],i;
while(gets(str)&&str[0]!='0')
{
ans[0]=ans[1]=1;
for(i=1;i<strlen(str);i++)
{
if(str[i]!='0'&&str[i-1]=='1'||str[i]>='1'&&str[i]<='6'&&str[i-1]=='2')
ans[i+1]=ans[i]+ans[i-1];
else if(str[i]=='0')
ans[i+1]=ans[i-1];
else
ans[i+1]=ans[i];
}
cout<<ans[strlen(str)]<<endl;
}
}