zoj 2202 ALPHACODE

Alphacode

Time Limit: 1 Second      Memory Limit: 32768 KB

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

25114
1111111111
3333333333
0

Sample Output

6
89
1

做完这一题之后才知道自己的逻辑有多么差了，还有考虑问题的角度是在是太次了

这题的关键在于对0的处理。如果当前字符是0那么考虑前面的字符，如果是0或者是大于2那么直接判断这个密码是不存在的

否则可知道当前字符串的组合密码与少0的字符串的组合密码是相同的，而且0必然是与前一个字符联系在一起。

哎，自己做的果然太次了，加油啊！

#include <iostream> #include <string> using namespace std; int code(string str, int l , int *n) { int x; if( l == 1) { *n = 1; return 1; } else { if(str.at(l-1) == '0') { if(str.at(l-2)=='0'||str.at(l-2)>'2') return x = 0; x = code(str,l-1,n); x = *n; *n = 0; return x; } if( str.at(l-2) > '2'||(str.at(l-2)=='2'&&str.at(l-1)>'6')||str.at(l-2)=='0') { x = code(str,l-1,n); *n = x; return x; } else { x = code(str,l-1,n); int t = x; x = 2**n + x-*n; *n = t; return x; } } } int main() { string str; while(cin>>str&&str!="0") { int n = 0 ; int r = code(str,str.length(),&n); cout<<r<<endl; } return 0; }

 

 

 

 

与上面的代码不同，这里采用非递归，但是基本逻辑思路是相同的

#include <iostream> using namespace std; int main() { char s[5000]; //int r[5000]; while(gets(s)) { int pre = 1,next=0; int n = 1; if(!strcmp(s,"0")) return 0; int l = strlen(s); for( int i = 1; i < l ; i++) { if(s[i] == '0') { if(s[i-1]=='0'||s[i-1]>'2') { pre = 0; break;} next = n; n = 0; } else { if( s[i-1] > '2'||(s[i-1]=='2'&&s[i-1]>'6')||s[i-1] == '0') { next = pre; n = pre; } else { next = pre; int t = next; next = 2*n + next-n; n = t; } } pre = next; } cout<<pre<<endl; } return 0; }

 

别人的代码，和我的不同，但是他开了两个大数组，不过逻辑却简单很多。学学习啊

#include <iostream> #include <cstring> using namespace std; #ifdef WIN32 #define for if(0); else for #endif int getnum(char x, char y) { int num = (x - '0') * 10 + y - '0'; return num; } int main() { char input[10000]; while (cin >> input) { if (strcmp(input, "0") == 0) break; int len = strlen(input); int* res = new int[len + 1]; for (int i = 0; i <= len; i ++) res[i] = 0; res[len] = 1; if (input[len - 1] == '0') res[len - 1] = 0; else res[len - 1] = 1; for (int i = len - 2; i >= 0; i --) { if (input[i] != '0') res[i] = res[i + 1]; int temp = getnum(input[i], input[i + 1]); if (temp >= 10 && temp <= 26) res[i] += res[i + 2]; } cout << res[0] << endl; delete []res; } return 0; }

 

测试数据

20
200
2002
210
3030
2303