谁能给我点意见我的超时_C语言

谁能给我点意见我的超时

Alphacode
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

25114
1111111111
3333333333
0

Sample Output

6
89
1

#include <stdio.h>
#include <iostream>
using namespace std;
int total(char *p);
int total(char *p)
{
int num,sum=0;
if (strlen(p) == 1)
return 1;
else if (strlen(p) == 2)
{
num = (*p - '0') * 10 + (*(p+1) - '0');
if (num <= 26)
sum = 2;
else
sum = 1;
}
else
{
num = (*p - '0') * 10 + (*(p+1) - '0');
if (num <= 26)
sum = total (p + 1) + total (p + 2);
else
sum = total (p + 1);
}
return sum;
}
int main()
{
char s[1000];
while (EOF != scanf("%s", s) && *s != '0')
printf("%d\n", total(s));
return 0;
}

[此贴子已经被作者于2007-11-2 16:02:49编辑过]

搜索更多相关的解决方案: 意见超时

----------------解决方案--------------------------------------------------------
超时是显然的
----------------解决方案--------------------------------------------------------

指点一下

[此贴子已经被作者于2007-11-2 13:26:01编辑过]