C实现 BinaryCode

Problem Statement
	Let's say you have a binary string such as the following: 011100011 One way to encrypt this string is to add to each digit the sum of its  adjacent digits. For example, the above string would become: 123210122 In particular, if P is the original string, and Q is the encrypted string, then Q[i] = P[i-1] + P[i] + P[i+1] for all digit positions i. Characters off the left and  right edges of the string are treated as zeroes. An encrypted string given to you in this format can be decoded as  follows (using 123210122 as an example): Assume P[0] = 0. Because Q[0] = P[0] + P[1] = 0       + P[1] = 1, we know that P[1] = 1. Because Q[1] = P[0] + P[1] +       P[2] = 0 + 1 + P[2] = 2, we know       that P[2]       = 1. Because Q[2] = P[1] + P[2] +       P[3] = 1 + 1 + P[3] = 3, we know       that P[3]       = 1. Repeating       these steps gives us P[4] = 0, P[5] = 0 , P[6] = 0, P[7] = 1,       and P[8]       = 1. We check       our work by noting that Q[8] = P[7] + P[8] = 1 + 1 = 2. Since this equation works out, we are       finished, and we have recovered one possible original string. Now we repeat the process, assuming the opposite about P[0] : Assume P[0] = 1. Because Q[0] = P[0] + P[1] = 1       + P[1] = 0, we know that P[1] = 0. Because Q[1] = P[0] + P[1] +       P[2] = 1 + 0 + P[2] = 2, we know       that P[2]       = 1. Now note       that Q[2]       = P[1] + P[2] + P[3] = 0 + 1 + P[3] = 3 , which leads us to the conclusion that P[3] = 2. However, this violates the fact that each       character in the original string must be '0' or '1'. Therefore, there       exists no such original string P where the       first digit is '1'. Note that this algorithm produces at most two decodings for any given  encrypted string. There can never be more than one possible way to decode a  string once the first binary digit is set. Given a String message, containing the encrypted string, return a  String[] with exactly two elements. The first element should contain the  decrypted string assuming the first character is '0'; the second element  should assume the first character is '1'. If one of the tests fails, return  the string "NONE" in its place. For the above example, you should  return {"011100011",  "NONE"}.
Definition
	Class: BinaryCode Method: decode Parameters: String Returns: String[] Method signature: String[] decode(String message) (be sure your method is public)
Constraints
-	message will  contain between 1 and 50 characters, inclusive.
-	Each character in message will be either '0',  '1', '2', or '3'.
Examples
0)
	"123210122"                      Returns: { "011100011",  "NONE" }                The example from above.
1)
	"11"                      Returns: { "01",  "10" }                We know that one of the digits must be '1', and the      other must be '0'. We return both cases.
2)
	"22111"                      Returns: { "NONE",  "11001" }                Since the first digit of the encrypted string is      '2', the first two digits of the original string must be '1'. Our test      fails when we try to assume that P[0] = 0.
3)
	"123210120"                      Returns: { "NONE",  "NONE" }                This is the same as the first example, but the      rightmost digit has been changed to something inconsistent with the rest      of the original string. No solutions are possible.
4)
	"3"                      Returns: { "NONE",  "NONE" }
5)
	"12221112222221112221111111112221111"                      Returns: { "01101001101101001101001001001101001",      "10110010110110010110010010010110010" }

This problem statement is the exclusive and proprietary property ofTopCoder, Inc. Any unauthorized use or reproduction of this information withoutthe prior written consent of TopCoder, Inc. is strictly prohibited. (c)2003,TopCoder, Inc. All rights reserved.

//代码如下

 

#include <stdlib.h>
#include <cstdlib>
#include <string.h>
#include <stdio.h>
#define N 50

void decode(char str[])
{
 int i,len;

 char cObj1[N],cObj2[N];
 len=strlen(str);
 
 //以0开头
 cObj1[0]=0;cObj1[1]=str[0]-'0';
    for(i=1;i<len-1;i++)
      {
       cObj1[i+1]=str[i]-'0'-cObj1[i-1]-cObj1[i];
       if(cObj1[i+1]<0||cObj1[i+1]>1) break;
      }
     

   if(i!=len-1)
       printf("NONE");
   else
    for(i=0;i<len;i++)
     {
         printf("%d",cObj1[i]);
        }
   
 printf("\n");
 
 
   //以1开头
  
   cObj2[0]=1;cObj2[1]=str[0]-'0'-1;;

   for(i=1;i<len-1;i++)
      {
       cObj2[i+1]=str[i]-'0'-cObj2[i-1]-cObj2[i];
       if(cObj2[i+1]<0||cObj2[i+1]>1) break;
      }   
 
   if(i!=len-1)
       printf("NONE");
   else
    for(i=0;i<len;i++)
     {
         printf("%d",cObj2[i]);
         }
   printf("\n");
}

main()
{
 int i;
 int cOriginLen;
 //char ch;
 char cOrigin[N];
 
 puts("输入：");
 gets(cOrigin);
 
 cOriginLen=strlen(cOrigin);
 //printf("数组长度为：%d\n",cOriginLen);
 
 for(i=0;i<cOriginLen;i++)
   {
    if(cOrigin[i]>'3'||cOrigin[i]<'0')
      {
       printf("输入错误！\n");
       printf("退出");
       system("pause");
       exit(0);      
       }  
   }
 
 
 decode(cOrigin);
 
 printf("退出");  
 system("pause"); 
 
}