URAL 1780 G - Gray Code 找规律-白红宇

URAL 1780 G - Gray Code 找规律

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发布时间：2019-06-21

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G - Gray Code

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://acm.hust.edu.cn/vjudge/contest/view.action?cid=87643#problem/G

Description

Denis, Vanya and Fedya gathered at their first team training. Fedya told them that he knew the algorithm for constructing a

Gray code.

Create a 2-bit list: {0, 1}.

Reflect this list and concatenate it with the original list: {0, 1, 1, 0}.

Prefix old entries with 0, and prefix new entries with 1: {00, 01, 11, 10}.

Repeat steps 2 and 3 until the length of all elements is equal to n.

The number

n is a

length of a Gray code. For example, the code of length 3 is: {000, 001, 011, 010, 110, 111, 101, 100}.

Denis ran the Fedya's algorithm and obtained a binary number

x at position

k (positions are numbered starting from zero). Vanya wrote down the numbers

k and

x in binary system. This story happened many years ago and now you hold the paper sheet with these numbers in your hands. Unfortunately, some digits are unreadable now. Could you determine the values of these digits using the readable digits?

Input

The first line contains a number k written in the binary system. Unreadable digits are denoted with symbol “?”. The second line contains a number x in the same format. The lengths of these numbers are equal and don't exceed 10 ⁵. The numbers may contain leading zeroes.

Output

If there is a unique way to restore the numbers k and x, output them, replacing the symbols “?” with “0” or “1”. If there are multiple ways to restore them, output “Ambiguity”. If Denis or Vanya certainly made a mistake in these numbers, output “Impossible”.

Sample Input

0?1

0?0

Sample Output

011

010

HINT

题意

题目给了一种序列的构造方式

然后给你一个k，判断第k个数是不是说给的x

题解：

找规律题，注意翻转= =

代码：

#include 
     
      #include 
      
       #include 
       
        #include 
        
         #include 
         
          #include 
          
           #include 
           
            #include 
            #include 
             
              #include 
              
               #include 
               
                #include 
                
                 #include 
                 
                  #include 
                  
                   #include 
                   
                    typedef unsigned char byte;#define pb push_back#define input_fast std::ios::sync_with_stdio(false);std::cin.tie(0)#define local freopen("in.txt","r",stdin)#define pi acos(-1)const char * fail = "Impossible";const char * more = "Ambiguity";using namespace std;const int maxn = 1e5 + 500;int dp[maxn][2];char s1[maxn] , s2[maxn];int main(int argc,char *argv[]){ scanf("%s%s",s1+1,s2+1); int Length = strlen(s1+1); s1[0] = '0'; for(int i = Length ; i > 0 ; -- i) { if (s2[i] == '0') { //同号 if (s1[i] == '0' && s1[i-1] == '0') continue; if (s1[i] == '0' && s1[i-1] == '1') { printf("%s\n",fail); return 0; } if(s1[i] == '0' && s1[i-1] == '?') { s1[i-1] = '0'; continue; } if (s1[i] == '1' && s1[i-1] == '0') { printf("%s\n",fail); return 0; } if (s1[i] == '1' && s1[i-1] == '1') continue; if (s1[i] == '1' && s1[i-1] == '?') { s1[i-1] = '1'; continue; } if (s1[i] == '?' && s1[i-1] == '0') { s1[i] = '0'; continue; } if (s1[i] == '?' && s1[i-1] == '1') { s1[i] = '1'; continue; } if (s1[i] == '?' && s1[i-1] == '?') continue; } if (s2[i] == '1') { //异号 if (s1[i] == '0' && s1[i-1] == '1') continue; if (s1[i] == '0' && s1[i-1] == '0') { printf("%s\n",fail); return 0; } if(s1[i] == '0' && s1[i-1] == '?') { s1[i-1] = '1'; continue; } if (s1[i] == '1' && s1[i-1] == '0') continue; if (s1[i] == '1' && s1[i-1] == '1') { printf("%s\n",fail); return 0; } if(s1[i] == '1' && s1[i-1] == '?') { s1[i-1] = '0'; continue; } if(s1[i] == '?' && s1[i-1] == '0') { s1[i] = '1'; continue; } if(s1[i] == '?' && s1[i-1] == '1') { s1[i] = '0'; continue; } if(s1[i] == '?' && s1[i-1] == '?') continue; } } for(int i=1;i<=Length;i++) if(s1[i]=='?') { if(s2[i]!='?') { if(s2[i]=='0') s1[i]=s1[i-1];else s1[i]='0'+1-(s1[i-1]-'0'); continue; } printf("%s\n",more); return 0; } for(int i=0;i

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