URAL 1780 G - Gray Code 找規律


G - Gray Code
Time Limit: 20 Sec

Memory Limit: 256 MB

題目連接

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.
  1. Create a 2-bit list: {0, 1}.
  2. Reflect this list and concatenate it with the original list: {0, 1, 1, 0}.
  3. Prefix old entries with 0, and prefix new entries with 1: {00, 01, 11, 10}.
  4. 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

 

題意

題目給了一種序列的構造方式

然后給你一個k,判斷第k個數是不是說給的x

題解

找規律題,注意翻轉= =

代碼:

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <vector>
#include <stack>
#include <map>
#include <set>
#include <queue>
#include <iomanip>
#include <string>
#include <ctime>
#include <list>
#include <bitset>
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')
        {
            //同號
            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')
        {
            //異號
            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<Length;i++) 
        if(s1[i]!=s1[i+1])
        {
            s2[i+1]='1';
        }
        else s2[i+1]='0';
    printf("%s\n%s\n",s1+1,s2+1);
    return 0;
}

 


注意!

本站转载的文章为个人学习借鉴使用,本站对版权不负任何法律责任。如果侵犯了您的隐私权益,请联系我们删除。



 
粤ICP备14056181号  © 2014-2020 ITdaan.com