HDU 4733 G(x) (2013成都网络赛，递推)

G(x)

Time Limit: 2000/500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 184    Accepted Submission(s): 44

Problem Description

For a binary number x with n digits (A_nA_n-1A_n-2 ... A₂A₁), we encode it as 
Where "" is bitwise XOR operation and "" indicates the largest integer which is not greater than x.
Due to some reasons, Mzry1992 encode his password P into G(P), and additionally, he encode P + 1 into G(P + 1) too, and write G(P) and G(P + 1) into his diary.
This story happened many years ago and now you hold the diary 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 has a number T (T <= 100) , indicating the number of test cases.
For every test case, it has 2 lines of same number of digits describe G(P) and G(P + 1), In every line, it only contains 1, 0 and ?. Unreadable digits are denoted with symbol ?, The length of every line in the input is up to 10⁵.

 

Output

For every case, you should output "Case #t: " at first, without quotes. The t is the case number starting from 1.
Then, if there is impossible to restore G(P) and G(P + 1), you should output "Impossible" in the second line.
Otherwise, if G(P) is unique, you should output restored G(P) and G(P +1) in the same format.
Otherwise, you should output "Ambiguous" and the number of possible G(P) in the second line.
The number may be very large so the answer should modulo 10^9 + 7.

 

Sample Input

3
10??
10??
0010
0110
1?01
0?01

 

Sample Output

Case #1:
Ambiguous 3
Case #2:
0010
0110
Case #3:
Impossible

Hint

In the first sample case, the three possible situations are:
1.
G(12) = 1010
G(13) = 1011
2.
G(13) = 1011
G(14) = 1001
3.
G(14) = 1001
G(15) = 1000

 

Source

2013 ACM/ICPC Asia Regional Chengdu Online

 

Recommend

liuyiding

 

很容易找出规律，

然后递推，枚举就解决了。

P和P+1 其实就是差后面一点

P: XXXXX  0   1    1     1

P+1: XXX  1   0    0     0

关键点就是枚举P的第一个0的位置。左边的P和P+1是一样的。

 /* ***********************************************
 Author        :kuangbin
 Created Time  :2013/9/14 星期六 14:16:53
 File Name     :2013成都网络赛\1006.cpp
 ************************************************ */

 #pragma comment(linker, "/STACK:1024000000,1024000000")
 #include <stdio.h>
 #include <string.h>
 #include <iostream>
 #include <algorithm>
 #include <vector>
 #include <queue>
 #include <set>
 #include <map>
 #include <string>
 #include <math.h>
 #include <stdlib.h>
 #include <time.h>
 using namespace std;

 const int MAXN = ;
 const int MOD = 1e9+;
 char str1[MAXN], str2[MAXN];
 int dp[MAXN][];
 int flag[MAXN][];

 char sss1[MAXN],sss2[MAXN];

 int main()
 {
     //freopen("in.txt","r",stdin);
     //freopen("out.txt","w",stdout);
     int T;
     int iCase = ;
     scanf("%d",&T);
     while(T--)
     {
         iCase++;
         printf("Case #%d:\n",iCase);
         scanf("%s%s",str1,str2);
         int n = strlen(str1);
         dp[][] = ;
         dp[][] = ;
         flag[][] = ;
         flag[][] = ;
         for(int i = ;i <= n;i++)
         {
             if(str1[i-] == '?' && str2[i-] == '?')
             {
                 dp[i][] = dp[i-][] + dp[i-][];
                 dp[i][] = dp[i-][] + dp[i-][];
                 int tmp = flag[i-][] + flag[i-][];
                 if(tmp == )
                     flag[i][] = flag[i][] = ;
                 else if(tmp == )
                     flag[i][] = flag[i][] = ;
                 else flag[i][] = flag[i][] = ;
             }
             else if(str1[i-] == '?' || str2[i-] == '?')
             {
                 if(str1[i-] == '' || str2[i-] =='')
                 {
                     dp[i][] = dp[i-][];
                     dp[i][] = dp[i-][];
                     flag[i][] = flag[i-][];
                     flag[i][] = flag[i-][];
                 }
                 else if(str1[i-] == '' || str2[i-] =='')
                 {
                     dp[i][] = dp[i-][];
                     dp[i][] = dp[i-][];
                     flag[i][] = flag[i-][];
                     flag[i][] = flag[i-][];
                 }
             }
             else if(str1[i-] != str2[i-])
             {
                 dp[i][] = dp[i][] = ;
                 flag[i][] = flag[i][] = ;
             }
             else
             {
                 if(str1[i-] =='')
                 {
                     dp[i][] = dp[i-][];
                     dp[i][] = dp[i-][];
                     flag[i][] = flag[i-][];
                     flag[i][] = flag[i-][];

                 }
                 else
                 {
                     dp[i][] = dp[i-][];
                     dp[i][] = dp[i-][];
                     flag[i][] = flag[i-][];
                     flag[i][] = flag[i-][];
                 }
             }
             if(dp[i][] >= MOD)dp[i][] -= MOD;
             if(dp[i][] >= MOD)dp[i][] -= MOD;
         }
         int ans = ;
         int flag_num  = ;
         int ss_id = -;
         for(int i = n;i > ;i--)
         {
             if(i+ <= n)
             {
                 if(str1[i+] == '' || str2[i+] == '')
                     break;
             }
             if(i+ <= n)
             {
                 if(str1[i] == '' || str2[i] == '')
                     continue;
             }
             if(i- >= )
             {
                 if(str1[i-] == str2[i-] && str1[i-] != '?')
                     continue;
             }
             char ch;
             if(str1[i-] == str2[i-] && str1[i-] == '?')
             {
                 ans += dp[i-][];
                 ans %= MOD;
                 ans += dp[i-][];
                 ans %= MOD;
                 flag_num += flag[i-][] + flag[i-][];
                 if(flag_num > )flag_num = ;
             }
             else
             {
             if(str1[i-] != '?')ch = str1[i-];
             else
             {
                 if(str2[i-] == '')ch = '';
                 else ch = '';
             }
             if(ch == '')
             {
                 ans += dp[i-][];
                 ans %= MOD;
                 flag_num += flag[i-][];
                 if(flag_num > )flag_num = ;
             }
             else
             {
                 ans += dp[i-][];
                 ans %= MOD;
                 flag_num += flag[i-][];
                 if(flag_num > )flag_num = ;
             }
             }
             if(flag_num ==  && ss_id == -)ss_id = i;
         }
         if(flag_num == )
         {
             printf("Impossible\n");
             continue;
         }
         if(flag_num > )
         {
             printf("Ambiguous %d\n",ans);
             continue;
         }
         if(flag_num == )
         {
             sss1[n] = sss2[n] = ;
             sss1[ss_id-] = '';
             sss2[ss_id-] = '';
             for(int i = ss_id;i < n;i++)
                 sss1[i] = '', sss2[i] = '';
             int last = ;
             for(int i = ss_id-;i >= ;i--)
             {
                 int now;
                 if(flag[i+][] > )now = ;
                 else now = ;
                 sss1[i] = sss2[i] = '' + now;

             }
             for(int i = n-; i > ;i--)
             {
                 if(sss1[i-] ==sss1[i])sss1[i] = '';
                 else sss1[i] = '';
                 if(sss2[i-] == sss2[i])sss2[i] = '';
                 else sss2[i] = '';
             }

             printf("%s\n%s\n",sss1,sss2);
             continue;
         }
         cout<<ans<<endl;

     }
     return ;
 }