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 (AnAn-1An-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。

 

 

 

  1 /* ***********************************************
  2 Author        :kuangbin
  3 Created Time  :2013/9/14 星期六 14:16:53
  4 File Name     :2013成都网络赛\1006.cpp
  5 ************************************************ */
  6 
  7 #pragma comment(linker, "/STACK:1024000000,1024000000")
  8 #include <stdio.h>
  9 #include <string.h>
 10 #include <iostream>
 11 #include <algorithm>
 12 #include <vector>
 13 #include <queue>
 14 #include <set>
 15 #include <map>
 16 #include <string>
 17 #include <math.h>
 18 #include <stdlib.h>
 19 #include <time.h>
 20 using namespace std;
 21 
 22 const int MAXN = 100010;
 23 const int MOD = 1e9+7;
 24 char str1[MAXN], str2[MAXN];
 25 int dp[MAXN][2];
 26 int flag[MAXN][2];
 27 
 28 char sss1[MAXN],sss2[MAXN];
 29 
 30 int main()
 31 {
 32     //freopen("in.txt","r",stdin);
 33     //freopen("out.txt","w",stdout);
 34     int T;    
 35     int iCase = 0;
 36     scanf("%d",&T);
 37     while(T--)
 38     {
 39         iCase++;
 40         printf("Case #%d:\n",iCase);
 41         scanf("%s%s",str1,str2);
 42         int n = strlen(str1);
 43         dp[0][0] = 1;
 44         dp[0][1] = 0;
 45         flag[0][0] = 1;
 46         flag[0][1] = 0;
 47         for(int i = 1;i <= n;i++)
 48         {
 49             if(str1[i-1] == '?' && str2[i-1] == '?')
 50             {
 51                 dp[i][1] = dp[i-1][0] + dp[i-1][1];
 52                 dp[i][0] = dp[i-1][0] + dp[i-1][1];
 53                 int tmp = flag[i-1][0] + flag[i-1][1];
 54                 if(tmp == 0)
 55                     flag[i][0] = flag[i][1] = 0;
 56                 else if(tmp == 1)
 57                     flag[i][0] = flag[i][1] = 1;
 58                 else flag[i][0] = flag[i][1] = 2;
 59             }
 60             else if(str1[i-1] == '?' || str2[i-1] == '?')
 61             {
 62                 if(str1[i-1] == '0' || str2[i-1] =='0')
 63                 {
 64                     dp[i][1] = dp[i-1][1];
 65                     dp[i][0] = dp[i-1][0];
 66                     flag[i][0] = flag[i-1][0];
 67                     flag[i][1] = flag[i-1][1];
 68                 }
 69                 else if(str1[i-1] == '1' || str2[i-1] =='1')
 70                 {
 71                     dp[i][1] = dp[i-1][0];
 72                     dp[i][0] = dp[i-1][1];
 73                     flag[i][1] = flag[i-1][0];
 74                     flag[i][0] = flag[i-1][1];
 75                 }
 76             }
 77             else if(str1[i-1] != str2[i-1])
 78             {
 79                 dp[i][0] = dp[i][1] = 0;
 80                 flag[i][0] = flag[i][1] = 0;
 81             }
 82             else
 83             {
 84                 if(str1[i-1] =='0')
 85                 {
 86                     dp[i][0] = dp[i-1][0];
 87                     dp[i][1] = dp[i-1][1];
 88                     flag[i][0] = flag[i-1][0];
 89                     flag[i][1] = flag[i-1][1];
 90 
 91                 }
 92                 else
 93                 {
 94                     dp[i][0] = dp[i-1][1];
 95                     dp[i][1] = dp[i-1][0];
 96                     flag[i][1] = flag[i-1][0];
 97                     flag[i][0] = flag[i-1][1];
 98                 }
 99             }
100             if(dp[i][0] >= MOD)dp[i][0] -= MOD;
101             if(dp[i][1] >= MOD)dp[i][1] -= MOD;
102         }
103         int ans = 0;
104         int flag_num  = 0;
105         int ss_id = -1;
106         for(int i = n;i > 0;i--)
107         {
108             if(i+2 <= n)
109             {
110                 if(str1[i+1] == '1' || str2[i+1] == '1')
111                     break;
112             }
113             if(i+1 <= n)
114             {
115                 if(str1[i] == '0' || str2[i] == '0')
116                     continue;
117             }
118             if(i-1 >= 0)
119             {
120                 if(str1[i-1] == str2[i-1] && str1[i-1] != '?')
121                     continue;
122             }
123             char ch;
124             if(str1[i-1] == str2[i-1] && str1[i-1] == '?')
125             {
126                 ans += dp[i-1][0];
127                 ans %= MOD;
128                 ans += dp[i-1][1];
129                 ans %= MOD;
130                 flag_num += flag[i-1][0] + flag[i-1][1];
131                 if(flag_num > 2)flag_num = 2;
132             }
133             else
134             {
135             if(str1[i-1] != '?')ch = str1[i-1];
136             else
137             {
138                 if(str2[i-1] == '0')ch = '1';
139                 else ch = '0';
140             }
141             if(ch == '0')
142             {
143                 ans += dp[i-1][0];
144                 ans %= MOD;
145                 flag_num += flag[i-1][0];
146                 if(flag_num > 2)flag_num = 2;
147             }
148             else
149             {
150                 ans += dp[i-1][1];
151                 ans %= MOD;
152                 flag_num += flag[i-1][1];
153                 if(flag_num > 2)flag_num = 2;
154             }
155             }
156             if(flag_num == 1 && ss_id == -1)ss_id = i;
157         }
158         if(flag_num == 0)
159         {
160             printf("Impossible\n");
161             continue;
162         }
163         if(flag_num > 1)
164         {
165             printf("Ambiguous %d\n",ans);
166             continue;
167         }
168         if(flag_num == 1)
169         {
170             sss1[n] = sss2[n] = 0;
171             sss1[ss_id-1] = '0';
172             sss2[ss_id-1] = '1';
173             for(int i = ss_id;i < n;i++)
174                 sss1[i] = '1', sss2[i] = '0';
175             int last = 0;
176             for(int i = ss_id-2;i >= 0;i--)
177             {
178                 int now;
179                 if(flag[i+1][0] > 0)now = 0;
180                 else now = 1;
181                 sss1[i] = sss2[i] = '0' + now;
182 
183             }
184             for(int i = n-1; i > 0;i--)
185             {
186                 if(sss1[i-1] ==sss1[i])sss1[i] = '0';
187                 else sss1[i] = '1';
188                 if(sss2[i-1] == sss2[i])sss2[i] = '0';
189                 else sss2[i] = '1';
190             }
191 
192             printf("%s\n%s\n",sss1,sss2);
193             continue;
194         }
195         cout<<ans<<endl;
196 
197 
198     }
199     return 0;
200 }

 

 

 

 

 

 

posted on 2013-09-15 00:22  kuangbin  阅读(1716)  评论(2)  编辑  收藏  举报

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