10000ms
10000ms
65536KB
 
64-bit integer IO format: %lld      Java class name: Main
Submit Status PID: 29371
Alice and Bob invented a new game again, as they usually did.
The rule of the new game is quite simple. There is an n*n matrix and at the very beginning of the game, some of ceil contain stone, and others contain nothing. Then they take turns (Alice first) to play the game.
Each time one person should choose two empty ceils which are on the same line, and the number of their columns’ difference is exactly 2(which means the absolute value of difference is 2). And then the people should put stones on each of the two ceil. The game ends when one people cannot do the operation above and thus lose the game.
Here\’s the problem: Who will win the game if both use the best strategy? Find it out quickly, before they get bored of the game again!
 
 
The first line contains a single integer T, indicating the number of test cases.
The first line of each test case contains an integer n (1<=n<=1000), and n lines follow. Each line contains n integers 0 or 1, which means there exist a stone or not (1 means exist).
 
For each test case, output the case number first, then “Alice” if Alice will win otherwise output “Bob”.
 
2
3
0 0 0
0 0 0
1 0 0
4
0 0 0 1
1 0 1 0
1 1 1 1
1 0 1 0
 
Case 1: Bob
Case 2: Alice

 

这道题，是一道博弈题。

每一行是可以单独考虑的，这个是很好理解的。

奇数和偶数行也是可以单独考虑的。推一推。

“1”的存在，用来分割他们进行讨论。

题目的转化为 n 堆石头子，每次从每一堆取出2个连续堆的方案。

这里就简单了。

  1.  1 #include<iostream>
  2.  2 #include<cstdio>
  3.  3 #include<cstring>
  4.  4 #include<cstdlib>
  5.  5 using namespace std;
  6.  6 
  7.  7 int SG[1003];
  8.  8 bool use[1003];
  9.  9 void prepare()
  10. 10 {
  11. 11     int i,j;
  12. 12     SG[0]=0;SG[1]=0;
  13. 13     for(i=2;i<=1000;i++)
  14. 14     {
  15. 15         memset(use,false,sizeof(use));
  16. 16         for(j=1;j<i;j++)
  17. 17         {
  18. 18             use[ (SG[j-1] ^ SG[i-j-1]) ]=true;
  19. 19         }
  20. 20         for(j=0;;j++)
  21. 21         if(use[j]==false)
  22. 22         {
  23. 23             SG[i]=j;
  24. 24             break;
  25. 25         }
  26. 26     }
  27. 27 }
  28. 28 int main()
  29. 29 {
  30. 30     int T,n,ans[2],t;
  31. 31     int i,j,x,XOR;
  32. 32     prepare();
  33. 33     scanf("%d",&T);
  34. 34     for(t=1;t<=T;t++)
  35. 35     {
  36. 36         scanf("%d",&n);
  37. 37         XOR=0;
  38. 38         for(i=1;i<=n;i++)
  39. 39         {
  40. 40             ans[0]=0;ans[1]=0;
  41. 41             for(j=1;j<=n;j++)
  42. 42             {
  43. 43                 scanf("%d",&x);
  44. 44                 if(x==0)
  45. 45                     ans[j%2]++;
  46. 46                 else
  47. 47                 {
  48. 48                     XOR=XOR^SG[ans[j%2]];
  49. 49                     ans[j%2]=0;
  50. 50                 }
  51. 51             }
  52. 52             XOR=XOR^SG[ans[0]];
  53. 53             XOR=XOR^SG[ans[1]];
  54. 54         }
  55. 55         printf("Case %d: ",t);
  56. 56         if(XOR==0)
  57. 57             printf("Bob\n");
  58. 58         else printf("Alice\n");
  59. 59     }
  60. 60     return 0;
  61. 61 }

 

 

SG[ ] 使用map。

  1.  1 #include<iostream>
  2.  2 #include<cstdio>
  3.  3 #include<cstring>
  4.  4 #include<cstdlib>
  5.  5 #include<map>
  6.  6 using namespace std;
  7.  7 
  8.  8 int SG[1003];
  9.  9 map<int,int>Q;
  10. 10 void prepare()
  11. 11 {
  12. 12     int i,j,k,s;
  13. 13     SG[0]=0;SG[1]=0;
  14. 14     for(i=2;i<=1000;i++)
  15. 15     {
  16. 16         while(!Q.empty())
  17. 17         {
  18. 18             Q.clear();
  19. 19         }
  20. 20         for(s=0,j=1;j<i;j++)
  21. 21         {
  22. 22            k=(SG[j-1] ^ SG[i-j-1]);
  23. 23            Q[k]=s++;
  24. 24         }
  25. 25         for(j=0;;j++)
  26. 26         if(Q.find(j)==Q.end())
  27. 27         {
  28. 28             SG[i]=j;
  29. 29             break;
  30. 30         }
  31. 31     }
  32. 32 }
  33. 33 int main()
  34. 34 {
  35. 35     int T,n,ans[2],t;
  36. 36     int i,j,x,XOR;
  37. 37     prepare();
  38. 38     scanf("%d",&T);
  39. 39     for(t=1;t<=T;t++)
  40. 40     {
  41. 41         scanf("%d",&n);
  42. 42         XOR=0;
  43. 43         for(i=1;i<=n;i++)
  44. 44         {
  45. 45             ans[0]=0;ans[1]=0;
  46. 46             for(j=1;j<=n;j++)
  47. 47             {
  48. 48                 scanf("%d",&x);
  49. 49                 if(x==0)
  50. 50                     ans[j%2]++;
  51. 51                 else
  52. 52                 {
  53. 53                     XOR=XOR^SG[ans[j%2]];
  54. 54                     ans[j%2]=0;
  55. 55                 }
  56. 56             }
  57. 57             XOR=XOR^SG[ans[0]];
  58. 58             XOR=XOR^SG[ans[1]];
  59. 59         }
  60. 60         printf("Case %d: ",t);
  61. 61         if(XOR==0)
  62. 62             printf("Bob\n");
  63. 63         else printf("Alice\n");
  64. 64     }
  65. 65     return 0;
  66. 66 }

 

 

 

 

 

 

 

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