hdu 4315 Climbing the Hill（阶梯博弈转nim博弈）

Climbing the Hill

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 919    Accepted Submission(s): 411

Problem Description

Alice and Bob are playing a game called "Climbing the Hill". The game board consists of cells arranged vertically, as the figure below, while the top cell indicates the top of hill. There are several persons at different cells, and there is one special people, that is, the king. Two persons can't occupy the same cell, except the hilltop. At one move, the player can choose any person, who is not at the hilltop, to climb up any number of cells. But the person can't jump over another one which is above him. Alice and Bob move the persons alternatively, and the player who move the king to the hilltop will win.

Alice always move first. Assume they play optimally. Who will win the game?

Input

There are several test cases. The first line of each test case contains two integers N and k (1 <= N <= 1000, 1 <= k <= N), indicating that there are N persons on the hill, and the king is the k-th nearest to the top. N different positive integers followed in the second line, indicating the positions of all persons. (The hilltop is No.0 cell, the cell below is No.1, and so on.) These N integers are ordered increasingly, more than 0 and less than 100000.

Output

If Alice can win, output "Alice". If not, output "Bob".

Sample Input

3 3 1 2 4 2 1 100 200

Sample Output

Bob Alice

Hint

The figure illustrates the first test case. The gray cell indicates the hilltop. The circles indicate the persons, while the red one indicates the king. The first player Alice can move the person on cell 1 or cell 4 one step up, but it is not allowed to move the person on cell 2.

Author

TJU

Source

2012 Multi-University Training Contest 2

代码：

 1 #include<cstring>
 2 #include<cstdio>
 3 #include<cstdlib>
 4 int aa[1005];
 5 int n,k;
 6 int main()
 7 {
 8   while(scanf("%d%d",&n,&k)!=EOF)
 9   {
10        for(int i=0;i<n;i++)
11       scanf("%d",&aa[i]);
12        int ans=0;
13      if(k==1)
14      {
15        puts("Alice");
16        continue;
17      }
18      if(n%2==0)  //偶数堆
19      {
20        for(int i=1;i<n;i+=2)
21          ans^=(aa[i]-aa[i-1]-1);
22      }
23      else
24      {
25          if(k==2) ans=aa[0]-1;
26          else     ans=aa[0];
27          for(int i=2;i<n;i+=2)
28            ans^=(aa[i]-aa[i-1]-1);
29      }
30      if(ans)  puts("Alice");
31      else puts("Bob");
32   }
33   return 0;
34 }