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  • 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
     
    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 }
    View Code
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  • 原文地址:https://www.cnblogs.com/gongxijun/p/4049798.html
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