HDU 3389 Game (阶梯博弈)

Posted 2020-07-31 Ritchie丶

Game

Time Limit: 1000MS

Memory Limit: 32768KB

64bit IO Format: %I64d & %I64u

Submit Status

Description

Bob and Alice are playing a new game. There are n boxes which have been numbered from 1 to n. Each box is either empty or contains several cards. Bob and Alice move the cards in turn. In each turn the corresponding player should choose a non-empty box A and choose another box B that B<A && (A+B)%2=1 && (A+B)%3=0. Then, take an arbitrary number (but not zero) of cards from box A to box B. The last one who can do a legal move wins. Alice is the first player. Please predict who will win the game.

 

Input

The first line contains an integer T (T<=100) indicating the number of test cases. The first line of each test case contains an integer n (1<=n<=10000). The second line has n integers which will not be bigger than 100. The i-th integer indicates the number of cards in the i-th box.

 

Output

For each test case, print the case number and the winner‘s name in a single line. Follow the format of the sample output.

 

Sample Input

2

2

1 2

7

1 3 3 2 2 1 2

 

Sample Output

Case 1: Alice

Case 2: Bob

 

Source

The 5th Guangting Cup Central China Invitational Programming Contest

题意：有t组数据。每组数据有n个盒子，这n个盒子编号为12345678......。(注意不是从0开始的)

　　　每个盒子中有一定量的卡片。每次取编号为B和编号为A的盒子， 要求满足

　　　B<A && (A+B)%2=1 && (A+B)%3=0，把A中的任意数量的卡片转移给B，谁不能再转移了谁输。

题解：阶梯博弈，只需要考虑步数为奇数的盒子，步数为偶数的盒子不需要考虑。

　　　在本题中编号为1,3,4的盒子不能转移卡片，其余盒子均可转移。例如：

　　　2->1,   5->4,   6->3,   7->2   ,8->1,   9->6...

　　　其本质为有n级阶梯，我们在%3的余数中进行转移0->0,   1->2,   2->1;最后的结果

　　　一定是1或者3或者4,这些盒子中卡片转移的步数的奇偶性是一定的。为什么这么说呢？

　　　因为即使有些盒子例如编号11的盒子，有11->4和11->10->8->1两种选择，但是这

　　　两种选择的步数的奇偶性是相同的，都是奇数，所以奇偶性是一定的。

　　　所以我们把这个阶梯博弈转化为尼姆博弈就行了，对步数为奇数的盒子进行尼姆博弈。

　　　在纸上多写几个数或者用打表的方法可以发现如下规律：

　　　盒子编号模6为0,2,5的位置的移动步数为奇，其余为偶。

　　　推到这里就很好实现了。

#include <iostream>
#include <cstdio>
using namespace std;
int main()
{
    int t,cas=1;
    cin>>t;
    while(t--)
    {
        int n,data,ans=0;
        cin>>n;
        for(int i=1;i<=n;i++)
        {
            cin>>data;
            if(i%6==2||i%6==5||i%6==0)
            ans^=data;
        }
        if(ans)
        printf("Case %d: Alice\n",cas++);
        else
        printf("Case %d: Bob\n",cas++);
    }
    return 0;
}