HDU 3389 Game （阶梯博弈）

Game

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 217    Accepted Submission(s): 152

Problem 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

 

 

Author

hanshuai@whu

 

 

Source

The 5th Guangting Cup Central China Invitational Programming Contest

 

 

Recommend

notonlysuccess

 

 

题意:

  1-N带编号的盒子,当编号满足A>B && A非空 && (A + B) % 3 == 0 && (A + B) % 2 == 1则可以从A中取任意石头到B中.谁不能取了谁就输.

思路: 其本质为阶梯博弈;

　　阶梯博弈:博弈在一列阶梯上进行,每个阶梯上放着自然数个点,两个人进行阶梯博弈...      每一步则是将一个集体上的若干个点( >=1 )移到前面去,最后没有点可以移动的人输;

 在本题中 1,3,4 的状态不能转移到其他状态; 其他每个状态皆可转移; 且位置特定, 如  2->1 , 5->4, 6->3, 7->2 , 8->1 9->6.....

其本质我们有N级阶梯,现在要在 %3 的余数间转移, 0->0, 1->2, 2->1; 其最后的结果为1, 3, 4; 那么他们的转移的步数的奇偶性也会确定;

我们只要选择步数为奇数的位置做nim博弈就行了;而可以通过打表归纳证明得出模6为0、2、5的位置移动步数为奇，其余为偶;

 

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>

using namespace std;

int main(){

    //freopen("input.txt","r",stdin);

    int t,n,cases=0;
    scanf("%d",&t);
    while(t--){
        scanf("%d",&n);
        int ans=0,x;
        for(int i=1;i<=n;i++){
            scanf("%d",&x);
            if(i%6==0 || i%6==2 || i%6==5)
                ans^=x;
        }
        printf("Case %d: ",++cases);
        if(ans!=0)
            puts("Alice");
        else
            puts("Bob");
    }
    return 0;
}