HDOJ 题目3389 Game（阶梯博弈）

最新推荐文章于 2022-04-18 00:12:30 发布

原创最新推荐文章于 2022-04-18 00:12:30 发布 · 939 阅读

0 ·

CC 4.0 BY-SA版权

技巧同时被 2 个专栏收录

57 篇文章

订阅专栏

博弈

41 篇文章

订阅专栏

本文介绍了一个游戏中的胜负预测算法，游戏中两名玩家轮流从编号盒子中按特定规则转移卡片，最终无法行动者为败。文章提供了AC代码实现，并通过实例演示了如何将问题转化为阶梯状尼姆博弈（Staircase Nim），利用异或运算找到获胜方。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Game

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 367 Accepted Submission(s): 244

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 | We have carefully selected several similar problems for you: 3387 3388 3390 3391 3393

题意：有n个盒子编号为1-n，每个盒子里面有若干物品
当编号满足a>b && (a+b)%3==0 && (a+b)%2==1时
可以从a盒子中拿>=1个物品到b盒子中
思路：找规律后转换为Staircase Nim
满足此条件：(a+b)%3==0 && (a+b)%2==1的即是3个倍数并且是奇数：3、9、15、21，即%6==3
每六个拆开后会发现如下3组可以传递的关系：
1<--2<--7<--8<--13<--14
3<--6<--9<--12<--15<--18
4<--5<--10<--11<--16<--17
即3组Staircase Nim，3组结果异或即可。

虽然似懂非懂，先珍藏一下再说吧

ac代码

#include<stdio.h>
int main()
{
	int t,c=0;
	scanf("%d",&t);
	while(t--)
	{
		int n,i,num[100000],s1=0,s2=0,s3=0,s;
		scanf("%d",&n);
		for(i=1;i<=n;i++)
			scanf("%d",&num[i]);
		for(i=2;i<=n;i+=6)
			s1^=num[i];
		for(i=6;i<=n;i+=6)
			s2^=num[i];
		for(i=5;i<=n;i+=6)
			s3^=num[i];
		s=s1^s2^s3;
		if(s)
			printf("Case %d: Alice\n",++c);
		else
			printf("Case %d: Bob\n",++c);
	}
}