Stone（巴什博弈/反巴什博弈）

最新推荐文章于 2024-04-18 07:59:15 发布

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本文探讨了一种两人轮流取物的游戏，其中每轮取物数量受限，目标是避免首先取到特定数量的物品。通过将问题转化为巴什博弈的变种，我们分析了最优策略，并提供了C++代码实现来确定先手或后手的优势。对于给定的物品总数n和最大取物数k，当(n-1)%(k+1)为0时，后手（Jiang）获胜；否则，先手（Tang）获胜。

Tang and Jiang are good friends. To decide whose treat it is for dinner, they are playing a game. Specifically, Tang and Jiang will alternatively write numbers (integers) on a white board. Tang writes first, then Jiang, then again Tang, etc… Moreover, assuming that the number written in the previous round is X, the next person who plays should write a number Y such that 1 <= Y - X <= k. The person who writes a number no smaller than N first will lose the game. Note that in the first round, Tang can write a number only within range [1, k] (both inclusive). You can assume that Tang and Jiang will always be playing optimally, as they are both very smart students.

Input
There are multiple test cases. For each test case, there will be one line of input having two integers N (0 < N <= 10^8) and k (0 < k <= 100). Input terminates when both N and k are zero.

Output
For each case, print the winner’s name in a single line.

Sample Input
1 1
30 3
10 2
0 0

Sample Output
Jiang
Tang
Jiang

巴什博弈：
只有一堆n个物品，两个人轮流从这堆物品中取物，规定每次至少取一个，最多取m个。最后取光者得胜。
结论：如果n%(m+1)==0，则后手赢，否则先手赢。
这题就是一个有n-1个物品的巴什博弈，至于为啥是n-1个物品，因为大于等于n的时候胜负就判断出来了。
反巴什博弈：只有一堆n个物品，两个人轮流从这堆物品中取物，规定每次至少取一个，最多取m个。最后取光者输。
结论：如果n%(m+1)==1，则后手赢，否则先手赢。

下面是代码：

#include<iostream>
using namespace std;
int main()
{
	int n,k;
	while(cin>>n>>k)
	{
		if(n==0&&k==0) break;
		if((n-1)%(k+1)==0) cout<<"Jiang"<<endl;
		else cout<<"Tang"<<endl;
	}
	return 0;
}