HDU 4764 Stone

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最新推荐文章于 2022-10-16 04:57:01 发布

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分类专栏：博弈

本文链接：https://blog.youkuaiyun.com/Houheshuai/article/details/49934503

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本文探讨了游戏开发中涉及的策略与算法实现，包括游戏引擎、AI处理、音视频技术等多个方面，深入分析了如何优化游戏性能和用户体验。

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Stone

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1163 Accepted Submission(s): 812

Problem Description

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

Sample Output

  
  
   
   Jiang
Tang
Jiang

Source

2013 ACM/ICPC Asia Regional Changchun Online

http://acm.hdu.edu.cn/showproblem.php?pid=4764

分析：

题目等价于:

一堆石子有 N 个，两个人轮流取，每次能取的范围是 [ 1 , k ] 谁最后一个取谁就输了。
Nim 的变形。

首先看下另一个问题：http://blog.youkuaiyun.com/houheshuai/article/details/44871687

这个问题将最后的条件改变了：最后取完石子的将会输。

我们先拿出一个石子放到一边，还剩 n-1 个，将这 n-1 个石子转化为上面链接中的问题，

那么先手必败的状态为： (n - 1) % ( k + 1) == 0

我们将拿出的那个石子放回去，再将游戏的最后条件变回来，那先手必败的状态仍然是：

(n - 1) % ( k + 1) == 0

代码：

#include <stdio.h>
int main()
{
    int n, k;
    while (~scanf("%d%d", &n, &k))
    {
        if (n == 0 && k == 0) break;
        if ((n - 1) % (k + 1) == 0) printf("Jiang\n");
        else printf("Tang\n");
    }
    return 0;
}