sdut 2161 Simple Game(山东省第二届acm省赛A题)（博弈论Nim+Bush?）

艾尔森

于 2017-05-02 19:15:00 发布

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分类专栏： acm省赛文章标签： c++ acm 算法

本文链接：https://blog.youkuaiyun.com/l769255844/article/details/71087520

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这是一个简单的两人博弈游戏，玩家每次可以选择1到3堆石子并取走任意数量（非零且不超过该堆数量）。游戏目标是避免在自己的回合结束时没有石子剩下。题目要求判断先手玩家是否有必胜策略。通过分析发现，当所有堆石子的二进制表示中，每一位1的数量均为4的倍数时，先手玩家必败。因此，计算所有堆石子数量的异或值，若异或结果为0，则先手必胜，否则先手必败。

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Simple Game

Time Limit: 1000MS Memory Limit: 65536KB

Problem Description

Here is a simple game. In this game, there are several piles of stones and two players. The two players play in turn. In each turn, one can choose at least one pile and at most three piles to take away arbitrary number of stones from the pile (Of course the number of stones, which is taken away, cannot be zero and cannot be larger than the number of stones in the chosen pile). If after a player ′ s turn, there is no stone left, the player is the loser.
Suppose that the two players are all very smart. Your job is to tell whether the player who plays first can win the game or not.