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[DhtmlXQ_adddate]2019-05-03 21:59:03[/DhtmlXQ_adddate]
[DhtmlXQ_editdate]2019-05-03 21:59:03[/DhtmlXQ_editdate]
[DhtmlXQ_title]双炮禁双炮[/DhtmlXQ_title]
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[DhtmlXQ_event]竹香斋象戏谱[/DhtmlXQ_event]
[DhtmlXQ_group]散局[/DhtmlXQ_group]
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[DhtmlXQ_result]红胜[/DhtmlXQ_result]
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[DhtmlXQ_remark]zqhuang[/DhtmlXQ_remark]
[DhtmlXQ_author]zqhuang[/DhtmlXQ_author]
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[DhtmlXQ_comment0]【zqhuang按】这是《竹香斋象戏谱》的一则数学游戏排局。||||这个数学问题以及它的推广我在中学时都有所总结。限于篇幅这里不做详细介绍，只简略给出取胜策略。||第一步：把单次可行步数（兵，炮之间间隔）1，4，6都写成二进制数，每个数一行，对齐个位：||001||100||110||||第二步：看是否每一列上1的个数都是偶数个。如果是的话，先行者负，如果不是，后行者负。在这里第2列，第3列都只有一个1，所以是后行者负。||||||第三步：在后行者负的局面下，先行者取胜策略是：尽量使得每一列1的个数变成偶数个。（这样轮到对方走子时成为先行者负的局面。）为了做到这一点，只要找出最先出现奇数个1的列（该例中是第二列），随便选一在该列是1的行（因为该列上是奇数个1，所以至少有一行满足条件）。该例中只有第三行在第二列是1。于是就改变第三行：把它变成101（也就是十进制的5）这样即使得每一列上1的个数都是偶数个了。||||所以正解是：炮三进一（把十进制的6，也就是二进制的110，变成十进制的5，也就是二进制的101）。||||以后黑不管如何应，红总可以用上述分析方法应对。[/DhtmlXQ_comment0]
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