趣味数学题+++++++++游戏+++++++++

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博弈题，只有两种情况，先取必胜或者后取必胜，只要笔的总数是奇数，最后肯定有一个人是奇数，另一个人是偶数，以下讨论中，笔的总数为奇数。
这里影响胜负的因素只有两个，一是当前桌子上还剩了多少只笔，二是自己手中拿的笔的数量的奇偶性。
如果当前桌子上还剩a只笔，自己手中拿的笔的数量的奇偶性为b(偶数时b=0,奇数时b=1)，若先取必胜，记X(a, b)=1，否则，记X(a, b)=0。
显然有，
X(1, 0) = 1 (还剩1只笔，手中有偶数只，取走这支笔就获胜了，必胜)
X(1, 1) = 0 (还剩1只笔，手中有奇数只，只能取走这支笔，必输)
X(2, 0) = 1
X(2, 1) = 1
X(3, 0) = 1
X(3, 1) = 1
X(4, 0) = 1
X(4, 1) = 0

我们来看看当a>3的时候，X(a, b)的递推情况，由于笔的总数是奇数，
若a为偶数，此时对方手中拿的笔的数量的奇偶性是1-b，
此时，若X(a-3, 1-b)、X(a-2, 1-b)、X(a-1, 1-b)都等于1，则X(a, b)=0，否则X(a, b)=1

若a为奇数，此时对方手中拿的笔的数量的奇偶性是b，
此时，若X(a-3, b)、X(a-2, b)、X(a-1, b)都等于1，则X(a, b)=0，否则X(a, b)=1

有了递推关系，我们可以用程序来求一些，看看规律：
X(1, 1) = 0, X(2, 1) = 1, X(3, 1) = 1 => X(4, 0) = 1
X(1, 0) = 1, X(2, 0) = 1, X(3, 0) = 1 => X(4, 1) = 0
X(2, 0) = 1, X(3, 0) = 1, X(4, 0) = 1 => X(5, 0) = 0
X(2, 1) = 1, X(3, 1) = 1, X(4, 1) = 0 => X(5, 1) = 1
X(3, 1) = 1, X(4, 1) = 0, X(5, 1) = 1 => X(6, 0) = 1
X(3, 0) = 1, X(4, 0) = 1, X(5, 0) = 0 => X(6, 1) = 1
X(4, 0) = 1, X(5, 0) = 0, X(6, 0) = 1 => X(7, 0) = 1
X(4, 1) = 0, X(5, 1) = 1, X(6, 1) = 1 => X(7, 1) = 1
X(5, 1) = 1, X(6, 1) = 1, X(7, 1) = 1 => X(8, 0) = 0
X(5, 0) = 0, X(6, 0) = 1, X(7, 0) = 1 => X(8, 1) = 1
X(6, 0) = 1, X(7, 0) = 1, X(8, 0) = 0 => X(9, 0) = 1
X(6, 1) = 1, X(7, 1) = 1, X(8, 1) = 1 => X(9, 1) = 0
X(7, 1) = 1, X(8, 1) = 1, X(9, 1) = 0 => X(10, 0) = 1
X(7, 0) = 1, X(8, 0) = 0, X(9, 0) = 1 => X(10, 1) = 1
X(8, 0) = 0, X(9, 0) = 1, X(10, 0) = 1 => X(11, 0) = 1
X(8, 1) = 1, X(9, 1) = 0, X(10, 1) = 1 => X(11, 1) = 1
X(9, 1) = 0, X(10, 1) = 1, X(11, 1) = 1 => X(12, 0) = 1
X(9, 0) = 1, X(10, 0) = 1, X(11, 0) = 1 => X(12, 1) = 0
X(10, 0) = 1, X(11, 0) = 1, X(12, 0) = 1 => X(13, 0) = 0
X(10, 1) = 1, X(11, 1) = 1, X(12, 1) = 0 => X(13, 1) = 1
X(11, 1) = 1, X(12, 1) = 0, X(13, 1) = 1 => X(14, 0) = 1
X(11, 0) = 1, X(12, 0) = 1, X(13, 0) = 0 => X(14, 1) = 1
X(12, 0) = 1, X(13, 0) = 0, X(14, 0) = 1 => X(15, 0) = 1
X(12, 1) = 0, X(13, 1) = 1, X(14, 1) = 1 => X(15, 1) = 1

X(15, 0) = 1，所以先取必胜。
找每一行让对方必败的策略，设取Y(a, b)只笔，找每一行最后一个数为1时，前3个数哪个为0，写程序算出取的支数。
Y(1, 0) = 1
Y(2, 0) = 1
Y(2, 1) = 2
Y(3, 0) = 3
Y(3, 1) = 2
Y(4, 0) = 3
Y(5, 1) = 1
Y(6, 0) = 2
Y(6, 1) = 1
Y(7, 0) = 2
Y(7, 1) = 3
Y(8, 1) = 3
Y(9, 0) = 1
Y(10, 0) = 1
Y(10, 1) = 2
Y(11, 0) = 3
Y(11, 1) = 2
Y(12, 0) = 3
Y(13, 1) = 1
Y(14, 0) = 2
Y(14, 1) = 1
Y(15, 0) = 2
Y(15, 1) = 3
按照这个策略，先取的必胜，
例如
Y(15, 0) = 2，表示桌上还剩15只笔，自己手中有偶数只笔，也就是最开始的时候，应该取2只。
Y(8, 1) = 3，表示桌上还剩8只笔，自己手中有奇数只笔的时候，应该取3只。

大家能看出这个策略有什么规律吗？我没看出来。
方案太多了.
如果你是甲方.
例1:(留最后四个要你先拿.你拿一个最后三个给乙,所以你拿的是奇数赢了)剩十一个你要观察乙的行动拿设计到让他抓完最后三个两个或一个,然后你拿最后四个中的一个!
不论怎样，一定要对手拿最后一次的时候是4，这样他拿什么我们必胜。但是我们给对手剩下5，6的话我们就惨了，所以我们倒数第二次取一定要保证取的时候还有7个(有5，6更好，直接取到4）。这样就必须保证对手取后应为9，10，11，他给我们留下8我们就惨了，这样我们就可以给他留下8了，这样一来你开始看他怎么取，控制他取后剩下9，10，11，就好了，取三个，应该没问题。