难度:+++
在纸上随便画几棵树(例如下图),两人轮流砍。每次可以在任何一个节点处砍掉一枝。可以砍掉整个一棵树。如果一个节点有几个树枝,只能砍掉一枝。最后无树可砍的人负。
就下图而言,先砍能不能获胜?
共11根枝,对乙来说,只要最后剩2棵树,偶数根枝就胜。乙需在第一次砍时,和甲奇偶错开,以后则相同。 甲若先砍树2共5枝,乙就砍掉3-2以上2枝; 甲若先砍树3共4枝,乙就砍掉2-2以上3枝; 甲若先砍树2-2以上共3枝,乙就砍掉树3共4枝; 甲若先砍树1共2枝,乙就砍掉3-2-2以上1枝; 甲若先砍树3-2以上共2枝,乙就砍掉树2共5枝; 只剩甲先砍1枝的可能了。 若甲砍树1或2上的1枝,乙就砍掉树3共4枝; 若甲砍树3上的1枝,乙就砍掉树1共2枝。 |
不总是想砍几枝就能砍几枝的,比如中间一棵树就砍不下偶数枝来。 |
I think the answer is yes. Make sure after your turn, there are even number of brunches left. Cut the lower branch of the middle tree first. |
但确实有小疏漏。对甲只砍1枝的情况,有一种情形需修正一下: 若甲砍树1上的1枝或树2上的2-1枝,乙就砍掉树3共4枝; 若甲砍树3上的1枝,乙就砍掉树1共2枝; 若甲砍树2上的2-2-?任1枝,乙应先砍掉2-2-?对应的另1枝,然后同前第一条。 本贴由[勇敢的辛]最后编辑于:2006-3-31 21:10:45 |
Cut one branch from either tree 1 or tree 3. Leave odd number of branches before the middle tree is touched. Once the middle tree is cut, leave even number of branches unless it is the last cut. |
最后两个不对: "若甲砍树3上的1枝,乙就砍掉树1共2枝; 若甲砍树2上的2-2-?任1枝,乙应先砍掉2-2-?对应的另1枝" |
Yes, You will win if you cut the third tree first. I think the tree with only one branch can be cut |
Yes. Cut the middle tree first. It doesn't matter whether the second person cut a tree or a branch, the first person should have the last cut. ( I'm afraid I missed some points, three + shouldn't be so easy). |
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