Age,carelessbess...But,this time I think I'm right.

We can draw 26 parallel lines with 4m gap. there 25 spaces.

It's clear that 4m 16m first lost, 8m,12m first won.

Induction: 3(k)+1,first lost,3(k)or 3(k)+2 first won.

let check 3(k+1)+1,

nomatter the Fst plant where, if tree is in gap,Snd near it plant the tree out gap killing 3 gaps,if Fst Put tree out gap ,Snd put the tre in gap killing 3 gaps.

Snd always can let one sid of two trees has 3t gaps (t=0,ok),the other side has (3x+1)

t,x<=k,By induction, one side Fst lost, the other Fst won.So, nomatter Fsd choice which side Fst will be lost.

The proof is for all whole numbers.

 

所有跟帖： 

• 回複：Age,carelessbess...But,this time I think I'm right. -wushuihe- ♂ (180 bytes) () 03/18/2011 postreply 09:47:01

• 第一回合後方總能使一邊為連續的3X+1另一邊為連續的3T. X,T<=K -jinjing- ♀ (195 bytes) () 03/18/2011 postreply 15:52:04

• you are right, I prove it again. -jinjing- ♀ (283 bytes) () 03/20/2011 postreply 21:35:08

• you are way far from understanding the problem -guest007- ♂ (37 bytes) () 03/21/2011 postreply 07:25:01

• Do you have answer? The key point is 整化. -jinjing- ♀ (0 bytes) () 03/21/2011 postreply 07:57:00