再，再改进一下。

都结巴了。
Let me define losing hand as whenever you create this hand to your opponent, you win.

1. 任何时候，不能留给对手有两堆数目相等的情况。
That’s right-
I will summarize your conclusion one step further – 0-X-X is a losing hand.

2. 如果有一堆变成1的时候，留给对手让两堆数目相差是1.
That’s incorrect(practically correct though). 1-X-(X+1) is NOT necessarily a losing hand. For example 1-3-4. You are just one step close to the common solution.
-- 1, x,x+1 (x>4, otherwise, pick to 1,4,5 ; 1,3,4;1,2,3
)

3. 如果任何一堆是2的时候。保持另两堆的数目相差等于2.
That’s incorrect (practically correct though). 2-X-X+2 is NOT necessary a losing hand. For example 2-6-8. You are just one step close to the common solution.
2,x,x+2 (x>6, otherwise, pick to 2,6,4; 2,5,3;

4. 任何一堆都大于2的时候，保持另两堆的数目相差大于2.
That’s incorrect for obvious reasons (3-4-7; 3-5-6). It might be more difficult than you think.
-- 3,x,x+3 (x>7，）
此题无解，或者说无穷多解