Description: There are two piles, the quantity, can be different. The game starts by two people. The game regulations, each time there are two different tissues, one is to take any more stones in any one of them; second, it can take the same number of stones while two piles. Finally, put all the bits all the people. Now give the number of the initial two stones, if the turn is to take, suppose both parties take the best strategy, ask the last person to be a victor or a loser. Assume the N, 2N-1 (n> 2) layout first. Proof in the back. Then, for any given layout A, B (assuming a 2). If b = 2A-1, then the first pickup is negative. If B> 2a-1, the first pickup takes out B-2A-1 from B, and the remaining constitutes A, 2A-1, and the first pickup. If B <2A-1, the first pickup takes 2A-1-B from both sides, constitutes A- (2A-1-B), B- (2A-1-B), and the simplification is (B- A 1), 2 (B-A 1) -1, namely N, 2N-1 layout, first picking. The following proves that N, 2N-1 (N> 2) layout is negative. When n = 3, it is 3, 5 layout, and the first pickup is negative. Assume that A, 2A-1 (a 0) from 2N-1, the layout changes to N, 2N-1-K first assumes there is an integer A, when the person is taken away from the above layout at the same time, it can Turn into layout J, 2J-1. Then J = N-A (1); 2N-1-K-a = 2J-1 (2), from (1) (2) to solve A = K. That is, when the passers are taken from both sides, the J, 2J-1 layout is constituted, and J 2) layout
