**Problem:**

Given 100 stones, two players alternate to take stones out. One can take any number from 1 to 15; however, one cannot take any number that was already taken. If in the end of the game, there is k stones left, but 1 - k have all been previously taken, one can take k stones. The one who takes the last stone wins. How can the first player always win?

**My Idea**

Use recursion (or dynamic programming). Base case 1, where player 1 has a winning strategy.

Reducing: for n stones left, if palyer 1 takes m1 stones, he has to ensure that for all options player 2 has (m2), he has a winning strategy. Thus the problem is reduced to (n - m1 - m2).

**Follow Up Question**:

If one uses DP, the potential number of tables to be filled is large (2^15), since the available options left depend on the history, which has 2^15 possibilities.

How can you optimize?

I don't have a great answer to the follow up question。。。