Here I assume the glasses can be empty and that the player who makes the last move wins. We'll call a pair of glasses with $2k$ and $2k+1$ coins good. It is not hard to see that the first player wins with the following strategy: - In the first turn, remove $100$ cents from the glass containing $200$. Now the glasses can be arranged into $50$ good pairs: $(100,101),(102,103),\ldots,(198,199)$. - In subsequent turns, the first player just maintains that there are always $50$ good pairs. For example, if the second player moves $123\to 50$, the first player moves $122\to 51$, essentially changing the good pair $(122,123)$ to $(50,51)$. Clearly the first player can always play, and as the game always end, the first player must win.

I think your solution is wrong. The second player can play with this strategy too

Eagleka94 wrote: I think your solution is wrong. The second player can play with this strategy too No, this method doesn't work for the second player. - First, the initial position of $101,102,\ldots,200$ cannot be paired into $50$ good pairs. Note that a good pair is $(2k,2k+1)$ and not $(2k-1,2k)$. - If you were to define a good pair as a pair of the form $(2k-1,2k)$ then the first player can reduce a glass to $0$ cents and the second player cannot create a glass with $-1$ cents.