This is also here. Solution is provided here.

B takes coins as well A. HAHAHAHA ITS EASY QUESTION

@Above I think that's not true. If $B$ does the same as $A$, $B$ can loose. Example: $A:(2010,2010)\rightarrow (1,2010)$ $B:(1,2010)\rightarrow (1,1)$ $A:(1,1)\rightarrow (0,0)$ and $A$ wins.

$B$ wins. Claim: If one of them gets $(1,2)$, then he wins. Proof:$(1,2)\rightarrow (1,1),(0,1),(0,2)\rightarrow (0,0)$ and the person who wrote $(1,2)$ wins. Claim: If one of them gets $(3,3)$, then he wins. Proof:$(3,3)\rightarrow (0,3)\rightarrow (0,0)$ and $(3,3)\rightarrow (1,3),(2,3),(2,2)\rightarrow (1,2)$ and we know that the person who writes $(1,2)$ wins. Let's prove that the person who writes $(3k,3k)$ and $(3k+1,3k+2)$ wins by induction. Let it be true for $k=1,2,...,m$ and let's prove that it's also true for $m+1$. Let $X$ write $(3m+3,3m+3)$. If $Y:(3m+3,3m+3)\rightarrow (3n,3m+3)$, then $X:(3n,3m+3)\rightarrow (3n,3n)$ and wins. If $Y:(3m+3,3m+3)\rightarrow (3n+1,3m+3)$, then $X:(3n+1,3m+3)\rightarrow (3n+1,3n+2)$ and wins. If $Y:(3m+3,3m+3)\rightarrow (3n+2,3m+3)$, then $X:(3n+2,3m+3)\rightarrow (3n+2,3n+1)$ and wins. If $Y:(3m+3,3m+3)\rightarrow (3m+2,3m+2)$, then $X:(3m+2,3m+2)\rightarrow (3m+1,3m+2)$ and wins. Let $X$ write $(3m+4,3m+5)$. If $Y:(3m+4,3m+5)\rightarrow (3n,3m+\{4,5\})$, then $X:(3n,3m+\{4,5\})\rightarrow (3n,3n)$ and wins. If $Y:(3m+4,3m+5)\rightarrow (3n+1,3m+\{4,5\})$ and $n<m+1$, then $X:(3n+1,3m+\{4,5\})\rightarrow (3n+1,3n+2)$ and wins. If $Y:(3m+4,3m+5)\rightarrow (3m+4,3m+4)$, then $X:(3m+4,3m+4)\rightarrow (3m+3,3m+3)$ and wins. If $Y:(3m+4,3m+5)\rightarrow (3n+2,3m+\{4,5\})$, then $X:(3n+2,3m+\{4,5\})\rightarrow (3n+2,3n+1)$ and wins. If $Y:(3m+4,3m+5)\rightarrow (3m+3,3m+4)$, then $X:(3m+3,3m+4)\rightarrow (3m+3,3m+3)$ and wins. Since $3|2010$, $B$ wins.