i think the winner is affected by minimum x satisfied (x-1) + x >= k that is odd or not

I think player 2 wins if n%(k+1) == 0, else player 1 wins.

Auto comment: topic has been updated by a2dalek (previous revision, new revision, compare).

I don't know proof but solution look like this series. If a(n) = 1, then first player win, otherwise second player win.

Inspired by Sakhiya07's result, I give a proof.

Note that whatever in 2k - 1-th turn, first player take, the second player could make the sum of 2k - 1-th turn and 2k-th turn to be $$$2 k$$$ or $$$2 k + 1$$$. so we may write $$$n$$$ as:

if $$$m = 2k, 2k + 1$$$, then since at 2k - 1-th turn, the first play take at most $$$2 k - 1$$$ stones, the second player will win.

Now, if $$$2k \leq m \leq 2k + k$$$, we may write $$$n$$$ as:

Similar, we know the second player will win.

Thus, if $$$k^2 + k \leq n \leq k^2 + 2k$$$, the second player will win.

Similar, whatever in 2k - 2-th turn, second player take, the second player could make the sum of 2k- 2-th turn and 2k - 1-th turn to be $$$2 k - 1$$$ or $$$2 k$$$, we can write $$$n$$$ as:

easy to easy if $$$2k + 1 \leq m \leq 2k + 1 + k$$$, the first player will win.

Thus, if $$$k ^ 2 \leq n < k^2 + k$$$, the first player will win.

Code is easy:

bool f(long long n) {
	long long k = std::sqrt(n + 0.1);
	return n < (k + 1) * k;
}