| 시간 제한 | 메모리 제한 | 제출 | 정답 | 맞힌 사람 | 정답 비율 |
|---|---|---|---|---|---|
| 2 초 | 2048 MB | 259 | 142 | 133 | 59.910% |
Bessie and Elsie have discovered a row of $N$ cakes $(2 \leq N \leq 5\cdot 10^5, N \text{ even})$ cakes, with sizes $a_1,a_2,\dots,a_N$ in that order (1ドル\le a_i\le 10^9$).
Each cow wants to eat as much as possible. However, being very civilized cows, they decided to play a game to split it! The game proceeds with both cows alternating turns. Each turn consists of one of the following:
When only one cake remains, Bessie eats it, and Elsie eats all cakes in her stash. If both cows play optimally to maximize the amount of cake they eat and Bessie plays first, how much cake will each cow eat?
Each input consists of $T$ (1ドル\le T\le 10$) independent test cases. It is guaranteed that the sum of all $N$ within an input does not exceed 10ドル^6$.
Each test case is formatted as follows. The first line contains $N$. The next line contains $N$ space-separated integers, $a_1,a_2,\ldots,a_N$.
For each test case, output a line containing $b$ and $e,ドル representing the amounts of cake Bessie and Elsie respectively will consume if both cows play optimally.
2 4 40 30 20 10 4 10 20 30 40
60 40 60 40
For the first test case, under optimal play,
Bessie will eat 30ドル+20+10=60$ cake, while Elsie will eat 40ドル$ cake.
The second test case is the reverse of the first, so the answer is the same.