C. Game on Integer
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Time Limit: 1.0 s
Memory Limit: 64.0 MB
Description
Roy and Hridoy play a game on a multiset of N
integers on the board. They maintain a running total S
, initially 0
. Roy moves first and they alternate turns. On each turn the current player may either
- stop the game (only if both players have already made at least one move), fixing the final total S,
- or choose and remove one remaining integer
x
from the board and updateS
as follows:- on Roy’s turn:
S ← S + x
- on Hridoy’s turn:
S ← S – x
- on Roy’s turn:
Roy aims to maximize the final total S
; Hridoy aims to minimize it. Assuming both play optimally, determine the value that Roy can guarantee.
Input
The first line contains a single integer \(T\) (\(1 ≤ T ≤ 10^4\)) — the number of test cases.
Each test case consists of two lines:
The first line contains a single integer \(N\) (\(2 ≤ N ≤ 2 * 10^5\))— the number of integers on the board.
The second line contains \(N\) integers \(A₁, A₂, …, A_N\) ( \(−10^9 ≤ A_i ≤ 10^9\) ).
It is guaranteed that sum of \(N\) over all test cases does not exceed \(2 * 10^5\).
Output
- For each test case, print one integer — maximum the final total
S
that Roy can guarantee under optimal play.
Sample
Input | Output |
---|---|
|
|
Test case 1
:
N = 4, Board = {3, 4, 0, -6}, S starts at 0.
Roy’s move: he removes 4, S = 0 + 4 = 4. Remaining {3, 0, -6}.
Hridoy’s move: he removes 3, S = 4 – 3 = 1. Remaining {0, - 6}.
Now stopping is allowed.
Roy can either stop (getting S = 1) or continue. He continues and removes 0, S = 1 + 0 = 1. Remaining {-6}.
Hridoy stopping gives S = 1; any further pick would make S lager, so he stops.
Final answer for this test case: 1
Test case 3
:
N = 5, Board = {-4, -6, -2, -2, -2}, S starts at 0.
Roy’s move: he removes -2, S = 0 + (-2) = -2. Remaining {-4, -6, -2, -2}.
Hridoy’s move: he removes -2, S = -2 - (-2) = 0. Remaining {-4, -6, -2}.
Now stopping is allowed.
- Roy can either stop (getting S = 0) or continue. Any further pick would make S smaller, so he stops.
So final S equal to 0.
Educational Round 1
- Status
- Done
- Rule
- ACM/ICPC
- Problem
- 6
- Start at
- 2025-07-14 15:30
- End at
- 2025-07-14 18:00
- Duration
- 2.5 hour(s)
- Host
- Partic.
- 127