1025. Divisor Game
Alice and Bob take turns playing a game, with Alice starting first.
Initially, there is a number n on the chalkboard. On each player's turn, that player makes a move consisting of:
- Choosing any integer x with 0 < x < n and n % x == 0.
- Replacing the number n on the chalkboard with n - x.
Also, if a player cannot make a move, they lose the game.
Return true if and only if Alice wins the game, assuming both players play optimally.
Example 1:
Input: n = 2
Output: true
Explanation: Alice chooses 1, and Bob has no more moves.
Example 2:
Input: n = 3
Output: false
Explanation: Alice chooses 1, Bob chooses 1, and Alice has no more moves.
Constraints:
- 1 <= n <= 1000
From: LeetCode
Link: 1025. Divisor Game
Solution:
Ideas:
The result only depends on whether n is even or odd.
- If n is even, Alice can always choose an odd divisor x, so n - x becomes odd for Bob.
- If n is odd, every divisor x of n is also odd, so n - x becomes even for Bob.
So:
- starting with an even number is a winning position
- starting with an odd number is a losing position
Therefore, Alice wins iff n is even.
Code:
c
bool divisorGame(int n) {
return (n % 2 == 0);
}