1025. Divisor Game


我在讨论区里面写了 自己的解法,直接粘贴过来吧。虽然一开始使用DP解决了这个问题,但是既然标签难度是Easy的,所以估计还有更容易的解法。

classical dp solution would be:

so the code would be

class Solution:
    def divisorGame(self, N: int) -> bool:
        dp = [-1] * (N + 1)

        def search(n):
            if dp[n] != -1:
                return dp[n]

            ans = 0
            for x in range(1, n):
                if n % x != 0: continue
                if search(n - x) == 0:
                    ans = 1
            dp[n] = ans
            return ans
        ans = search(N)
        return ans

But if you watch dp closely, looks like there is a pattern. You will win at even number, and lose at odd number. So you would come accorss a intuition and guess a simple solution return N % 2 == 0 . Fortunately the guess is correct.

The explanation is followed. Let's start with base situation

  1. N==2. apparently it's true.
  2. if N is even number, then N % 2 == 0, we can get 2 stones. So the case is N-2, which is still a even number, and we still win.
  3. if N is odd number, and assume there is a x st. N % x == 0. x must be a odd number. So the case is N-x, which is a even number, and we lose.


class Solution:
    def divisorGame(self, N: int) -> bool:
        return N % 2 == 0