3021 Alice and Bob Playing Flower Game
Автор: Fogy Free
Загружено: 2025-08-28
Просмотров: 19
3021 Alice and Bob Playing Flower Game. This Java code solves the flower game by recognizing a key mathematical insight:
Alice wins exactly when the total flowers is odd.
The first solution can be made into a one liner on the return, but is easier to see how it works as it is, the second solution uses Bit Operations.
Key Insight:
Alice wins if she picks the last flower.
Each turn, a player picks one flower from either lane.
The total number of flowers is x + y.
Alice starts, so if the total number of flowers is odd, Alice will pick the last flower (since the number of turns is odd, and she starts).
If the total is even, Bob will pick the last flower.
So, for each pair (x, y), Alice wins if x + y is odd.
Game Theory: Alice goes first, so she wins when total moves is odd (meaning total flowers is odd)
Odd Total Rule: An odd sum comes from adding an odd + even number
Example: 3 + 2 = 5 (odd), 2 + 4 = 6 (even)
Counting Strategy:
✅ Count odd numbers in range [1,n]: (n + 1) / 2
✅ Count even numbers in range [1,n]: n / 2
✅ Winning pairs = (odd from first range × even from second) + (even from first × odd from second)
✅ Bit Operations: Uses signed right shift 1 instead of / 2 for slightly faster division by 2
The brilliance is converting a complex game into simple counting!
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