Available question text
Two-pile take-1-per-turn game reduces to parity: Alice wins iff x+y is odd (since exactly one flower is removed per move and Alice goes first), so the task is to count pairs (x,y) in [1..n]×[1..m] with opposite parity — computable from the counts of odd/even numbers in each range.
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Leetcode 3021. Alice and Bob Playing Flower Game
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Source snapshot: Nov 12, 2026
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