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`用int[A/B]表示A/B商的整數部分 1到2000中能被6或8整除的共有 int[2000/6]+int[2000/8]-int[2000/24]=333+250-83=500個 在1到2000中隨機的取整數,取到的整數不能被6或8整除的概率為 1-500/2000=1-1/4=3/4
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