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算法：数学方法

Posted by milksea at 2007-01-31 04:38:32 on Problem 2000

解不等式
(for k = 1 to i)∑k = i * (i + 1) / 2 ≤ n，
注意到i是满足上式的最小正整数，得到
i = √(8 * n + 1) - 1（取整数）
于是有：
金币数
= (for k = 1 to i)∑k^2 + (n - i) * (i + 1)
= (i + 1) * (6 * n - 2 * i - i^2) / 6

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Re:更正 (244) milksea 2007-01-31 04:40:18

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