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Tag Archives: 位运算
POJ 2081 —— Recaman’s Sequence
说有一个序列，a[0] = 0 ;对于 m > 0，a = a − m，如果 a 是正的并且没有出现在序列中，否则 a = a + m。问序列的第 k 项是多少。 貌似不复杂，不过得用位操作来检查元素是否存在过，不能直接查找。
POJ 2453 —— An Easy Problem
给你一个数 找出比他大的最小的一个数 满足二进制中1的个数和他相同