C - Summarize to the Power of Two 解题报告

C++编程：求使序列变为好序列的最小删除数

最新推荐文章于 2019-08-16 12:45:04 发布

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最新推荐文章于 2019-08-16 12:45:04 发布 · 695 阅读

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这是一个关于计算机科学中的算法问题，目标是确定为了使给定序列变成一个“好序列”（序列中任意元素都能与其他元素相加得到一个2的幂次）所需的最少元素删除数。通过预处理2的幂次并在输入序列中查找匹配项来解决此问题，同时考虑重复元素的情况。

A sequence a1,a2,…,ana1,a2,…,an is called good if, for each element aiai, there exists an element ajaj (i≠ji≠j) such that ai+ajai+aj is a power of two (that is, 2d2d for some non-negative integer dd).

For example, the following sequences are good:

[5,3,11][5,3,11] (for example, for a1=5a1=5 we can choose a2=3a2=3. Note that their sum is a power of two. Similarly, such an element can be found for a2a2 and a3a3),

[1,1,1,1023][1,1,1,1023],

[7,39,89,25,89][7,39,89,25,89],

[][].

Note that, by definition, an empty sequence (with a length of 00) is good.

For example, the following sequences are not good:

[16][16] (for a1=16a1=16, it is impossible to find another element ajaj such that their sum is a power of two),

[4,16][4,16] (for a1=4a1=4, it is