4kyu Twice linear

題目背景：

Task
Consider a sequence u where u is defined as follows:

1. The number u(0) = 1 is the first one in u. 2. For each x in u, then y = 2 * x + 1 and z = 3 * x + 1 must be in u too. 3. There are no other numbers in u.

Given parameter n the function dbl_linear (or dblLinear…) returns the element u(n) of the ordered (with <) sequence u (so, there are no duplicates).

Example

u = [1, 3, 4, 7, 9, 10, 13, 15, 19, 21, 22, 27, ...] dbl_linear(10) should return 22

題目分析：
本道題目的思路很清晰，就是不斷迭代數(shù)組中的元素，以及該元素 u 的 2 * u + 1 和 3 * u + 1，我們需要做的是如何把這些元素按照從小到大的順序排序。簡(jiǎn)單的思路就是借用隊(duì)列去維護(hù) 2 * u + 1，及 3 * u + 1 的非記錄元素，每個(gè)隊(duì)列里面的數(shù)據(jù)是從小到大的排序，每一次都是比對(duì)兩個(gè)隊(duì)列的隊(duì)首元素，然后次數(shù) +1。

AC代碼：

#include <queue>class DoubleLinear { public:static int dblLinear(int n); };int DoubleLinear::dblLinear(int n) {std::queue<int> q2;std::queue<int> q3;int tmp = 1, cnt = 0;while( cnt != n ) {q2.push(tmp * 2 + 1);q3.push(tmp * 3 + 1);if ( q2.front() < q3.front() ) {tmp = q2.front();q2.pop();}else if ( q2.front() > q3.front() ) {tmp = q3.front();q3.pop();}else {tmp = q2.front();q2.pop();q3.pop();}cnt++; }return tmp; }

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