Leetcode 526. Beautiful Arrangement

题目

526. Beautiful Arrangement

Difficulty: Middle
Total Accepted: 31,338
Total Submissions: 59,005

Suppose you have N integers from 1 to N. We define a beautiful arrangement as an array that is constructed by these N numbers successfully if one of the following is true for the ith position (1 <= i <= N) in this array:

The number at the ith position is divisible by i.
i is divisible by the number at the ith position.
Now given N, how many beautiful arrangements can you construct?

Example 1:

Input: 2
Output: 2

Explanation: 

The first beautiful arrangement is [1, 2]:

Number at the 1st position (i=1) is 1, and 1 is divisible by i (i=1).

Number at the 2nd position (i=2) is 2, and 2 is divisible by i (i=2).

The second beautiful arrangement is [2, 1]:

Number at the 1st position (i=1) is 2, and 2 is divisible by i (i=1).

Number at the 2nd position (i=2) is 1, and i (i=2) is divisible by 1.

Note:

N is a positive integer and will not exceed 15.

解题报告

AC 截图

题目大意

给定 1～N 的 N 个连续自然数，要求将其填入一个序列，使得对于序列上的每个位置，都有位置序号可以整除或者被当前数字整除。

解题思路

正常的思路：

动态规划或者直接暴力遍历。

奇怪的思路：

在编译阶段算出结果，然后直接调用结果即可。

题解

结果截图

无：Leetcode Bug

C++ 代码

static const auto runfirst = []() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(nullptr);
    std::cout.tie(nullptr);
    return nullptr;
}();

class Solution {
    int count = 0;
public:
    int countArrangement(int N) {
        bool* visited = new bool[N + 1];
        calculate(N, 1, visited);
        return count;
    }
    
    void calculate(int N, int pos, bool* visited) {
        if (pos > N)
            ++count;
        for (int i = 1; i <= N; ++i) {
            if (!visited[i] && (pos % i == 0 || i % pos == 0)) {
                visited[i] = true;
                calculate(N, pos + 1, visited);
                visited[i] = false;
            }
        }
    }
};

奇怪的解法（打表简化版）（最优解）

AC 截图

C++ 代码

static const auto runfirst = []() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(nullptr);
    std::cout.tie(nullptr);
    return nullptr;
}();

static const int result[16] = {
    1, 1, 2, 3, 8, 10, 36, 41,
    132, 250, 700, 750, 4010, 4237,
    10680, 24679
};

class Solution {
public:
    int countArrangement(int N) {
        return result[N];
    }
};