题目描述

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?

样例

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.

算法1

(DFS)

C++ 代码

class Solution {
public:

    int count = 0;
    int countArrangement(int N) {
        vector<bool> used(N + 1, false);
        dfs(0, N, used);    
        return count;
    }

    void dfs(int pos, int N,  vector<bool>& used) {
        if (pos == N) {
            count ++;
            return;
        }
        int curpos = pos + 1;
        for (int i = 1; i <= N; ++i) {
            if (i % curpos == 0 || curpos % i == 0) {
                if (!used[i]) {
                    used[i] = true;
                    dfs(pos + 1, N used);
                    used[i] = false;
                }
            }
        }
    }

};