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.
思路：
回溯法。。。
int counts(int n,vector<int>& intvec)
{
if (n <= ) return ;
int res = ;
for (int i = ; i < n;i++)
{
if (intvec[i]%n == || n %intvec[i] ==)
{
swap(intvec[i], intvec[n - ]);
res += counts(n - , intvec);
swap(intvec[i], intvec[n - ]);
}
}
return res;
}
int countArrangement(int N)
{
vector<int> intvec;
for (int i = ; i < N; i++)intvec.push_back(i + );
return counts(N, intvec);
}
参考：
https://discuss.leetcode.com/topic/79921/my-c-elegant-solution-with-back-tracking