一、热身 [Cloned] K - next_permutation

原题：

Now our hero finds the door to the BEelzebub feng5166. He opens the door and finds feng5166 is about to kill our pretty Princess. But now the BEelzebub has to beat our hero first. feng5166 says, "I have three question for you, if you can work them out, I will release the Princess, or you will be my dinner, too." Ignatius says confidently, "OK, at last, I will save the Princess." 

"Now I will show you the first problem." feng5166 says, "Given a sequence of number 1 to N, we define that 1,2,3...N-1,N is the smallest sequence among all the sequence which can be composed with number 1 to N(each number can be and should be use only once in this problem). So it's easy to see the second smallest sequence is 1,2,3...N,N-1. Now I will give you two numbers, N and M. You should tell me the Mth smallest sequence which is composed with number 1 to N. It's easy, isn't is? Hahahahaha......" 
Can you help Ignatius to solve this problem? 

题意：

输入N和K，要找出1,2,3，....，N组成的第K字典序小的序列

题解：

这个题想了一些时间，本来以为是数学问题找排序。结果发现头文件里有next_permutation函数，可以直接找出全排列，强大强大。找第k个直接排序k次，然后输出

代码：AC

#include<iostream>
#include<algorithm>
using namespace std;
int arr[1111];
int main()
{
	int N,K;
	while(cin>>N>>K)
	{
		int i;
		for(i=0;i<N;i++)
		{
			arr[i]=i+1;
		}
		K--;
		while(K--)
		{
			next_permutation(arr,arr+N);
		}
		for(i=0;i<N-1;i++)
		{
			cout<<arr[i]<<" ";
		}
		cout<<arr[i]<<endl;
	}
	return 0;
}