Ignatius and the Princess II

题目来自杭电：http://acm.hdu.edu.cn/showproblem.php?pid=1027
Ignatius and the Princess II

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5768 Accepted Submission(s): 3394

Problem Description
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?

Input
The input contains several test cases. Each test case consists of two numbers, N and M(1<=N<=1000, 1<=M<=10000). You may assume that there is always a sequence satisfied the BEelzebub’s demand. The input is terminated by the end of file.

Output
For each test case, you only have to output the sequence satisfied the BEelzebub’s demand. When output a sequence, you should print a space between two numbers, but do not output any spaces after the last number.

Sample Input
6 4
11 8

Sample Output
1 2 3 5 6 4
1 2 3 4 5 6 7 9 8 11 10

代码：

#include 
#include 
#include 
const int MAX = 1000 + 10;
using namespace std;


int main()
{
    int n,m;
    int i;
    int a[MAX];

    while( cin >> n >> m)
    {
        //初始化
        for(i = 1; i <= n; i++)
            a[i] = i;
        //排序
        for(i = 1; i < m; i++)
            next_permutation(a + 1,a + n + 1);
        //输出
        for(i = 1; i < n; i++)
            cout << a[i] << " ";
        cout << a[i] << endl;

    }
    return 0;
}

小结：
组合数学问题，题目乍看上去很难，如果不用STL或者DFS，解决起来确实不是很容易，这里用到了STL ，用到了一个函数next_permutation