HDU - 1027 Ignatius and the Princess II STL—全排列

HDU - 1027 Ignatius and the Princess II
题目描述：
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

题目大意：就是给出两个数n和m，n代表这个一个序列1到n,找出这个序列中的第m小的序列，并打印出该序列。

思路：一开始真的不知道有next_permutation函数，自己拿着草稿纸推了好久好久
next_permutation（）：是按照字典序产生的全排列，从当前字典序依次增大到最大字典序
知道有这个大学。。。so easy！！！

#include <iostream>
#include <algorithm>

using namespace std;
int a[10010];
int N,M;

int main(){
    while(~scanf("%d%d",&N,&M)){
        for(int i = 1 ; i <= N ; ++i){
            a[i] = i;
        }
        M--;
        while(M--){
            next_permutation(a+1,a+1+N);
        }
        for(int i = 1 ; i < N ; ++i) printf("%d ",a[i]);
        printf("%d\n",a[N]);
    }
    return 0;
}