链接：https://vjudge.net/problem/HDU-1027

Task

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:

1
2

6 4
11 8

output:

1
2

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

Solution

这种定义序的方式和next_permutation的序一样，因此初始化完后直接调用M-1次。 注：next_permutation(first,last)的作用是，对[first,last)间的元素，根据operator<定义一个字典序，返回当前排列在字典序中的下一个排列。若下一个排列存在则返回true，不存在则将元素排序（变为字典序的第一个排列）并返回false

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#include<cstdio>
#include<algorithm>
using namespace std;
int main(){
    int N,M;
    int seq[1000]={0};
    while(scanf("%d %d",&N,&M)!=EOF){
        // init with sequence
        for(int i=0; i<N; ++i)
            seq[i]=i+1;
        // permute M-1 times
        M--;
        while(M--)
            next_permutation(seq,seq+N);
        for(int i=0; i<N-1; ++i)
            printf("%d ",seq[i]);
        printf("%d\n",seq[N-1]);
    }
    return 0;
}