Subset sequence

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3441    Accepted Submission(s): 1740

Problem Description

Consider the aggregate An= { 1, 2, …, n }. For example, A1={1}, A3={1,2,3}. A subset sequence is defined as a array of a non-empty subset. Sort all the subset sequece of An in lexicography order. Your task is to find the m-th one.

Input

The input contains several test cases. Each test case consists of two numbers n and m ( 0< n<= 20, 0< m<= the total number of the subset sequence of An ).

Output

For each test case, you should output the m-th subset sequence of An in one line.

Sample Input

1 1
2 1
2 2
2 3
2 4
3 10

Sample Output

1
1
1 2
2
2 1
2 3 1

题意：给出n和m，然后有很多个集合{1,2，，，，n}的非空子集，按照一定方式排列，例如n==3时，

1

1    2

1    2    3

1  3

1    3    2

2

2    1

2    1    3

2    3

2    3    1

3

3    1

3    1    2

3  2

3  2    1

然后输出第m个序列。

结题思路，仔细观察可以看出，1、2、3开头的集合个数是一样多的，因此那m/(每组个数)可以得出第一个数字，然后依次类推即可

#include<cstdio>
#include<iostream>
#define LL long long int
using namespace std;
int main(){
    int n,a[21],i;
    LL m,t;
    LL p[21]={0,1};
    for(i=2;i<=20;i++){
        p[i]=p[i-1]*(i-1)+1;//0、1、2、5、16
    }
    while(cin>>n>>m){
        for(i=0;i<=20;i++)
            a[i]=i;
        while(n--&&m){
            t=m/p[n+1]+((m%p[n+1])?1:0);
            cout<<a[t];//当前的首数字
            for(i=t;i<=n;i++)
                a[i]=a[i+1];//回到n-1个数字，更新输出数组
            m-=((t-1)*p[i]+1);//去掉前面的小于t开头的组合，且将去掉一个仅有t的集合
            printf(m==0?"\n":" ");
        }
    }
    return 0;
}