Subset sequence

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 8188 Accepted Submission(s): 3735

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

題意

對於一個序列$A_n=\{1,2,3...n\}$，$A_1=\{1\},A_3=\{1,2,3\}$，我們稱一個非空子集元素的排列為一個子集序列。對所有的子集序列按字典順序排序。求出第m個子序列

思路

首先要求出$A_n$有多少子集： $n=1$時，子集個數為1個 $n=2$時，子集個數為4個 $n=3$時，子集個數為15個 子集的個數滿足$F(n)=(F(n-1)+1)\times n$

AC程式碼

/*
* @Author: WZY
* @School: HPU
* @Date:   2018-11-11 15:02:40
* @Last Modified by:   WZY
* @Last Modified time: 2018-11-11 16:06:05
*/
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <math.h>
#include <limits.h>
#include <map>
#include <stack>
#include <queue>
#include <vector>
#include <set>
#include <string>
#include <time.h>
#define ll long long
#define ull unsigned long long
#define ms(a,b) memset(a,b,sizeof(a))
#define pi acos(-1.0)
#define INF 0x7f7f7f7f
#define lson o<<1
#define rson o<<1|1
#define bug cout<<"---------"<<endl
#define debug(...) cerr<<"["<<#__VA_ARGS__":"<<(__VA_ARGS__)<<"]"<<"\n"
const double E=exp(1);
const ll maxn=1e6+10;
const int mod=1e9+7;
using namespace std;
// F[n]=(F[n-1]+1)*n
// 計算n對應的子集的個數
ll F(ll n)
{
	if(n<=1)
		return n;
	return (F(n-1)+1)*n;
}
int main(int argc, char const *argv[])
{
	ios::sync_with_stdio(false);
	#ifndef ONLINE_JUDGE
	    freopen("in.txt", "r", stdin);
	    freopen("out.txt", "w", stdout);
	    double _begin_time = clock();
	#endif
	ll n,m;
	ll a[30];	
	while(cin>>n>>m)
	{
		for(int i=1;i<30;i++)
			a[i]=i;
		int flag=0;
		while(m>0)
		{
			ll k=F(n-1)+1;	//記錄每行有多少個集合
			int res=m/k+1;	//記錄在第幾行
			if(m%k==0)		//在上一行
				res--;
			if(flag)
				cout<<" ";
			cout<<a[res];
			for(int i=res;i<=n;i++)
				a[i]=a[i+1];
			m-=k*(res-1)+1;
			n--;
			flag++;
		}
		cout<<endl;
	}
	#ifndef ONLINE_JUDGE
	    double _end_time = clock();
	    printf("time = %lf ms.", _end_time - _begin_time);
	#endif
	return 0;
}