Subset sequence

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4587    Accepted Submission(s): 2228


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

設f[i]為1到n的排成的序列的總數，則可由打表發現：f[i]=n*(f[i-1]+1)，這裡其實蘊含著遞迴的思想。我們想要知道序列的具體位置，由規律性：zushu=ceil(m*1.0/(f[n-1]+1))，然後就可以取出首位元素了，剩下的序列可由遞迴求解，同時，原序列需要更新，元素被取出。然後就要更新內層的組數了：m=m-(zushu-1)*(f[n-1]+1)-1，因為每個序列的最小的那一個就一個元素，其後沒有元素，所以需要-1。n還需自減一次。

/*------------------Header Files------------------*/
#include <iostream>
#include <cstring>
#include <string>
#include <cstdio>
#include <algorithm>
#include <cstdlib>
#include <ctype.h>
#include <cmath>
#include <stack>
#include <queue>
#include <deque>
#include <map>
#include <vector>
#include <limits.h>
using namespace std;
/*------------------Definitions-------------------*/
#define LL long long
#define PI acos(-1.0)
#define INF 0x3F3F3F3F
#define MOD 10E9+7
#define MAX 500050
/*---------------------Work-----------------------*/

void work()
{
	int n;
	LL m,f[25];
	f[0]=0;
	for(int i=1;i<=20;i++)
		f[i]=i*(f[i-1]+1);
	while(scanf("%d%I64d",&n,&m)==2)
	{
		int num[25];
		for(int i=1;i<=n;i++)
			num[i]=i;
		bool flag=true;
		while(m>0)
		{
			int zushu=ceil(m*1.0/(f[n-1]+1)); //計算所屬組數
			if(!flag) printf(" ");
			flag=false;
			printf("%d",num[zushu]);
			for(int i=zushu;i<n;i++) //合併掉這個位置
				num[i]=num[i+1];
			m=m-(zushu-1)*(f[n-1]+1)-1; //-1是因為第一個只有一個數字
			n--;
		}
		printf("\n");
	}
}
/*------------------Main Function------------------*/
int main()
{
	//freopen("test.txt","r",stdin);
	//freopen("cowtour.out","w",stdout);
	//freopen("cowtour.in","r",stdin);
	work();
	return 0;
}