HDU 2062 Subset sequence

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本文介绍了一种算法，用于解决给定整数集合An的所有非空子集序列按字典序排序后找到第m个的问题。输入包含多个测试用例，每个用例包括两个整数n和m。

Subset sequence

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3005 Accepted Submission(s): 1563

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

Sample Output

Author

Source

校庆杯Warm Up

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#include<iostream>
#include<stdio.h>
using namespace std;
int main()
{
    int n,s[21];
    __int64 m;
    __int64 g[21];
    g[1]=1;
    for(int i=2;i<=21;i++)
     g[i]=(i-1)*g[i-1]+1;
    while(scanf("%d%I64d",&n,&m)!=EOF)
    {
       int t;
       for(int i=0;i<=21;i++)
        s[i]=i;
       while(n)
       {
           t=m/g[n]+(m%g[n]>0?1:0);
           if(t>0)
           {
              printf("%d",s[t]);
              for(int i=t;i<=n;i++)
               s[i]=s[i+1];
              m-=((t-1)*g[n]+1);
              printf(m==0?"\n":" ");
           }
           n--;
       }
    }
    return 0;
}