hduoj 2062Subset sequence

Subset sequence

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4123    Accepted Submission(s): 2019

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

 

 

Author

LL

 

 

Source

校庆杯Warm Up

 

 

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学习网址：http://blog.csdn.net/lianqi15571/article/details/8877014

#include<cstdio>
#include<cstring>
#include<iostream>
#include<stack>
#include<set>
#include<map>
#include<queue>
#include<algorithm>
using namespace std;
long long c[21]={0};//c[n]表示长度为n，第一位固定，剩下数字排列形成数列的个数
//易知c[n]=(n-1)*c[n-1]+1
//含义：比如第一位是A1  剩下A2A3。。An 共n-1个数可以固定在第二位，再加上一个空集
bool s[22];
int main(){
    //freopen("D:\\INPUT.txt","r",stdin);
    int n,i,j;
    long long m;
    for(i=1;i<21;i++){
        c[i]=(i-1)*c[i-1]+1;
        //cout<<c[i]<<endl;
    }
    long long t,count;
    while(scanf("%d %lld",&n,&m)!=EOF){
        memset(s,false,sizeof(s));
        j=n;
        queue<int> q;
        while(m){
          count=0;
          t=m/c[j]-(m%c[j]==0?1:0);
          m-=t*c[j];
          m--;//去掉一个空集!!
          j--;
          for(i=1;i<=n;i++){
            if(!s[i]){
                count++;
                if(count==t+1){
                    break;
                }
            }
          }
          s[i]=true;
          q.push(i);
        }
        printf("%d",q.front());
        q.pop();
        while(!q.empty()){
            printf(" %d",q.front());
            q.pop();
        }
        printf("\n");
    }
    return 0;
}