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[]={};//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[];
 int main(){
     //freopen("D:\\INPUT.txt","r",stdin);
     int n,i,j;
     long long m;
     for(i=;i<;i++){
         c[i]=(i-)*c[i-]+;
         //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=;
           t=m/c[j]-(m%c[j]==?:);
           m-=t*c[j];
           m--;//去掉一个空集!!
           j--;
           for(i=;i<=n;i++){
             if(!s[i]){
                 count++;
                 if(count==t+){
                     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 ;
 }