【康拓逆展开】HDU2062Subset sequence

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2062

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
   
   
   
   
   
   
   
   


   
   
   
   

可以用vector实现康拓逆展开;

代码：

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<vector>
#define LL long long
using namespace std;
int ans[30];
int len=0;
LL fac[23];
/*      康拓逆展开已AC
void Cantor(int n,LL m)
{
    int N=n;
    int i,j,vis[23]={0};
    while(n--&&m){
        LL t=m/fac[n+1];
        if(m%fac[n+1]) t+=1;
        LL k=t;
        for(i=1;i<=N;i++){
            if(!vis[i]){
                k--;
                if(k==0) break;

            }
        }
        vis[i]=1;
        ans[len++]=i;
        m-=((t-1)*fac[n+1]+1);
    }
}
*/
//  用vector实现康拓逆展开
void Cantor(int n,LL m)
{
    vector<int>q;
    for(int i=1;i<=20;i++){
        q.push_back(i);
    }
    while(n--&&m){
        LL t=m/fac[n+1];
        if(m%fac[n+1]) t+=1;
        ans[len++]=q[t-1];
        q.erase(q.begin()+t-1);
        m-=((t-1)*fac[n+1]+1);
    }
}
void f()
{
    fac[0]=0;
    fac[1]=1;       //  注意数据类型;
    for(int i=2;i<=21;i++){
        fac[i]=fac[i-1]*(i-1)+1;
    }
}
int main()
{
    f();
    LL n,m;
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    cin.sync_with_stdio(false);
    while(cin>>n>>m){
        len=0;
        Cantor(n,m);
        for(int i=0;i<len;i++){
            if(i<len-1) cout<<ans[i]<<' ';
            else cout<<ans[i]<<endl;
        }
    }
    return 0;
}