Subset sequence

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3441    Accepted Submission(s): 1740


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 題意：給出n和m，然後有很多個集合{1,2，，，，n}的非空子集，按照一定方式排列，例如n==3時， 1 1   2 1   2    3 13 1   3    2 2 2   1 2   1    3 2   3 2   3    1 3 3   1 3   1    2 32 32    1 然後輸出第m個序列。 結題思路，仔細觀察可以看出，1、2、3開頭的集合個數是一樣多的，因此那m/(每組個數)可以得出第一個數字，然後依次類推即可

#include<cstdio>
#include<iostream>
#define LL long long int
using namespace std;
int main(){
    int n,a[21],i;
    LL m,t;
    LL p[21]={0,1};
    for(i=2;i<=20;i++){
        p[i]=p[i-1]*(i-1)+1;//0、1、2、5、16
    }
    while(cin>>n>>m){
        for(i=0;i<=20;i++)
            a[i]=i;
        while(n--&&m){
            t=m/p[n+1]+((m%p[n+1])?1:0);
            cout<<a[t];//當前的首數字
            for(i=t;i<=n;i++)
                a[i]=a[i+1];//回到n-1個數字，更新輸出陣列
            m-=((t-1)*p[i]+1);//去掉前面的小於t開頭的組合，且將去掉一個僅有t的集合
            printf(m==0?"\n":" ");
        }
    }
    return 0;
}