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
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学习网址：http://blog.csdn.net/lianqi15571/article/details/8877014
1 #include<cstdio> 2 #include<cstring> 3 #include<iostream> 4 #include<stack> 5 #include<set> 6 #include<map> 7 #include<queue> 8 #include<algorithm> 9 using namespace std; 10 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 11 //易知c[n]=(n-1)*c[n-1]+1 12 //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 13 bool s[22]; 14 int main(){ 15 //freopen("D:\INPUT.txt","r",stdin); 16 int n,i,j; 17 long long m; 18 for(i=1;i<21;i++){ 19 c[i]=(i-1)*c[i-1]+1; 20 //cout<<c[i]<<endl; 21 } 22 long long t,count; 23 while(scanf("%d %lld",&n,&m)!=EOF){ 24 memset(s,false,sizeof(s)); 25 j=n; 26 queue<int> q; 27 while(m){ 28 count=0; 29 t=m/c[j]-(m%c[j]==0?1:0); 30 m-=t*c[j]; 31 m--;//去掉一个空集!! 32 j--; 33 for(i=1;i<=n;i++){ 34 if(!s[i]){ 35 count++; 36 if(count==t+1){ 37 break; 38 } 39 } 40 } 41 s[i]=true; 42 q.push(i); 43 } 44 printf("%d",q.front()); 45 q.pop(); 46 while(!q.empty()){ 47 printf(" %d",q.front()); 48 q.pop(); 49 } 50 printf(" "); 51 } 52 return 0; 53 }