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
题意:给出n和m,然后有很多个集合{1,2,,,,n}的非空子集,按照一定方式排列,例如n==3时,
1
1 2
1 2 3
1 3
1 3 2
2
2 1
2 1 3
2 3
2 3 1
3
3 1
3 1 2
3 2
3 2 1
然后输出第m个序列。
结题思路,仔细观察可以看出,1、2、3开头的集合个数是一样多的,因此那m/(每组个数)可以得出第一个数字,然后依次类推即可
#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];//当前的首数字 for(i=t;i<=n;i++) a[i]=a[i+1];//回到n-1个数字,更新输出数组 m-=((t-1)*p[i]+1);//去掉前面的小于t开头的组合,且将去掉一个仅有t的集合 printf(m==0?"\n":" "); } } return 0; }