HDOJ 2062 Subset sequence

Subset sequence

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4329    Accepted Submission(s): 2101

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
   
   
   
   

 

考虑一个集合 An = { 1, 2, ..., n}。比如，A1={1}，A3={1,2,3}。我们称一个非空子集元素的排列为一个子集序列。对所有的子序列按字典顺序排序。你的任务就是给出第m个子序列。

我们可以发现f[n] = n * (f[n-1] + 1);f[i]表示有i个元素时，

排列的总数序列一共分为n组，每组有(f[n-1] + 1)个元素

思路：求n个元素时序列首元素，序列变为n-1，

求n-1个元素时序列首元素......

用t = ceil(m/(f[n-1]+1)),即可求得所求序列在所有序列中是第几组

，也就是当前第一个元素在序列数组a中的位置

用数组a表示序列数组[1,2,...,n]（需要动态更新，每次求出t之后，

都要删除t位置的元素）

在更新之后，序列数组总长度变为n-1，我们要求一下所求序列的新位置

m = m - (t-1)*(f[n-1]+1) - 1（前面有t-1组，每组f[n-1]个元素）

#include
#include

int main(){
    int n;
    long long m, f[25];
    f[0] = 0;
    for (int i =1; i < 21; i++)
        f[i] = i * (f[i - 1] + 1);  
    while (scanf("%d%I64d", &n, &m) != EOF){
        int first = 1, a[25];
        for (int i = 1; i <= n; i++)
            a[i] = i;
        while (m > 0){
            int t = ceil(m * 1.0 / (f[n - 1] + 1));  //ceil函数是返回大于或者等于指定表达式的最小整数
            if (!first)
                printf(" ");
            first = 0;
            printf("%d", a[t]);
            for (int i = t; i < n; i++)
                a[i] = a[i + 1];
            m = m - (t - 1) * (f[n - 1] + 1) - 1;
            n--;
        }
        printf("\n");
    }
    return 0;
}