HDUOJ 2062 Subset sequence_2062. subset sequence-优快云博客

本文链接：https://blog.youkuaiyun.com/qq_45114168/article/details/123306287

本文解析了如何解决一个关于计算给定整数集合An的非空子集按字典序排序后的第M个子集问题，涉及动态规划和计数原理。通过实例和代码实现，展示了求解步骤和关键算法思路。

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HDUOJ 2062 Subset sequence

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

题目的意思简单来说就是：给出一个N代表有一个集合，集合有1~N中的数字组成。集合的所有子集，

非空集，按照字典序排序，给出一个M，求第M个集合的元素。

看了好多歌题解才明白意思，这个大佬讲的还挺清楚

参考题解

代码：

#include <iostream>
#include <stdio.h>
#include <string.h>

using namespace std;
int ans[50];
long long power[30];
long long num[30];  //num[i]用来记录N个数的时候有多少个子集
int vis[30];
int cnt;
void solve(int N,long long M) {
    cnt = 0;
    memset(vis,0,sizeof(vis));
    int rest = N;
    while(true) {
         //cout<<"M = "<<M<<endl;
         int temp = M/num[rest];   ///计算答案在第几组
         if(M%num[rest] != 0)
            temp++;
         int ccount = 0;
         for(int i = 1; i <= N; i++) {
            if(vis[i]==0) {
                ccount++;
                if(ccount == temp) {
                    vis[i] = 1;
                    ans[cnt++] = i;
                    rest--;       ///数字减少一个
                    break;
                }
            }
         }
         M = M-(temp-1)*num[rest+1];   ///计算它是该组的第几个
         M--;
         if(M == 0)
            break;
    }
}
int main() {
    int N;
    long long M,Mul = 1;
    power[0] = 1;
    for(int i = 1; i <= 20; i++) {
        power[i] = Mul*i;
        Mul = Mul*i;
    }

    for(int i = 1; i <= 20; i++) {
        num[i] = 0;
        for(int j = 1; j <= i; j++) {
            num[i] = num[i] + power[i]/power[i-j];
        }
        num[i] = num[i]/i;
    }
    while(~scanf("%d%lld",&N,&M)) {
        solve(N,M);
        printf("%d",ans[0]);
        for(int i = 1; i < cnt; i++) {
            printf(" %d",ans[i]);
        }
        printf("\n");
    }
    return 0;
}