[HDOJ] 2026.Max Sum

2026.Max Sum (c++)

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

Sample Output

题意：给定一个集合{1，2，3，4….n}，求它内部的不同元素的字典序下的第m个字典序。
思路：
找规律吧
例：n=3，m=10时，有
{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}

n表示有多少组，g(n)表示每一组的个数，f(n)表示总的个数
g(n) = f(n) / n
->f(n) = n[f(n-1) + 1]
->g(n) = n[f(n-1) + 1] / n = f(n-1) + 1
->f(n-1) = (n-1) * g(n-1)
->g(n) = (n-1) * g(n-1) + 1

观察得：
是第几组则首元素是几，返回第一个元素；
第二元素通过判断在这一组的第几个，也就是第几组再输出；

#include <algorithm>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <vector>
using namespace std;

int main()
{
  int n, a[25];
  long long int m, c[21];
  memset(c, 0, sizeof(c));
  for (int i = 1; i <= 25; i++)
    c[i] = (i - 1) * c[i - 1] + 1; // c[25]保存当n=i时，每一组的个数
  while (cin >> n >> m)
  {
    for (int i = 1; i <= n; i++)
      a[i] = i;
    while (n > 0 && m > 0)
    {
      int temp = (m - 1) / c[n] + 1; // 第几组=输出的数字
      if (temp > 0)
      {
        printf("%d", a[temp]);
        for (int i = temp; i <= n; i++)
        { ///删掉这个数字
          a[i] = a[i + 1];
        }
        m -= ((temp - 1) * c[n] + 1); //再该组的第几项
        printf(m == 0 ? "\n" : " ");
      }
      n--;
    }
  }
  return 0;
}

2026.Max Sum (c++)

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