Question

我必须计算集合的幂集，其中可能包含更多元素10^5。我尝试了一个算法和下面的代码，但它失败了（我认为导致pow(2, size)的大值。）

void printPowerSet(int *set, int set_size)
{
unsigned int pow_set_size = pow(2, set_size);
int counter, j,sum=0;
 for(counter = 0; counter < pow_set_size; counter++)
{
  for(j = 0; j < set_size; j++)
   {
      if(counter & (1<<j))
     std::cout<<set[i]<<" ";
   }
   std::cout<<sum;
   sum=0;
   printf("\n");
}
}

是否还有其他算法或如何修复此算法（如果可能的话）？
或
Can you suggest me how to do it i.e. finding subset of large set.

正如答案所指出的，我似乎陷入了困境      X-Y problem。基本上，我需要任何集合的所有子集的总和。现在，如果你能建议我解决问题的任何其他方法。
    谢谢。

Answer 1

这是一种算法，可以打印出任何适合计算机内存的电源组。

如果有足够的时间，它将打印一组长度为10 ^ 5的幂集。

然而，“足够的时间”将达到几万亿亿亿年。

C ++ 14

#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

#include <iostream>

void printPowerset (const vector<int>& original_set)
{
    auto print_set = [&original_set](auto first, auto last) -> ostream&
    {
        cout << '(';
        auto sep = "";
        for ( ; first != last  ; ++first, sep = ",")
        {
            cout << sep << original_set[(*first) - 1];
        }
        return cout << ')';
    };

    const int n = original_set.size();
    std::vector<int> index_stack(n + 1, 0);
    int k = 0;


    while(1){
        if (index_stack[k]<n){
            index_stack[k+1] = index_stack[k] + 1;
            k++;
        }

        else{
            index_stack[k-1]++;
            k--;
        }

        if (k==0)
            break;

        print_set(begin(index_stack) + 1, begin(index_stack) + 1 + k);
    }
    print_set(begin(index_stack), begin(index_stack)) << endl;
}

int main(){

    auto nums = vector<int> { 2, 4, 6, 8 };
    printPowerset(nums);

    nums = vector<int> { 2, 4, 6, 8, 10, 12, 14, 16, 18, 20 };
    printPowerset(nums);

    return 0;
}

预期结果：

第一组电源（4项）：

(2)(2,4)(2,4,6)(2,4,6,8)(2,4,8)(2,6)(2,6,8)(2,8)(4)(4,6)(4,6,8)(4,8)(6)(6,8)(8)()

第二组电源（10项）

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Answer 2

这是不可能的，因为你指出自己的权力集包含pow(2, size)没有元素。您需要打印整个集合，只能通过使用backtracking生成集合来完成，没有更好的。

要找到给定集合的所有子集的总和，这是一个更简单的问题。

如果元素的数量为n，则数组中的每个元素在幂集中出现2^(n-1)次。

伪代码：

int sum = 0;

for(int i = 0;i < n;i++){
    sum += ( set[i] * ( 1 << (n-1) ) );    //pow(2, n-1) == 1 << (n-1)
}

cout << sum << endl;

对于Big Integer Library的较大值，您需要n。

Answer 3

由于集合中的元素数量可以达到10^5，因此集合的大小将达到2^(10^5)，这是巨大的。您可以只打印它，或者如果要存储它，请将其存储在文件中。

大集的权力集

3 个答案: