求一个数组的子集

递归法：参考博文：https://www.cnblogs.com/microgrape/archive/2011/05/11/2043814.html

#include "stdafx.h"

#include <iostream>
#include <vector>


using namespace std;
void all_subset(int s[], int n, vector<bool> &contains, int index)
{
if (index == n)
{
for (int i = 0; i < contains.size(); i++)
{
if (contains[i])
cout << s[i] << " ";
}
cout << endl;
}
else
{
contains[index] = false;
all_subset(s, n, contains, index+1);
contains[index] = true;
all_subset(s, n, contains, index + 1);
}
}
int main()
{
int s[] = { 1, 2, 3, 4, 5 };
int size = sizeof(s) / sizeof(int);
vector<bool> contains(5, false);
all_subset(s, 5, contains, 0);
system("pause");
return 0;

}

方法2：假设数组长度为5，则总共有2^5=32中子集，分别用00000-11111表示

例如 {1,2,3,4,5}

00000都没有

11111表示1 2 3 4 5

10101表示1 3 5

#include "stdafx.h"
#include <iostream>
#include <vector>


using namespace std;


int main()
{
int s[] = { 1, 2, 3, 4, 5 };
int size = sizeof(s) / sizeof(int);
int n = 1 << size;
for (int i = 0; i < n; i++)
{
for (int j = 0; j < size; j++)
{
if ((1 << j)&i)
cout << s[j] << " ";
}
cout << endl;
}


system("pause");
return 0;
}