Seeking a subset of the set of

The first method is to continue to add elements, then consider the deletion of one element, a New elements come together

 

1 void print_subset(int n,int A[],int cur) {
2     for (int i=0;i<cur;i++)
3         printf("%d ",A[i]);
4     printf("\n");
5     int s = cur ? A[cur-1]+1 : 0;
6     for (int i=s;i<n;i++) {
7         A[cur] = i;
8         print_subset(n,A,cur+1);
9     }

 

The second method is better understood, that is, for each element we have two cases (take and not take)

 

 1 void print_Subset(int n,int B[],int cur) {
 2     if (cur == n) {
 3         for (int i=0;i<cur;i++) {
 4            if (B[i])
 5                printf("%d ",i);
 6         }
 7         printf("\n");
 8         return ;
 9     }
10     B[cur] = 1;
11     print_Subset(n,B,cur+1);
12     B[cur] = 0;
13     print_Subset(n,B,cur+1);
14 }

 

The third method Binary!

Binary method is very simple, in fact, the binary 0 and 1 as a "switch." 1 for selecting, 0 is not selected.

In fact, every binary actually corresponds to the subject of our next map array (My personal understanding is this)

 

Seek length n and is continuous from a subset of the array of 0 ~ n-1

1 void print_subset(int n,int s) {
2     for (int i=0;i<n;i++) {
3         if (s & (1<<i))
4             printf("%d ",i);
5     }
6     printf("\n");
7 }

 

A subset of the array seek any length of n    if understand I said earlier, is the subject of the next binary array of maps, this is very simple

1 int a[3] = {5, 4, 6};
2 
3 void print_subset(int n,int s) {
4     for (int i=0;i<n;i++) {
5         if (s & (1<<i))
6             printf("%d ",a[i]);
7     }
8     printf("\n");
9 }