Thread: I need help!!!

Code:

Hello everybody.

I have a problem and i need help.

Here is a set {a,b,c,d,e,f,.......,w,x,y,z}
I want to make a program to read a number  and to print all possible subsets from 'a' to digit that the chosen number shows.For example:

%Please give number.
Given number is 2 and these are the subsets of {a,b} :
{}
{a}
{b}
{a,b}

%Please give number.
Given number is 4 and these are the subsets of {a,b,c,d} :
{}
{a}
{b}
{a,b}
{c}
{a,c}
{b,c}
{a,b,c}
{d}
{a,d}
{b,d}
{c,d}
{a,b,d}
{a,c,d}
{b,c,d}
{a,b,c,d}

Thanks

I try this but nothing

Code:

#include<stdio.h>
int main(void){
    int A[26],B[26],i,j,k,l,max,o,z,x,c,v;
    
    for(i=0;i<26;i++){B[i]='a'+i;}
    printf("Give num");
    scanf("%d",&max);
    for(i=0;i<max;i++){
                       printf("%c\n",B[i]);
                       for(j=0;j<i;j++){
                                        A[j]=0;
                                        }
                       for(j=0;j<i;j++){
                                        A[j]=B[j];
                                        }
                       for(j=0;j<=i-1;j++){
                                          
                                          for(k=j;k<=i-1;k++){
                                                             printf("%c",A[k]);
                                                             for(o=k;o<=i-1;o++){
                                                                                 printf("%c",A[o]);
                                                             for(z=o;z<=i-1;z++){
                                                                                 printf("%c",A[z]);
                                                             }
                                                             }
                                                             printf("%c\n",B[i]);
                                                             A[k]=B[k+1];
                                          
                                                             }
                                                             printf("%c\n",B[i]);
                                                             A[o]=B[o+1];                   
                                          }                  
                       }
getchar();                       
getchar();
}

Thanks
1.Each time, i want to print the string and not the number
2.I can not understand absolute what you mean

Code:

take number n
build number m thats binary representation has n 1: 0000111...1
and make a loop from 0 to m
for each i get the binary representation of the number and build the subset

Thanks Everybody!!!

Code:

2, 1
^^
00 - Empty set. Zero Array[2]'s and Zero Array[1]'s. 
01 - Zero Array[2]'s, One Array[1], (no b's, one a)
10 - One Array[2], Zero Array[1]'s (one b, no a's)
11 - One Array[2], and One Array[1] (one b, one a)

What do you mean Zero Array[2]'s and One Array[1], (no b's, one a)
Do we use two types of Arrays?
Can you show me what included in Array?

Code:

Here's a rough draft that goes backwards and always displays a fixed length field, but it should help:

Thanks a lot Deckard it works.

Thank Everyone
Mixing all your ideas i made this

Code:

#include<stdio.h>
int power(int,int);
int main(void){
    int A[27],i,j,max,k,n;
    scanf("%d",&max);
    for(i=0;i<max;i++){
                       A[i]=0;
                       }
     printf("{ }\n");
     n=power(2,max)-1;             
    for(i=0;i<n;i++){
                     A[max-1]=A[max-1]+1;
                     for(j=max;j>=0;j--){
                                         if(A[j]==2){
                                                     A[j]=0;
                                                     A[j-1]=A[j-1]+1;
                                                     }
                                         }
                     putchar('{');                    
                     for(j=0;j<max;j++){                                        
                                        if(A[j]==1){
                                                    k='a'+j;
                                                    printf("%c,",k);
                                                    }
                                        }
                     printf("\b}\n");
                     }    
}


int power(int base, int n)
{ int p;
  for (p=1 ; n > 0 ; n--)                /* base^n equals to */
    p *= base;         /* base * base * ... * base (n times) */
  return p;                                 /* Return result */
}