Sorry for not understanding the problem at first...
I hope this solution will find the all the subsets from set S . The sum of
elements in subset is equivalent to certain k.
fun(int a[],int k){
int i,j;
int k1;
for(i=0;i<10;i++){
k1=a[i];
j=i+1;
while(j<10){ //if we have 10
elements in set S;
k1=k1+a[j];
if(k>k1){
j++;
}else if(k==k1){
printf("True\t");
while(k1!=0){
printf("%d\t",a[j]);
k1=k1-a[j];
j--;
}
printf("\n");
}else if(k<k1) {
break;
}
}
}
}
---
Regards
Peeyush Bishnoi
On 4/2/07, Peeyush Bishnoi <[EMAIL PROTECTED]> wrote:
>
> Following is the Solution for sum of subsets with complexity O(n).
> suppose we have to find out the 2 numbers from array(Set) whose sum is
> equivalent to certain no: k
>
> int main(){
> int a[10]={1,2,3,4,5,6,7,8,9,10};
> int sum =15; //no. to find for
> int i=0,j;
> front=i;
> back=10
> while(front<back){
> if(sum>a[front]+a[back]) front++;
> else if(sum<a[front]+a[back]) back--;
> else if(sum=a[front]+a[back]){
> printf("%d\t",a[front]); front++;
> printf("%d\n",a[back]); back-- ;
> } else {
> break;
> }
> }
> return 0;
> }
>
> In same way we can make for 3 , 4 numbers from array eq. to certain k.
>
> If you have any doubt please ask .
>
> Regards ,
> Peeyush Bishnoi
>
> On 4/2/07, pramod <[EMAIL PROTECTED] > wrote:
> >
> >
> > Dor,
> >
> > If I understand the problem correctly, we don't know what are all the
> > elements in S (that's what we need to find).
> > So how are you going to pick 'k' first and how do you know of 'x'
> > belonging to S?
> >
> >
> > > >
> >
>
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