sorry a lil bit modification in the above answer..
*int ref[] = {1,2,3};*
*void printcombination(int n,int index,int i)*
*{*
* static int a[100];*
* int j;*
* if (n == 0)*
* {*
* for(j=0;j<index;j++)*
* printf("%d ",a[j]);*
* printf("\n");*
* }*
* else if(n>0)*
* {*
* for(j=i;j<3;j++)*
* {*
* a[index]=ref[j];*
* printcombination(n-ref[j],index+1,j);*
* }*
* }*
*} *
*main()*
*{*
* int n;*
* printf("enter value of n :");*
* scanf("%d",&n);*
* printcombination(n,0,0);*
*}*
On Thu, Jun 23, 2011 at 11:17 AM, Piyush Sinha <ecstasy.piy...@gmail.com>wrote:
> pass one more argument to the function "*int index*" and instead of
> starting the loop from *i = 0 to N, *make it start from *i = index to N *and
> then call
>
> *printcombinations(a,sum-a[i],level+1,index+1);
> *I think it will work then...
> On Thu, Jun 23, 2011 at 10:48 AM, ross <jagadish1...@gmail.com> wrote:
>
>> Given an array and a sum S output all combinations of elements that
>> sum to S.
>> eg: 1 2 3
>>
>> sum = 3
>> 1+1+1,
>> 2+1
>> 3
>>
>> I came up with the foll algorithm, but it outputs 2+1 and 1+2 again.
>> (does not handle repetitions)
>>
>> printcombinations(int a[],int sum,int level) {
>> if(sum==0) { print array}
>> else if (sum>0) {
>> for ( i = 0 to N ) {
>> array[level]=a[i];
>> printcombinations(a,sum-a[i],level+1);
>> }
>> }
>> }
>>
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>
>
> --
> *Piyush Sinha*
> *IIIT, Allahabad*
> *+91-8792136657*
> *+91-7483122727*
> *https://www.facebook.com/profile.php?id=100000655377926 *
>
>
--
*Piyush Sinha*
*IIIT, Allahabad*
*+91-8792136657*
*+91-7483122727*
*https://www.facebook.com/profile.php?id=100000655377926 *
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