void subsetsum(int b[],int a[],int i,int sum,int j)
{
int s=0,c[N],k=0;
if(i<N-1)
{
if(sum==summ)
{
for(k=0;k<j;k++)
{
if(b[k]==0)
continue;
printf("%d ",b[k]);
}
printf("\n");
k=0;
}
else
{
subsetsum(b,a,i+1,sum,j);
for(k=0;k<j;k++)
c[k]=b[k];
c[k]=a[i+1];
if(sum<summ)
{
s=sum+a[i+1];
subsetsum(c,a,i+1,s,j+1);
}
}
}
}
int main()
{
int a[N]={1,3,2,4,5,1,7},b[N],i=0;
printf("enter the value of summ\n");
scanf("%d",&summ);
b[0]=0;
subsetsum(b,a,-1,0,1);
getch();
}
int main()
{
frequency_t a[] = { {6, 4}, {9, 3}, {3, 1} };
int N = sizeof(a)/sizeof(*a);
int sum = 27;
int* result = (int*) malloc(sizeof(int) * sum);
subsetsum(a, N, sum, result, 0);
return 0;
}
/*
a : pointer to an array of frequency_t elements.
N : size of the above array.
sum : sum we are looking for.
result : array where selected numbers will be stored.
index : caller must initially setup result and index.
*/
void subsetsum(frequency_t* a, int N, int sum, int* result, int index)
{
if(N == 0) return;
int psum = sum - a[0].item;
//we cannot use an element if its frequency count is <= zero.
if(a[0].count <= 0) goto try_with_next_element;
result[index] = a[0].item;
//if psum is zero, we found a combination which adds up to sum, so print it. otherwise decrease the frequency
//count of the current element and make a recursive call.
if(psum == 0)
printsubset(result, index + 1);
else
{
--a[0].count;
subsetsum(a, N, psum, result, index + 1);
++a[0].count;
}
try_with_next_element:
subsetsum(a + 1, N - 1, sum, result, index);
}
int main()
{
int a[]={1,2,3};
printf("%lld",subsetsum(a,sizeof(a)/sizeof(*a)));
getchar();
}
